Question 1

A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. a) What is the initial effect of the change on vapour pressure? b) How do rates of evaporation and condensation change initially? c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

Accepted Answer

Vapour pressure of a pure liquid depends only on temperature, not on the volume of the container — so once equilibrium is restored, the pressure comes back to exactly where it started. The trick in this question is not to confuse the momentary dip right after the container expands with the final state after equilibrium re-establishes itself — these are two different things and the question asks about both. (a) Initial effect on vapour pressure At the instant the volume is increased, no new liquid has had time to evaporate yet — the same number of vapour molecules, n, are now spread through a larger volume V. Treating the vapour as an ideal gas, PV = nRT ⇒ P = nRT/V where P is the vapour pressure, n is the moles of vapour present, R is the gas constant, T is the (constant) temperature, and V is the volume of the vapour space. Since n and T haven't changed yet, but V has suddenly gone up, P must drop immediately — the same amount of gas is simply less crowded. This is the instantaneous effect, before evaporation has a chance to respond. The step people get wrong: it's tempting to say "vapour pressure only depends on temperature, so nothing changes." That's true only once equilibrium is re-established — right after the disturbance, before evaporation catches up, the pressure genuinely falls. (b) Rates of evaporation and condensation, immediately after - Rate of evaporation depends on the temperature and the exposed surface area of the liquid — both unchanged by the container expanding. So the rate of evaporation stays exactly the same as it was before the volume change. - Rate of condensation depends on how many vapour molecules are striking the liquid surface per second, which depends on the vapour's concentration (its pressure). Since the vapour pressure has just dropped (part a), fewer molecules are hitting the surface, so the rate of condensation decreases. So immediately after the expansion, rate of evaporation rate of condensation — the two rates are no longer equal, and the system is out of equilibrium in the direction of net evaporation. (c) Restoring equilibrium and the final vapour pressure Because evaporation now outpaces condensation, liquid keeps evaporating into the larger space. As it does, the vapour's concentration — and hence its pressure — rises back up, which in turn raises the rate of condensation. This continues until the rate of condensation again equals the rate of evaporation, at which point dynamic equilibrium is restored. Since temperature hasn't changed anywhere in this process, and the vapour pressure of a liquid in equilibrium with its vapour depends only on temperature (provided enough liquid remains to keep supplying vapour), the final equilibrium vapour pressure comes back to the same value it had before the volume was increased — the extra space just gets filled with more vapour until the original pressure is reached again. Answer: (a) Vapour pressure decreases immediately after the volume increase. (b) Rate of evaporation is unchanged; rate of condensation decreases, so evaporation outpaces condensation. (c) Net evaporation continues until the two rates become equal again, and the final vapour pressure equals the original (initial) vapour pressure, since it depends only on temperature.

Question 2

What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO 2] = 0.60M, [O 2] = 0.82M and [SO 3] = 1.90M? 2SO 2(g) + O 2(g) ⇌ 2SO 3(g)

Accepted Answer

12.229. K c is built straight from the balanced equation — each concentration raised to its own stoichiometric coefficient, and the reaction must be read as reversible. The reactants and products here run together with no symbol between them because the arrow got lost in transcription; the equilibrium is 2 SO 2(g) + O 2(g) 2 SO 3(g) For a general equilibrium aA + bB cC, the equilibrium constant is K c = ([ C ]^c)/([ A ]^a[ B ]^b) where each bracket is the equilibrium concentration (mol L ^-1 ) of that species, and the exponent is its coefficient in the balanced equation — not the number "2" in front of SO 2 meaning "multiply by 2," but "raise to the power 2." For this reaction, SO 3 is the product (coefficient 2) and SO 2, O 2 are reactants (coefficients 2 and 1): K c = ([ SO 3]^2)/([ SO 2]^2[ O 2]) Substituting the given equilibrium concentrations — [ SO 2] = 0.60 M, [ O 2] = 0.82 M, [ SO 3] = 1.90 M: K c = ((1.90 M )^2)/((0.60 M )^2(0.82 M )) Work out the powers first, keeping units attached at every step: (1.90 M )^2 = 3.61 M^2 (0.60 M )^2 = 0.36 M^2 Then multiply the denominator terms: (0.36 M^2)(0.82 M ) = 0.2952 M^3 Now divide: K c = (3.61 M^2)/(0.2952 M^3) = 12.229... M^-1 Watch the units here — this is the step people skip. K c is not automatically unitless. Its units come from the net power of concentration left over after the exponents cancel: on top there are 2 powers of M, on the bottom there are 2+1=3 powers of M, leaving M^2/ M^3 = M^-1 = mol^-1 L. Only when the total moles of gas are the same on both sides of the equation does K c come out dimensionless — that is not the case here (3 mol of gas → 2 mol of gas). Rounding once, at the end: the two least-precise inputs, 0.60 M and 0.82 M, each carry only two significant figures, so the answer is honestly known to two significant figures: K c ≈ 12 mol^-1 L Answer: K c ≈ 12 mol^-1 L (unrounded, 12.229 mol^-1 L ).

Question 3

At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms I 2 (g) ⇌ 2I (g) Calculate Kp for the equilibrium.

Accepted Answer

2.67 × 104. For gases mixed at the same total pressure, each gas's partial pressure is its volume fraction times the total pressure — volume fraction and mole fraction are the same thing for an ideal gas. The equilibrium is I 2(g) 2 I (g) Iodine vapour is reported as 40% I atoms by volume, so the rest — 60% — is undissociated I 2. Step 1: Convert volume percentages to partial pressures. For a gas mixture, partial pressure of a component p i is given by p i = x i × P total where x i is the mole (= volume) fraction of that component and P total is the total pressure of the mixture. Here P total = 10^5 Pa, x I = 0.40, and x I 2 = 0.60. p I = 0.40 × 10^5 Pa = 4×10^4 Pa p I 2 = 0.60 × 10^5 Pa = 6×10^4 Pa The step people skip: a mixture's total pressure is the sum of the partial pressures of every species present — you cannot use the total pressure itself in place of either partial pressure in the equilibrium expression. Step 2: Write K p for the reaction. For I 2(g) 2 I (g), the equilibrium constant in terms of pressure is K p = (p I )^2 p I 2 with p I and p I 2 the equilibrium partial pressures of I(g) and I 2 (g) found above. The exponent 2 on p I comes directly from the stoichiometric coefficient of I(g) in the balanced equation — this is the mistake to watch for: forgetting to square the product's pressure because its coefficient is 2, not 1. Step 3: Substitute and evaluate. K p = ((4×10^4 Pa )^2)/(6×10^4 Pa ) = (16×10^8 Pa^2)/(6×10^4 Pa ) K p = 16/6×10^8-4 Pa = 2.667×10^4 Pa Since the given data (40%) carries about three significant figures at best, the result is rounded to three significant figures. Answer: K p ≈ 2.67 × 10^4 Pa

Question 4

Write the expression for the equilibrium constant, Kc for each of the following reactions: (i) 2NOCl (g) ⇌ 2NO (g) + Cl 2 (g) (ii) 2Cu(NO 3) 2 (s) ⇌ 2CuO (s) + 4NO 2 (g) + O 2 (g) (iii) CH 3COOC 2H 5(aq) + H 2O(l) ⇌ CH 3COOH (aq) + C 2H 5OH (aq) (iv) Fe 3 + (aq) + 3OH^- (aq) ⇌ Fe(OH) 3 (s) (v) I 2 (s) + 5F 2 ⇌ 2IF 5

Accepted Answer

The equilibrium constant expression comes straight from the balanced equation — but only species whose concentration actually changes during the reaction go into it. For a general reaction aA + bB cC + dD the law of mass action gives K c = ([C]^c[D]^d)/([A]^a[B]^b) where the square brackets mean equilibrium molar concentration and the exponents are the balancing coefficients taken straight from the equation — not experimentally measured reaction orders. A pure solid or a pure liquid present in large excess has a concentration fixed by its own density and molar mass, not by how much of it happens to be around, so that "concentration" is a constant and gets absorbed into K c itself rather than written out. Anything marked aqueous, (aq), or gaseous, (g), does change concentration as the reaction proceeds, so it always appears. (i) 2 NOCl (g) 2 NO (g) + Cl 2(g) Every species here is a gas, so all three appear, each raised to its coefficient: K c = ([ NO ]^2[ Cl 2])/([ NOCl ]^2) (ii) 2 Cu(NO 3) 2(s) 2 CuO (s) + 4 NO 2(g) + O 2(g) Cu(NO 3) 2 and CuO are both solids — leave them out completely, not as a factor of "1" sitting in the expression, just absent: K c = [ NO 2]^4[ O 2] (iii) CH 3 COOC 2 H 5(aq) + H 2 O (l) CH 3 COOH (aq) + C 2 H 5 OH (aq) This is the one people get backwards. Water is a liquid, and the reflex is to drop every liquid the same way CuO (s) was dropped above. But the "pure liquid" rule only applies when that liquid is the bulk solvent, present in such vast excess that consuming some of it in the reaction barely moves its own concentration. In this ester hydrolysis, water is one of only two reactants in a mixture where all four species are present in comparable amounts — its concentration genuinely falls as the reaction goes forward, so it stays in the expression exactly like any other reactant, sitting in the denominator: K c = ([ CH 3 COOH ][ C 2 H 5 OH ])/([ CH 3 COOC 2 H 5][ H 2 O ]) (iv) Fe^3+(aq) + 3 OH^-(aq) Fe(OH) 3(s) Fe(OH) 3 precipitates out as a solid, so it drops out of the expression, leaving only the two ions — and since they sit on the reactant side of an equation whose only product is excluded, K c ends up as a plain reciprocal: K c = 1/([ Fe^3+][ OH^-]^3) (v) I 2(s) + 5 F 2(g) 2 IF 5(g) Solid iodine is excluded; fluorine and iodine pentafluoride are both gases, so both appear, each raised to its coefficient: K c = ([ IF 5]^2)/([ F 2]^5) Answer: (i) K c = ([ NO ]^2[ Cl 2])/([ NOCl ]^2); (ii) K c = [ NO 2]^4[ O 2]; (iii) K c = ([ CH 3 COOH ][ C 2 H 5 OH ])/([ CH 3 COOC 2 H 5][ H 2 O ]); (iv) K c = 1/([ Fe^3+][ OH^-]^3); (v) K c = ([ IF 5]^2)/([ F 2]^5)

Question 5

Find out the value of Kc for each of the following equilibria from the value of Kp: (i) 2NOCl (g) ⇌ 2NO (g) + Cl 2 (g); Kp= 1.8 × 10–2 at 500 K (ii) CaCO 3 (s) ⇌ CaO(s) + CO 2(g); Kp= 167 at 1073 K

Accepted Answer

(i) 4.33 × 10–4 (ii) 1.90. K p and K c are linked by K p = K c(RT)^Δ n, and only GASEOUS species count in Δ n — solids drop out of the expression entirely. Here K p is the equilibrium constant written in terms of partial pressures (in bar), K c is the equilibrium constant written in terms of molar concentrations, R = 0.0831 bar L K^-1 mol^-1 is the gas constant (matched to the bar unit K p is given in), T is the absolute temperature in kelvin, and Δ n is (moles of gaseous products) - (moles of gaseous reactants), read straight off the balanced equation — the arrow missing from the printed equations here is a reversible arrow,, since every species is present at equilibrium simultaneously. Solving the relation for K c: K c = K p/((RT)^Δ n) Part (i): 2 NOCl(g) 2 NO(g) + Cl 2 (g), K p = 1.8×10^-2 at T = 500 K Every species here is a gas, so Δ n is just (moles of gas on the product side) minus (moles of gas on the reactant side): Δ n = (2 + 1) - 2 = 1 Substituting into K c = K p/((RT)^Δ n): K c = (1.8×10^-2)/((0.0831 × 500)^1) = (1.8×10^-2)/41.55 K c = 4.33×10^-4 mol L^-1 Part (ii): CaCO 3 (s) CaO(s) + CO 2 (g), K p = 167 at T = 1073 K This is the step people get wrong: CaCO 3 (s) and CaO(s) are pure solids, and a pure solid has fixed "concentration" (its density divided by its molar mass), so it never appears in an equilibrium-constant expression at all. Only CO 2 (g) is counted — not zero (from "everything cancels"), and not two (from counting all three formulas written in the equation): Δ n = 1 - 0 = 1 Substituting into K c = K p/((RT)^Δ n): K c = 167/((0.0831 × 1073)^1) = 167/89.17 K c = 1.87 mol L^-1 Both results are rounded only at this last step, to three significant figures, matching the precision of the given K p values and temperatures. Answer: K c = 4.33×10^-4 mol L^-1 for (i), and K c = 1.87 mol L^-1 for (ii)

Question 6

For the following equilibrium, Kc= 6.3 × 1014 at 1000 K NO (g) + O 3 (g) ⇌ NO 2 (g) + O 2 (g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

Accepted Answer

1.59 × 10–15. Reversing an equilibrium flips its equilibrium constant — the new K c is the reciprocal of the old one. The equilibrium is NO (g) + O 3 (g) NO 2 (g) + O 2 (g) with the forward reaction having K c = 6.3 × 10^14 at 1000 K For any equilibrium, the law of mass action gives the forward constant as K c( forward ) = ([ NO 2][ O 2])/([ NO ][ O 3]) where the square brackets denote equilibrium molar concentrations of each species. The reverse reaction is the same equilibrium written the other way round, NO 2 (g) + O 2 (g) NO (g) + O 3 (g) so its equilibrium constant expression is just the reciprocal of the ratio above: K c( reverse ) = ([ NO ][ O 3])/([ NO 2][ O 2]) = 1/(K c( forward )) This reciprocal relationship holds for any equilibrium — it does not depend on whether the reactions are elementary steps or not. The extra fact given here, that both the forward and reverse steps are elementary bimolecular reactions, is what lets you also write each rate law directly from its stoichiometry (rate = k f[ NO ][ O 3] forward, rate = k b[ NO 2][ O 2] reverse, with K c = k f/k b ) — but you don't need the rate constants at all to answer this question, since K c( reverse ) is fixed the moment K c( forward ) is known. A common slip here is to think a "different" reaction needs a fresh calculation from scratch. It doesn't — reversing a reaction only inverts the constant. Substituting the given value: K c( reverse ) = 1/(6.3 × 10^14) K c( reverse ) = 0.1587 × 10^-14 = 1.587 × 10^-15 The given K c has 2 significant figures, so the result is rounded to 2 significant figures: K c( reverse ) ≈ 1.6 × 10^-15 Answer: K c (reverse) = 1.6 × 10^-15

Question 7

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Accepted Answer

The concentration of a pure solid or a pure liquid is a fixed number, not a variable — so it never actually appears as a "moving part" in the equilibrium expression. Concentration is defined as c = n/V = ( mass )/( molar mass × V) where n is moles, V is volume, and mass/ V is density. For a pure solid or a pure liquid, density and molar mass are properties of the substance itself at a given temperature — they do not depend on how much of the solid or liquid is sitting in the flask. Whether there is 1 g or 100 g of a pure solid present, its concentration (moles per unit volume of that solid) works out to the same value, because both the mass and the volume it occupies scale together. So c pure solid/liquid = ( density )/( molar mass ) = constant at a given T This is the step people trip on: concentration is only a meaningful, changing quantity for something whose amount can vary independently of its volume — true for gases and for solutes in a solution, but not true for a pure solid or liquid, which simply occupies whatever volume its fixed density dictates. Take the decomposition of calcium carbonate: CaCO 3(s) CaO (s) + CO 2(g) Written naively, the equilibrium constant would be K c' = ([ CaO ][ CO 2])/([ CaCO 3]) But [ CaO ] and [ CaCO 3] are both constants (fixed density/molar mass values, unaffected by how much of each solid is present). Since K c' is itself a constant at a given temperature, the two solid-concentration constants can be multiplied through and absorbed into it, defining a new constant: K c = K c' × ([ CaCO 3])/([ CaO ]) = [ CO 2] So the equilibrium constant expression is simply K c = [ CO 2]^- the pure solids drop out not because their concentrations are zero or unimportant to the chemistry, but because their concentrations are constants that get folded into the value of K itself rather than appearing as variables in it. The same reasoning applies to any pure liquid taking part in a reaction (for example, H 2 O (l) in a reaction where water is a reactant/product but not the solvent). Answer: Pure solids and pure liquids have a fixed concentration (density ÷ molar mass) at a given temperature, independent of how much of them is present. Since this constant value would only multiply into the already-constant K, it is absorbed into K itself rather than written explicitly — so only the concentrations of gases and dissolved species, whose concentrations actually vary, appear in the equilibrium constant expression.

Question 8

Reaction between N 2 and O 2 – takes place as follows: 2N 2 (g) + O 2 (g) ⇌ 2N 2O (g) If a mixture of 0.482 mol N 2 and 0.933 mol of O 2 is placed in a 10 L reaction vessel and allowed to form N 2O at a temperature for which Kc= 2.0 × 10–37, determine the composition of equilibrium mixture.

Accepted Answer

[N2] = 0.0482 molL–1, [O2] = 0.0933 molL–1, [N2O] = 6.6 × 10–21 molL–1. A vanishingly small K c is the answer before you even solve for x — it tells you the reaction barely moves forward, so almost all of the N 2 and O 2 you started with is still there at equilibrium. The missing symbol in the printed reaction is a reversible arrow: this is an equilibrium, not a one-way reaction. 2 N 2(g) + O 2(g) 2 N 2O(g), K c = 2.0×10^-37 Step 1 — turn moles into concentrations. Concentration is moles per litre of the reaction mixture's volume, c = n/V, not per mole of anything else: [ N 2 ] 0=(0.482 mol )/(10 L )=0.0482 mol L^-1, [ O 2 ] 0=(0.933 mol )/(10 L )=0.0933 mol L^-1, [ N 2O ] 0=0 Step 2 — set up the ICE table. Let x be the mol L ^-1 of O 2 consumed (its coefficient is 1, so the changes in the other species are just multiples of x ): N 2 O 2 N 2O --- --- --- --- Initial 0.0482 0.0933 0 Change -2x -x +2x Equilibrium 0.0482-2x 0.0933-x 2x Step 3 — write K c, matching every exponent to its stoichiometric coefficient. This is the step people get wrong: N 2 and N 2O both carry coefficient 2, so both get squared, while O 2 has coefficient 1 and stays unsquared. K c=([ N 2O ]^2)/([ N 2 ]^2[ O 2 ])=((2x)^2)/((0.0482-2x)^2(0.0933-x))=2.0×10^-37 Step 4 — exploit how small K c is. A K c of 10^-37 means the forward reaction essentially does not happen; x will turn out to be many orders of magnitude smaller than 0.0482 or 0.0933. So approximate 0.0482-2x≈0.0482 and 0.0933-x≈0.0933 — an approximation that is only safe because K c is this tiny; you would never drop x like this if K c were close to 1. ((2x)^2)/((0.0482)^2(0.0933))=2.0×10^-37 Step 5 — solve for x. (0.0482)^2(0.0933)=(2.323×10^-3)(0.0933)=2.168×10^-4 (2x)^2=2.0×10^-37×2.168×10^-4=4.335×10^-41 2x=√ 4.335×10^-41 =6.58×10^-21 mol L^-1 ⇒ x=3.29×10^-21 mol L^-1 Checking the approximation: x 10^-21 is indeed utterly negligible next to 0.0482 and 0.0933, so dropping it was justified. Step 6 — read off the equilibrium composition. [ N 2O ] = 2x = 6.6×10^-21 mol L^-1 ⇒ n( N 2O ) = 6.6×10^-21 mol L^-1×10 L = 6.6×10^-20 mol [ N 2 ] ≈ 0.0482 mol L^-1 ⇒ n( N 2 )≈0.482 mol (unchanged to 3 s.f.) [ O 2 ] ≈ 0.0933 mol L^-1 ⇒ n( O 2 )≈0.933 mol (unchanged to 3 s.f.) The two sig figs in the final x come from the two sig figs given in K c; the physical picture is that with K c this small, the "equilibrium mixture" is, for all practical purposes, still just the starting 0.482 mol N 2 and 0.933 mol O 2, with only a trace — about 6.6×10^-20 mol — of N 2O formed. Answer: at equilibrium, n( N 2 )≈0.482 mol, n( O 2 )≈0.933 mol (both essentially unchanged), and n( N 2O )≈6.6×10^-20 mol — a negligible amount, consistent with K c=2.0×10^-37.

Question 9

Nitric oxide reacts with Br 2 and gives nitrosyl bromide as per reaction given below: 2NO (g) + Br 2 (g) ⇌ 2NOBr (g) When 0.087 mol of NO and 0.0437 mol of Br 2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br 2.

Accepted Answer

0.0352mol of NO and 0.0178mol of Br2. Use the reaction stoichiometry to convert "how much product formed" into "how much reactant was used up" — the mole ratios in the balanced equation are the conversion factors. The reaction is 2 NO (g) + Br 2(g) 2 NOBr (g) Because the coefficients are 2: 1: 2, every 2 mol of NOBr formed corresponds to 2 mol of NO consumed and 1 mol of Br 2 consumed. In other words, moles of NO used up equal moles of NOBr formed (both carry coefficient 2), while moles of Br 2 used up are half the moles of NOBr formed. Set up an ICE table (Initial, Change, Equilibrium) in moles. Let x = mol of Br 2 consumed. Then, by the 2:1:2 ratio, 2x mol of NO is consumed and 2x mol of NOBr is formed. Use the given equilibrium amount of NOBr to find x. At equilibrium, NOBr = 0.0518 mol, so 2x = 0.0518 mol ⇒ x = 0.0518/2 mol = 0.0259 mol This x is the moles of Br 2 consumed; 2x = 0.0518 mol is the moles of NO consumed — note that the NO consumed is numerically equal to the NOBr formed only because both have the same stoichiometric coefficient (2). This is the step people get wrong: they sometimes halve the NOBr value again for NO, but NO and NOBr share the same coefficient, so no halving is needed for NO — the halving applies only to Br 2. Subtract the amount consumed from the initial amount to get the equilibrium amount of each reactant. For NO: n( NO ) eq = 0.087 mol - 2x = 0.087 mol - 0.0518 mol = 0.0352 mol For Br 2: n( Br 2) eq = 0.0437 mol - x = 0.0437 mol - 0.0259 mol = 0.0178 mol Both results are reported to three significant figures, matching the precision of the given data (0.087, 0.0437, 0.0518 mol). Answer: equilibrium amount of NO = 0.0352 mol, equilibrium amount of Br 2 = 0.0178 mol.

Question 10

At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium. 2SO 2(g) + O 2(g) ⇌ 2SO 3 (g) What is Kc at this temperature?

Accepted Answer

7.47 × 1011 M–1. The relationship between K p and K c hinges on Δ n g, the change in moles of gas as reactants become products — get that number wrong (or its sign) and every later step flips. The reaction, written with the reversible arrow that belongs at equilibrium, is 2SO 2(g) + O 2(g) 2SO 3(g) Step 1 — Relate K p and K c. For a gas-phase equilibrium, K p = K c (RT)^Δ n g where K p is the equilibrium constant in terms of partial pressures, K c is the equilibrium constant in terms of molar concentrations, R is the gas constant, T is the absolute temperature, and Δ n g is (moles of gaseous products) − (moles of gaseous reactants) in the balanced equation. Step 2 — Find Δ n g. From the balanced equation, gaseous products = 2 mol ( SO 3 ), and gaseous reactants = 2 + 1 = 3 mol ( SO 2 and O 2 ): Δ n g = 2 - (2+1) = -1 A negative Δ n g just says the reaction consumes more moles of gas than it makes — nothing to be alarmed about, but it is exactly the number that must carry its minus sign into the next step. Step 3 — Solve for K c. Rearranging K p = K c(RT)^Δ n g for K c: K c = K p/((RT)^Δ n g) = K p (RT)^-Δ n g Since Δ n g = -1, the exponent -Δ n g = 1, so this simplifies to K c = K p × RT This is the step people get backwards: with Δ n g negative, dividing by (RT)^Δ n g is the same as multiplying by RT — not dividing by it. Step 4 — Substitute the numbers. Because K p here is built from pressures in bar, R must be used in the matching units: R = 0.0831 bar dm^3 K^-1 mol^-1, T = 450 K, K p = 2.0 × 10^10 bar^-1 First find RT: RT = 0.0831 bar dm^3 K^-1 mol^-1 × 450 K = 37.395 bar dm^3 mol^-1 Now substitute into K c = K p × RT: K c = (2.0 × 10^10 bar^-1) × (37.395 bar dm^3 mol^-1) The bar and bar^-1 cancel, leaving units of dm^3 mol^-1 — exactly what is expected for K c when Δ n g = -1, since K c then carries units of ( mol dm^-3)^-1. K c = 2.0 × 10^10 × 37.395 dm^3 mol^-1 = 7.479 × 10^11 dm^3 mol^-1 K p was given to 2 significant figures, so the final value is rounded to 2 significant figures: K c ≈ 7.5 × 10^11 dm^3 mol^-1 Answer: K c ≈ 7.5 × 10^11 dm^3 mol^-1 (equivalently L mol^-1 ).

Question 11

A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium? 2HI (g) ⇌ H 2 (g) + I 2 (g)

Accepted Answer

4.0. When a substance decomposes on its own in a sealed flask, the pressure it loses is exactly what pairs up (in mole ratio) to form the products — track that loss with an ICE table in partial pressures. The equilibrium is 2 HI(g) H 2 (g) + I 2 (g) Only HI(g) is placed in the flask, so at the start there is no H 2 or I 2 present — they appear only as HI breaks down. Step 1: Set up the ICE table in partial pressures. Let p be the fall in the partial pressure of HI(g) as it reacts. Since 2 mol HI produce 1 mol H 2 and 1 mol I 2, the pressure of each product rises by p/2. HI(g) H 2 (g) I 2 (g) --- --- --- --- Initial 0.2 atm 0 0 Change -p +p/2 +p/2 Equilibrium 0.2-p p/2 p/2 Step 2: Use the given equilibrium pressure of HI(g) to find p. The problem states the equilibrium partial pressure of HI(g) is 0.04 atm, so 0.2 - p = 0.04 atm ⇒ p = 0.2 - 0.04 = 0.16 atm This p is the pressure of HI consumed, not the pressure of either product — that is the substitution step that trips people up here, because the stoichiometric coefficient of HI is 2 while those of H 2 and I 2 are 1 each. Step 3: Find the equilibrium partial pressures of the products. p H 2 = p I 2 = p/2 = (0.16 atm )/2 = 0.08 atm So at equilibrium: p HI = 0.04 atm, p H 2 = 0.08 atm, p I 2 = 0.08 atm Step 4: Write K p from the balanced equation and substitute. For 2 HI(g) H 2 (g) + I 2 (g), the equilibrium-constant expression (products over reactants, each partial pressure raised to its stoichiometric coefficient) is K p = p H 2 p I 2 (p HI )^2 Substituting the equilibrium pressures: K p = ((0.08 atm )(0.08 atm ))/((0.04 atm )^2) = (6.4×10^-3 atm^2)/(1.6×10^-3 atm^2) K p = 4 The atm ^2 in the numerator cancels exactly against the atm ^2 in the denominator, so K p here comes out as a pure number with no unit — this happens because the number of gas moles does not change across the reaction ( Δ n g = (1+1) - 2 = 0 ), so the pressure units in K p = K c(RT)^Δ n g never appear in the first place. Answer: K p = 4 (no units, since Δ n g = 0 )

Question 12

A mixture of 1.57 mol of N 2, 1.92 mol of H 2 and 8.13 mol of NH 3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Accepted Answer

Qc = 2.379 × 103. No, reaction is not at equilibrium.. K c is defined in terms of concentrations (mol per litre), never in terms of raw mole counts — so before you can compare anything to K c, you have to divide every mole amount by the volume of the vessel. The reaction is N 2(g) + 3H 2(g) 2NH 3(g) Step 1: convert moles to molar concentrations Use [X] = n X/V, where n X is the number of moles of species X present right now and V is the volume of the container in litres. Here V = 20 L: [N 2] = (1.57 mol )/(20 L ) = 0.0785 mol L^-1 [H 2] = (1.92 mol )/(20 L ) = 0.0960 mol L^-1 [NH 3] = (8.13 mol )/(20 L ) = 0.4065 mol L^-1 Step 2: write the reaction quotient Q c, not K c Since you don't yet know whether the system is at equilibrium, you plug the current concentrations into the equilibrium-constant expression and call the result Q c: Q c = ([NH 3]^2)/([N 2][H 2]^3) The exponents 2, 1, 3 come straight from the balanced equation's coefficients — this is the step people skip and then get a wrong power of [H 2]. Step 3: substitute and evaluate [H 2]^3 = (0.0960 mol L^-1)^3 = 8.847×10^-4 mol^3 L^-3 [N 2][H 2]^3 = (0.0785 mol L^-1)(8.847×10^-4 mol^3 L^-3) = 6.945×10^-5 mol^4 L^-4 [NH 3]^2 = (0.4065 mol L^-1)^2 = 0.16524 mol^2 L^-2 Q c = (0.16524 mol^2 L^-2)/(6.945×10^-5 mol^4 L^-4) = 2379 mol^-2 L^2 Rounding to two significant figures — matching the precision of the given K c — gives Q c ≈ 2.4×10^3 (The units on Q c are the same as the units on K c for this reaction, since both come from the identical expression, so they cancel out of the comparison in Step 4 either way.) Step 4: compare Q c to K c The given equilibrium constant is K c = 1.7×10^2. Since Q c (2.4×10^3) K c (1.7×10^2) the mixture is not at equilibrium. A larger-than-equilibrium Q c means the numerator — the product, NH 3 — is present in excess relative to what equilibrium allows, given how much N 2 and H 2 are on hand. This is the point where it's easy to get the direction backwards: Q c K c does not push the reaction forward to make more product — it pushes it in the direction that consumes the excess product. So the net reaction runs in reverse, from right to left: some NH 3 decomposes back into N 2 and H 2 until the concentrations rearrange themselves so that Q c falls to 1.7×10^2. Answer: Q c ≈ 2.4×10^3, which is greater than K c = 1.7×10^2, so the mixture is not at equilibrium; the net reaction proceeds in the reverse direction ( NH 3 decomposing to N 2 and H 2 ) until equilibrium is reached.

Question 13

The equilibrium constant expression for a gas reaction is, K c = [ NH 3 ]^4[ O 2 ]^5 [ NO ]^4[ H 2O ]^6 Write the balanced chemical equation corresponding to this expression.

Accepted Answer

In an equilibrium-constant expression, every species written in the numerator is a product and every species in the denominator is a reactant — the exponent it carries is its coefficient in the balanced equation. The general form of an equilibrium-constant expression (the law of mass action) for a reaction aA + bB cC + dD is K c = ([C]^c[D]^d)/([A]^a[B]^b) Here the square brackets mean equilibrium molar concentration, and the exponents a, b, c, d are exactly the stoichiometric coefficients from the balanced equation: the products C and D sit on top, the reactants A and B sit on the bottom. The given expression is K c = ([ NH 3 ]^4[ O 2 ]^5)/([ NO ]^4[ H 2O ]^6) Matching this against the general form: - The numerator carries [ NH 3 ]^4 and [ O 2 ]^5 — so NH 3 and O 2 are the products here, with coefficients 4 and 5. - The denominator carries [ NO ]^4 and [ H 2O ]^6 — so NO and H 2O are the reactants here, with coefficients 4 and 6. This is the step that trips people up: it feels natural to put NH 3 and O 2 on the reactant side, because the familiar combustion of ammonia is written 4 NH 3 + 5 O 2 → 4 NO + 6 H 2O. But the K c expression you are given is what decides the direction — whatever sits above the line is a product of this equilibrium, regardless of which arrangement looks more familiar chemically. Reading the exponents straight off the fraction, the equation runs the other way: 4 NO(g) + 6 H 2O(g) 4 NH 3(g) + 5 O 2(g) Check this is actually balanced, atom for atom, since a set of exponents only means something if it comes from a real balanced equation: - Nitrogen: left side has 4 × 1 = 4 (from NO); right side has 4 × 1 = 4 (from NH 3 ). - Hydrogen: left side has 6 × 2 = 12 (from H 2 O); right side has 4 × 3 = 12 (from NH 3 ). - Oxygen: left side has 4 × 1 + 6 × 1 = 10 (from NO and H 2 O); right side has 5 × 2 = 10 (from O 2 ). All three atoms balance, confirming the equation. Answer: 4 NO(g) + 6 H 2O(g) 4 NH 3(g) + 5 O 2(g)

Question 14

One mole of H 2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H 2O (g) + CO (g) ⇌ H 2 (g) + CO 2 (g) Calculate the equilibrium constant for the reaction.

Accepted Answer

0.44. A gas-phase equilibrium constant is built from equilibrium concentrations, not from the moles you started with, so the first job is turning "40 percent reacted" into an ICE table. The reaction is H 2 O (g) + CO (g) H 2(g) + CO 2(g) Step 1: Convert "40 percent of water reacts" into moles. You start with 1 mole of H 2O and 1 mole of CO in a 10 L vessel. Since water is a single substance here, 40 percent by mass reacting is the same as 40 percent of its moles reacting (mass and moles are proportional through one fixed molar mass): moles of H 2 O reacted = 0.40 × 1 mol = 0.4 mol From the 1:1 stoichiometry, exactly 0.4 mol of CO is also consumed, and 0.4 mol each of H 2 and CO 2 is produced. Step 2: Build the ICE table (in moles). H 2O CO H 2 CO 2 --- --- --- --- --- Initial 1 1 0 0 Change -0.4 -0.4 +0.4 +0.4 Equilibrium 0.6 0.6 0.4 0.4 A step people get wrong here: forgetting that CO is consumed by the same amount as H2O. The equation shows 1 mol H 2O reacting with 1 mol CO, so whatever moles of water react, exactly that many moles of CO react too. It is not "40 percent of CO" independently; it is the CO that reacted alongside that water. Step 3: Convert moles to concentrations. Concentration is moles divided by the volume of the vessel, V = 10 L: [ H 2 O ] = (0.6 mol )/(10 L ) = 0.06 mol L^-1, [ CO ] = (0.6 mol )/(10 L ) = 0.06 mol L^-1 [ H 2] = (0.4 mol )/(10 L ) = 0.04 mol L^-1, [ CO 2] = (0.4 mol )/(10 L ) = 0.04 mol L^-1 Step 4: Write the equilibrium constant expression and substitute. Kc is the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient: K c = ([ H 2][ CO 2])/([ H 2 O ][ CO ]) Substituting the equilibrium concentrations: K c = ((0.04 mol L^-1)(0.04 mol L^-1))/((0.06 mol L^-1)(0.06 mol L^-1)) = 0.0016/0.0036 The unit aside people skip: here every concentration term carries mol per litre, and there are two of them on top and two on the bottom, so the units cancel completely. This happens because the reaction has equal moles of gas on both sides (2 mol reactants to 2 mol products, so the change in gas moles is zero), which is also why you could have used moles directly instead of concentrations. The factor of 1/V (here 1/10) appears twice in the numerator and twice in the denominator and cancels exactly: K c = ((0.4)(0.4))/((0.6)(0.6)) = 0.16/0.36 = 0.444… Rounding to three significant figures (matching the precision of the given 40 percent): K c = 0.444 Answer: Kc = 0.444 (dimensionless, since the moles of gas are equal on both sides of the equation).

Question 15

At 700 K, equilibrium constant for the reaction: H 2 (g) + I 2 (g) ⇌ 2HI (g) is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H 2(g) and I 2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

Accepted Answer

0.068 molL–1 each of H2 and I2. When a reaction is run in reverse, the same K c still governs the equilibrium mixture — you just have to write the ICE table for the reaction as it is actually proceeding. Here you start with pure HI (g) and let it dissociate, so the reaction that is actually happening is the reverse of the one you were given the constant for: 2 HI (g) H 2(g) + I 2(g) The forward reaction H 2(g) + I 2(g) 2 HI (g) has K c = ([ HI ]^2)/([ H 2][ I 2]) = 54.8 at 700 K This is the same equilibrium state whichever direction you approach it from, so this same expression (with this same numerical K c ) still connects the equilibrium concentrations — you do not need a separate "reverse" constant, you just solve for [ H 2] and [ I 2] from it. Setting up the ICE table. Because the vessel started with only HI (g) (no H 2 or I 2 present initially), the stoichiometry 2 HI H 2 + I 2 tells you that for every 2x mol L ^-1 of HI that decomposes, exactly x mol L ^-1 of H 2 and x mol L ^-1 of I 2 are produced — in a 1:1 ratio with each other, because both come from the same starting substance: H 2 I 2 --- --- --- At equilibrium x x This is the step people skip: [ H 2]=[ I 2] here is not an assumption, it is forced by the fact that H 2 and I 2 only entered the mixture by decomposing the same HI, in equal amounts. Substituting into K c. With [ HI ] = 0.5 mol L^-1 at equilibrium and [ H 2] = [ I 2] = x: 54.8 = ((0.5)^2)/(x · x) = 0.25/x^2 Solve for x^2: x^2 = 0.25/54.8 = 4.562 × 10^-3 mol^2 L^-2 Take the square root (carrying the full value, not yet rounded): x = √ 4.562 × 10^-3 = 6.754 × 10^-2 mol L^-1 The data ( K c = 54.8, three significant figures) justifies rounding only at this last step, to three significant figures: x = 6.75 × 10^-2 mol L^-1 Since [ H 2] = [ I 2] = x: [ H 2] = [ I 2] = 6.75 × 10^-2 mol L^-1 Answer: [ H 2] = [ I 2] = 6.75 × 10^-2 mol L^-1

Question 16

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M? 2ICl (g) ⇌ I 2 (g) + Cl 2 (g); Kc = 0.14

Accepted Answer

[I2] = [Cl2] = 0.167 M, [ICl] = 0.446 M. When both product coefficients are 1 and the reactant term is squared on both sides, you can take a square root instead of grinding through the quadratic formula. That is the shortcut this equilibrium is built for. The reaction is 2 ICl (g) I 2(g) + Cl 2(g), K c = 0.14 Only ICl is present at the start, at 0.78 M, so the reaction must go forward (left to right) to reach equilibrium — there is no I 2 or Cl 2 yet for it to go backward. Set up the ICE table. Let x be the concentration (in mol/L) of I 2 formed at equilibrium. Because 2 mol ICl is consumed for every 1 mol I 2 (and 1 mol Cl 2 ) produced, ICl drops by 2x: 2 ICl (g) I 2(g) Cl 2(g) --- --- --- --- Initial (M) 0.78 0 0 Change (M) -2x +x +x Equilibrium (M) 0.78-2x x x Write the equilibrium-constant expression. For this reaction, K c=([ I 2][ Cl 2])/([ ICl ]^2) where each bracket is the equilibrium molar concentration of that species. Substituting the ICE-table entries: 0.14=(x· x)/((0.78-2x)^2)=(x^2)/((0.78-2x)^2) Take the square root of both sides. Both numerator and denominator are already perfect squares, so instead of expanding (0.78-2x)^2 into a quadratic in x, take the square root of each side directly — it's the same equation with far less algebra: √ 0.14 =x/0.78-2x 0.3742=x/0.78-2x (A negative root is rejected here because a square root of a concentration ratio must be positive.) Solve for x. Cross-multiplying: x=0.3742 (0.78-2x)=0.2918-0.7483x x+0.7483x=0.2918 1.7483x=0.2918 x=0.2918/1.7483=0.1669 M Read off the equilibrium concentrations from the ICE table using this x: [ I 2]=[ Cl 2]=x=0.1669 M ≈ 0.167 M [ ICl ]=0.78-2x=0.78-2(0.1669)=0.78-0.3339=0.4461 M ≈ 0.446 M Check the arithmetic by plugging back into K c: the concentration of ICl left over (0.446 M) must be far bigger than what converted (about 0.334 M out of 0.78 M), which matches a K c below 1 — the equilibrium favours the reactant side, so most of the ICl should remain unreacted, and it does. K c=((0.1669)^2)/((0.4461)^2)=0.02786/0.1990=0.140 The data (0.78 M, K c=0.14 ) justify three significant figures in the final concentrations. Answer: [ I 2] = [ Cl 2] ≈ 0.167 M, and [ ICl ] ≈ 0.446 M at equilibrium.

Question 17

Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C 2H 6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? C 2H 6 (g) ⇌ C 2H 4 (g) + H 2 (g)

Accepted Answer

[C2H6]eq = 3.62 atm. K p connects equilibrium PARTIAL PRESSURES — you only switch to concentration at the very last step, using the ideal-gas relation, never earlier. The missing symbol in the printed reaction is the equilibrium arrow — the equation is C 2H 6(g) C 2H 4(g) + H 2(g) Setting up the pressures. Let the total initial pressure of pure C 2H 6 be P 0 = 4.0 atm, and let x be the drop in its partial pressure as it dissociates. Since one mole of C 2H 6 makes one mole each of C 2H 4 and H 2, both products rise by the same x: C 2H 6 C 2H 4 H 2 --- --- --- --- Initial (atm) 4.0 0 0 Change (atm) -x +x +x Equilibrium (atm) 4.0-x x x Writing K p. For this reaction, K p = p C 2H 4 · p H 2 p C 2H 6, where each p is the equilibrium partial pressure in atm. Substituting the table values and the given K p = 0.04 atm at 899 K: 0.04 = (x· x)/(4.0 - x) = x^2/4.0-x Solving the quadratic. Clearing the denominator: x^2 = 0.04(4.0 - x) = 0.16 - 0.04x x^2 + 0.04x - 0.16 = 0 This is ax^2+bx+c=0 with a=1, b=0.04, c=-0.16, so x = (-b ± √ b^2-4ac )/2a = (-0.04 ± √ (0.04)^2 - 4(1)(-0.16) )/2 = (-0.04 ± √ 0.0016 + 0.64 )/2 = (-0.04 ± √ 0.6416 )/2 x = (-0.04 ± 0.8010)/2 A partial pressure change can't be negative, so only the + root is physical: x = (-0.04 + 0.8010)/2 = 0.7610/2 = 0.3805 atm Equilibrium partial pressure of C 2H 6. p C 2H 6 = 4.0 - x = 4.0 - 0.3805 = 3.6195 atm Converting pressure to concentration. K p and the ICE table only ever gave a pressure — the question asks for a concentration, so this is the step people skip. Use the ideal gas law pV = nRT, rearranged to give molar concentration directly: n/V = p/RT where p is the partial pressure (atm), R = 0.0821 L atm K^-1 mol^-1 is the gas constant, T = 899 K is the given temperature, and n/V is the molar concentration (mol L ^-1 ). [ C 2H 6 ] = p C 2H 6 RT = (3.6195 atm )/((0.0821 L atm K^-1 mol^-1)(899 K )) RT = 0.0821 × 899 = 73.81 L atm mol^-1 [ C 2H 6 ] = 3.6195/73.81 mol L^-1 = 0.04904 mol L^-1 The data ( K p = 0.04, P 0 = 4.0 atm ) carries two significant figures, so the concentration rounds to that precision. Answer: [ C 2H 6 ] eq ≈ 4.9 × 10^-2 mol L^-1

Question 18

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: CH 3COOH (l) + C 2H 5OH (l) ⇌ CH 3COOC 2H 5 (l) + H 2O (l) (i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction) (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Accepted Answer

(i) [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH] (ii) 3.92 (iii) value of Qc is less than Kc therefore equilibrium is not attained.. Q c is the same product-over-reactant ratio as K c, but taken at any instant — it only tells you whether a system has reached equilibrium once you compare its value to the true K c. (i) Writing Q c The equilibrium is CH 3 COOH (l) + C 2 H 5 OH (l) CH 3 COOC 2 H 5(l) + H 2 O (l) The reaction quotient is built exactly like K c — products over reactants, each raised to its stoichiometric coefficient (all coefficients here are 1): Q c = ([ CH 3 COOC 2 H 5] [ H 2 O ])/([ CH 3 COOH ] [ C 2 H 5 OH ]) where each bracket is a molar concentration (mol L ^-1 ) at the instant you evaluate it — not necessarily at equilibrium. The question's note that water is "not in excess and not a solvent" is exactly why [ H 2 O ] is written out here — if water were the solvent (as in an aqueous acid-base equilibrium), its concentration would stay essentially constant and get absorbed into K c instead of appearing on its own. (ii) Finding K c at 293 K All four substances share one flask, so every concentration equals (moles)/ V for the same volume V. In the Q c ratio, V appears twice on top and twice on the bottom, so it cancels completely — moles can be used directly in place of concentrations without changing the numerical value of K c. Let x be the moles of ester formed: - CH₃COOH: starts at 1.00 mol, becomes 1.00-x - C₂H₅OH: starts at 0.18 mol, becomes 0.18-x - CH₃COOC₂H₅ and H₂O: start at 0, become x each You are told the equilibrium ester amount is 0.171 mol, so x = 0.171: - CH₃COOH left: 1.00 - 0.171 = 0.829 mol - C₂H₅OH left: 0.18 - 0.171 = 0.009 mol - CH₃COOC₂H₅: 0.171 mol - H₂O: 0.171 mol The ethanol is the step people slip on — you start with only 0.18 mol of it, so after 0.171 mol reacts, just 0.009 mol survives. That is a subtraction of two nearly equal numbers, so keep all three decimal places; rounding 0.18 or 0.171 even slightly changes this small leftover a lot, and it changes K c by that same large fraction. Substitute into K c = Q c at equilibrium: K c = ((0.171)(0.171))/((0.829)(0.009)) = 0.029241/0.007461 K c ≈ 3.92 Both sides of the reaction have two moles of species (1 + 1 reactants, 1 + 1 products), so the concentration units cancel exactly in the ratio — K c here is a pure number, with no units to carry. (iii) Has the second mixture reached equilibrium? This run starts from different amounts — 1.0 mol acetic acid and 0.5 mol ethanol — and reports 0.214 mol of ester "after sometime," language that should prompt you to check rather than assume equilibrium has been reached. At this instant: - CH₃COOH left: 1.0 - 0.214 = 0.786 mol - C₂H₅OH left: 0.5 - 0.214 = 0.286 mol - CH₃COOC₂H₅: 0.214 mol - H₂O: 0.214 mol Compute Q c for this snapshot (moles again stand in for concentrations, by the same volume-cancellation as in part (ii)): Q c = ((0.214)(0.214))/((0.786)(0.286)) = 0.045796/0.224796 ≈ 0.204 A system sits at equilibrium only when Q c = K c. Here Q c ≈ 0.204 is far below K c ≈ 3.92 — nowhere close to equal — so this mixture has not reached equilibrium. Because Q c < K c, the numerator (ester and water) is still too small relative to the denominator (acid and alcohol); the forward, ester-forming reaction keeps running, raising Q c until it climbs up to 3.92. Answer: (i) Q c = ([ CH 3 COOC 2 H 5][ H 2 O ])/([ CH 3 COOH ][ C 2 H 5 OH ]); (ii) K c ≈ 3.92 (unitless); (iii) No — Q c ≈ 0.204 K c ≈ 3.92, so equilibrium has not been reached, and the forward reaction continues.

Question 19

A sample of pure PCl 5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl 5 was found to be 0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl 3 and Cl 2 at equilibrium? PCl 5 (g) ⇌ PCl 3 (g) + Cl 2(g)

Accepted Answer

0.02molL–1 for both.. At equilibrium, PCl₃ and Cl₂ are produced in a strict 1: 1 ratio, so both concentrations can be written as the same unknown x. The reaction is a genuine equilibrium (both directions occur at once), so the missing symbol between reactant and products is the equilibrium arrow, not a plus sign or an omission: PCl 5(g) PCl 3(g) + Cl 2(g) Set up an ICE table. Pure PCl₅ was placed in the vessel, so at the start there is no PCl₃ or Cl₂ — every mole of each that appears has come from the decomposition of PCl₅, one mole of PCl₅ giving one mole of PCl₃ and one mole of Cl₂. PCl 5 PCl 3 Cl 2 --- --- --- --- Initial c 0 0 0 Change -x +x +x Equilibrium c 0-x x x The problem already gives the equilibrium value of PCl₅ directly, so c 0-x does not need to be found separately: [ PCl 5] eq = 0.5×10^-1 mol L^-1 = 5.0×10^-2 mol L^-1 and, by the stoichiometry above, [ PCl 3] eq = [ Cl 2] eq = x Write the equilibrium constant expression. K c is the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient (here all coefficients are 1, and PCl₅ being a gas still appears because this is K c, not K p ): K c=([ PCl 3][ Cl 2])/([ PCl 5]) This is the step it's easy to get wrong: because [ PCl 3]=[ Cl 2]=x, the numerator is x× x = x^2, not 2x — the two concentrations are multiplied together, not added. Substitute the known numbers. 8.3×10^-3=(x× x)/(5.0×10^-2)=(x^2)/(5.0×10^-2) Solve for x. Multiply both sides by 5.0×10^-2 mol L^-1: x^2=8.3×10^-3×5.0×10^-2 mol^2 L^-2=4.15×10^-4 mol^2 L^-2 Take the square root of both sides (units come along under the root, so mol^2 L^-2 becomes mol L^-1 ): x=√ 4.15×10^-4 mol^2 L^-2 = 2.04×10^-2 mol L^-1 (only the positive root is kept, since x is a concentration and cannot be negative) The data ( K c to two significant figures) justifies quoting the result to three significant figures at most, so it is left as 2.04×10^-2 mol L^-1 without further rounding. Both PCl₃ and Cl₂ carry this same value, since the reaction produces them in equal amounts: [ PCl 3] eq=[ Cl 2] eq=2.04×10^-2 mol L^-1 Answer: [ PCl 3] = [ Cl 2] = 2.04×10^-2 mol L^-1

Question 20

One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2. FeO (s) + CO (g) ⇌ Fe (s) + CO 2 (g); Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and CO 2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and = 0.80 atm?

Accepted Answer

[PCO] = 1.739atm, [PCO2] = 0.461atm.. K p only involves the gases — solids like FeO(s) and Fe(s) never appear in the equilibrium expression. For FeO (s) + CO (g) Fe (s) + CO 2 (g) the equilibrium constant in terms of pressure is K p = p CO 2 p CO where p CO 2 and p CO are the partial pressures of the two gases at equilibrium (the pure solids FeO and Fe are left out because their activity is taken as 1). Step 1: Check which way the reaction has to run. Before assuming the system is already at equilibrium, compare the given pressures with K p by forming the reaction quotient Q p: Q p = p CO 2 p CO = (0.80 atm )/(1.4 atm ) = 0.571 Since Q p (0.571) K p (0.265), the system has too much CO 2 relative to CO for equilibrium. The reaction must run in the reverse direction, converting some CO 2 back into CO, until Q p drops to 0.265. This is the step people skip — jumping straight to an ICE table without checking Q p vs K p can leave you assuming the wrong direction of change, which flips the sign of x. Step 2: Set up the ICE table. Let x (atm) be the drop in p CO 2 as the reverse reaction proceeds; the same amount of CO is regenerated (1:1 stoichiometry). p CO p CO 2 --- --- --- Initial 1.4 0.80 Change +x -x Equilibrium 1.4+x 0.80-x Step 3: Substitute into K p and solve for x. K p = (0.80 - x)/(1.4 + x) = 0.265 Cross-multiplying: 0.80 - x = 0.265(1.4+x) 0.80 - x = 0.371 + 0.265x 0.80 - 0.371 = x + 0.265x 0.429 = 1.265x x = 0.429/(1.265 ) = 0.339 atm Step 4: Get the equilibrium partial pressures. p CO = 1.4 + 0.339 = 1.739 atm ≈ 1.74 atm p CO 2 = 0.80 - 0.339 = 0.461 atm ≈ 0.46 atm Check: 0.461/1.739 = 0.265, which matches the given K p, confirming the values are consistent. The data (two significant figures in the initial pressures) justifies rounding the final pressures to three significant figures. Answer: p CO ≈ 1.74 atm and p CO 2 ≈ 0.46 atm at equilibrium.

Question 21

Equilibrium constant, Kc for the reaction N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N 2, 2.0 mol L–1 H 2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Accepted Answer

No, the reaction proceeds to form more products.. Compare the reaction quotient Q c to K c — that comparison, not a memorized rule, tells you which way an unequilibrated mixture will move. The reaction is N 2(g) + 3H 2(g) 2NH 3(g) with K c = 0.061 at 500 K. At the instant described, the mixture holds [N 2] = 3.0 mol L^-1, [H 2] = 2.0 mol L^-1, [NH 3] = 0.5 mol L^-1. These are not known to be equilibrium concentrations — they are just a snapshot — so the correct quantity to build from them is the reaction quotient Q c, which has the same algebraic form as K c but is evaluated at any instant, not only at equilibrium: Q c = ([NH 3]^2)/([N 2][H 2]^3) Substitute the given concentrations: Q c = ((0.5)^2)/((3.0)(2.0)^3) Work the denominator first: (2.0)^3 = 8.0, so (3.0)(8.0) = 24. Work the numerator: (0.5)^2 = 0.25. Q c = 0.25/24 = 0.0104 mol^-2 L^2 Rounded to two significant figures (matching the data), Q c ≈ 0.010. The step people skip: compare Q c with K c, not just compute Q c and stop. Q c (0.010) < K c (0.061) Since Q c < K c, the numerator ( [NH 3]^2 ) is too small, and the denominator ( [N 2][H 2]^3 ) is too large, relative to the equilibrium ratio. The system is not at equilibrium, and it must shift in the direction that raises [NH 3] while lowering [N 2] and [H 2]^- that is, the forward direction (more N 2 and H 2 convert to NH 3 ) — until Q c climbs up to meet K c = 0.061. Answer: The reaction is not at equilibrium, since Q c ≈ 0.010 mol^-2 L^2 ≠ K c = 0.061. Because Q c < K c, the reaction proceeds in the forward direction (toward more NH 3 ) to reach equilibrium.

Question 22

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) ⇌ Br 2 (g) + Cl 2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium?

Accepted Answer

3 × 10–4 molL–1. When K c is written as one perfect square over another, you can take a square root instead of solving a full quadratic. Here the missing symbol between reactants and products in the printed reaction is the equilibrium arrow — the reaction is reversible: 2 BrCl (g) Br 2(g) + Cl 2(g) Set up an ICE table (initial, change, equilibrium) in concentration, not moles. Let x = the concentration of Br 2 (in mol L ^-1 ) formed by the time equilibrium is reached. Because 2 mol of BrCl are consumed for every 1 mol of Br 2 produced (read straight off the balanced equation's coefficients), BrCl falls by 2x, while Br 2 and Cl 2 each rise by x: Initial (mol L ^-1 ): [ BrCl ] = 3.3×10^-3, [ Br 2] = 0, [ Cl 2] = 0 Change (mol L ^-1 ): [ BrCl ] = -2x, [ Br 2] = +x, [ Cl 2] = +x Equilibrium (mol L ^-1 ): [ BrCl ] = 3.3×10^-3-2x, [ Br 2] = x, [ Cl 2] = x K c is products over reactants, each raised to its stoichiometric coefficient — so BrCl's concentration is squared because its coefficient is 2: K c=([ Br 2][ Cl 2])/([ BrCl ]^2)=(x· x)/((3.3×10^-3-2x)^2)=x^2/((3.3×10^-3-2x)^2)=32 The right-hand side is a ratio of two squares, so both sides are perfect squares of the same quantities — take the positive square root of both sides instead of expanding a quadratic: x/(3.3×10^-3-2x)=√ 32 =5.657 Aside: this square-root shortcut only works because the numerator and denominator of K c are each a single term squared. If the stoichiometry gave, say, a 1:3 mix of products, you would be stuck with a genuine quadratic in x — always check the powers before reaching for a square root. Solve for x by clearing the fraction: x=5.657(3.3×10^-3-2x) x=1.867×10^-2×10^-1-11.314x (that is, 5.657×3.3×10^-3=1.867×10^-2 is wrong in magnitude if left as is — carrying the arithmetic through directly:) x = (5.657×3.3×10^-3)-11.314x = 1.867×10^-2 mol L^-1×10^-... Doing the multiplication carefully: 5.657×3.3×10^-3=1.867×10^-2×10^-1. To avoid this slip, keep the powers of ten attached to the number at every step: 5.657×(3.3×10^-3)=1.867×10^-2 is incorrect — the correct product is 1.867×10^-2×10^-1 Redone cleanly: 5.657 × 3.3 = 18.67, and 18.67×10^-3=1.867×10^-2. So: x = 1.867×10^-2... This is getting tangled — restart the last step numerically and keep every digit attached to its power of ten: x = (5.657×3.3×10^-3)-(5.657×2)x x = 1.867×10^-2 mol L^-1 ( this is wrong by a factor of 10 — the correct value is 1.867×10^-2×10^-1) Answer computed numerically and verified: 5.657×3.3×10^-3=0.018668=1.867×10^-2... Given the repeated slips above, the clean and verified arithmetic is: 5.657×(3.3×10^-3)=0.018668 x+11.314x=0.018668 12.314x=0.018668 x=0.018668/12.314=1.516×10^-3 mol L^-1 The question asks for [ BrCl ] at equilibrium, not for x itself — a common slip is to stop at x and report that as the final concentration. [ BrCl ] eq=3.3×10^-3-2x=3.3×10^-3-2(1.516×10^-3)=3.3×10^-3-3.032×10^-3 [ BrCl ] eq=0.268×10^-3 mol L^-1=2.68×10^-4 mol L^-1 Check by substituting back into the K c expression: ((1.516×10^-3)^2)/((2.68×10^-4)^2)=(2.298×10^-6)/(7.182×10^-8)=32.0, matching the given K c. Rounding to two significant figures, matching the precision of the given data (3.3×10^-3, K c=32): Answer: [ BrCl ] eq ≈ 2.7×10^-4 mol L^-1 (more precisely 2.68×10^-4 mol L^-1 )

Question 23

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO 2 in equilibrium with soild carbon has 90.55% CO by mass C (s) + CO 2 (g) ⇌ 2CO (g) Calculate Kc for this reaction at the above temperature.

Accepted Answer

0.149. Percentage by mass has to be turned into moles before it means anything about an equilibrium. The reaction is C (s) + CO 2 (g) 2 CO (g) Carbon is a pure solid, so it never appears in the equilibrium expression — only the two gases, CO and CO 2, do. Step 1 — Fix a basis and convert mass % to moles Take 100 g of the gas mixture. Then: - mass of CO = 90.55 g - mass of CO 2 = 100 − 90.55 = 9.45 g Using n = ( mass )/( molar mass ) (molar mass of CO = 28 g mol ^-1, of CO 2 = 44 g mol ^-1 ): n CO = (90.55 g )/(28 g mol^-1) = 3.234 mol n CO 2 = (9.45 g )/(44 g mol^-1) = 0.2148 mol Total moles of gas: n total = 3.234 + 0.2148 = 3.449 mol Step 2 — Mole fractions, then partial pressures Mole fraction, x i = n i n total: x CO = 3.234/3.449 = 0.9377 x CO 2 = 0.2148/3.449 = 0.0623 The people who get this wrong here plug the mass percentages straight in as pressures — you can't skip the mole-fraction step, because CO and CO 2 have different molar masses. Since the total pressure is given as 1 atm, each partial pressure is just p i = x i × P total: p CO = 0.9377 × 1 atm = 0.9377 atm p CO 2 = 0.0623 × 1 atm = 0.0623 atm Step 3 — Write K p and evaluate it For C (s) + CO 2 (g) 2 CO (g), the solid's activity is 1, so it is left out: K p = (p CO )^2 p CO 2 K p = ((0.9377 atm )^2)/(0.0623 atm ) = (0.8793 atm^2)/(0.0623 atm ) = 14.12 atm Step 4 — Convert K p to K c The relation is K p = K c(RT)^Δ n g, where Δ n g is the change in moles of gas only (the solid carbon does not count): Δ n g = ( mol gaseous products ) - ( mol gaseous reactants ) = 2 - 1 = 1 So K c = K p/((RT)^Δ n g) = K p/RT With R = 0.0821 L atm K^-1 mol^-1 and T = 1127 K: RT = 0.0821 L atm K^-1 mol^-1 × 1127 K = 92.53 L atm mol^-1 K c = (14.12 atm )/(92.53 L atm mol^-1) = 0.1526 mol L^-1 This is the step where units and the direction of the K p – K c relation trip people up: dividing (not multiplying) by RT here is correct only because Δ n g is positive — check the sign of Δ n g before deciding whether to multiply or divide. Rounding to three significant figures, matching the precision of the 90.55% data: Answer: K c ≈ 0.153 mol L^-1

Question 24

Calculate a) ∆G0 and b) the equilibrium constant for the formation of NO 2 from NO and O 2 at 298K NO (g) + ½ O 2 (g) ⇌ NO 2 (g) where ∆fG0 (NO 2) = 52.0 kJ/mol ∆fG0 (NO) = 87.0 kJ/mol ∆fG0 (O 2) = 0 kJ/mol

Accepted Answer

a) – 35.0kJ, b) 1.365 × 106. Δ G^° tells you which way a reaction goes, and it fixes the equilibrium constant through Δ G^° = -RT ln K. The reaction, written correctly with the reversible arrow that belongs between reactants and products, is NO (g) + 1/2 O 2(g) NO 2(g) Step 1 — Get Δ r G^° from the formation energies. The rule is: Δ r G^° = Σ Δ f G^°( products ) - Σ Δ f G^°( reactants ), where Δ f G^° is the standard Gibbs energy of formation of each species, and every species is weighted by its stoichiometric coefficient in the balanced equation — this is the step people skip, and the 1/2 in front of O 2 does not disappear just because Δ f G^°( O 2) = 0. Δ r G^° = Δ f G^°( NO 2) - [Δ f G^°( NO ) + 1/2Δ f G^°( O 2)] Δ r G^° = 52.0 kJ mol^-1 - [87.0 kJ mol^-1 + 1/2(0 kJ mol^-1)] Δ r G^° = 52.0 - 87.0 = -35.0 kJ mol^-1 The value is negative, so the forward reaction (NO and O 2 turning into NO 2 ) is thermodynamically favoured at 298 K under standard conditions. Step 2 — Convert Δ G^° into the equilibrium constant. Name the formula: Δ G^° = -RT ln K, where R = 8.314 J K^-1 mol^-1 is the gas constant, T is the absolute temperature in kelvin (298 K here — not 25, the Celsius reading), and K is the equilibrium constant. The unit trap in this step is mixing kJ and J: Δ G^° was found in kJ mol ^-1, but R is in J, so convert first. Δ r G^° = -35.0 kJ mol^-1 = -35000 J mol^-1 Substitute: -35000 J mol^-1 = -(8.314 J K^-1 mol^-1)(298 K ) ln K ln K = 35000/(8.314 × 298) = 35000/2477.572 = 14.13 Step 3 — Undo the natural log. K = e^14.13 K ≈ 1.37 × 10^6 A K this large (about a million times greater than 1) says the equilibrium mixture is overwhelmingly NO 2 once equilibrium is reached — consistent with the large negative Δ G^° found in Step 1. Answer: Δ r G^° = -35.0 kJ mol^-1, and the equilibrium constant K ≈ 1.37 × 10^6.

Question 25

Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume? (a) PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g) (b) CaO (s) + CO 2 (g) ⇌ CaCO 3 (s) (c) 3Fe (s) + 4H 2O (g) ⇌ Fe 3O 4 (s) + 4H 2 (g)

Accepted Answer

When you push down on the pressure of a gas system by opening up the volume, the equilibrium slides toward whichever side has the larger number of moles of gas — because that is the side whose total pressure drops the most, and the system reacts to soften the disturbance. This is Le Chatelier's principle applied to volume/pressure changes. Only gas-phase species carry weight in this count; a solid sitting in the reaction does not contribute moles to either side. The step people usually rush past is exactly that count — tallying every species instead of only the gaseous ones gives the wrong direction. For each part, write Δ n g = ( moles of gas on the product side ) - ( moles of gas on the reactant side ). Increasing the volume (dropping the pressure) always favours the side with more gas moles, so: - Δ n g 0: equilibrium shifts forward (toward products) — product moles increase. - Δ n g < 0: equilibrium shifts backward (toward reactants) — product moles decrease. - Δ n g = 0: neither side is favoured — the equilibrium position, and product moles, stay the same. (a) PCl 5 (g) PCl 3 (g) + Cl 2 (g) All three species are gases. Reactant side: 1 mole of gas. Product side: 1 mole PCl 3 + 1 mole Cl 2 = 2 moles of gas. Δ n g = 2 - 1 = 1 0 The product side carries more gas moles, so opening up the volume pushes the equilibrium forward: more PCl 5 dissociates, and the moles of PCl 3 and Cl 2 increase. (b) CaO (s) + CO 2 (g) CaCO 3 (s) Here CaO and CaCO 3 are solids and do not count. Reactant side: 1 mole of gas ( CO 2 ). Product side: 0 moles of gas, since CaCO 3 is a solid. Δ n g = 0 - 1 = -1 < 0 The reactant side carries more gas moles, so increasing the volume pushes the equilibrium backward — CaCO 3 decomposes back into CaO and CO 2. The moles of the solid product CaCO 3 decrease. This is the reverse of what a quick glance suggests: writing the product as "solid, so it's fixed" misses that the reaction still runs backward and consumes it. (c) 3Fe (s) + 4H 2O (g) Fe 3O 4 (s) + 4H 2 (g) Fe and Fe 3O 4 are solids and drop out of the count. Reactant side: 4 moles of gas ( H 2O ). Product side: 4 moles of gas ( H 2 ). Δ n g = 4 - 4 = 0 Neither side is favoured by the volume change, since both sides already carry the same number of gas moles — an increase in volume drops the partial pressure of H 2O and H 2 by the same factor, leaving the ratio in K p undisturbed. The equilibrium position does not shift, so the moles of H 2 remain the same. Answer: (a) increases (forward shift, since gas moles go from 1 to 2); (b) decreases (backward shift, since gas moles go from 1 to 0); (c) remains the same (no shift, since gas moles are 4 on each side).

Question 26

Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction. (i) COCl 2 (g) ⇌ CO (g) + Cl 2 (g) (ii) CH 4 (g) + 2S 2 (g) ⇌ CS 2 (g) + 2H 2S (g) (iii) CO 2 (g) + C (s) ⇌ 2CO (g) (iv) 2H 2 (g) + CO (g) ⇌ CH 3OH (g) (v) CaCO 3 (s) ⇌ CaO (s) + CO 2 (g) (vi) 4 NH 3 (g) + 5O 2 (g) ⇌ 4NO (g) + 6H 2O(g)

Accepted Answer

Increasing pressure pushes an equilibrium toward whichever side has fewer moles of gas — count only the gas-phase species, because solids don't compete for that shared volume. By Le Chatelier's principle, squeezing a gaseous system (raising the pressure, i.e. shrinking the volume) is a stress the system relieves by moving toward the side that takes up less volume — the side with the smaller total moles of gas, n g. Define Δ n g = n g( products ) - n g( reactants ) counting only species marked (g). If Δ n g = 0, the two sides occupy equal gas volume for equal extents of reaction, so squeezing the system does not favour either side — pressure has no effect. If Δ n g ≠ 0, pressure has an effect, and the shift is always toward the side with fewer gas moles: forward when Δ n g < 0 (products side is "smaller"), backward when Δ n g 0 (reactants side is "smaller"). The trap to watch for: a solid, (s), sits in the equilibrium but does not add to n g — leaving it out of the mole count is not an approximation, it's the correct count, since only gas particles fill the volume that pressure is squeezing. Each part below is written as an equilibrium (a reactant side and a product side written with nothing but a gap between them, as in this exercise, is the reversible reaction of this chapter, not two separate statements). (i) COCl 2(g) CO (g) + Cl 2(g) n g( reactants ) = 1, n g( products ) = 1 + 1 = 2. Δ n g = 2 - 1 = +1 ≠ 0 Affected. Products occupy more gas volume, so increasing pressure favours the smaller side — the reactant side. The equilibrium is pushed in the backward direction. (ii) CH 4(g) + 2 S 2(g) CS 2(g) + 2 H 2 S (g) n g( reactants ) = 1 + 2 = 3, n g( products ) = 1 + 2 = 3. Δ n g = 3 - 3 = 0 Not affected by pressure — both sides already occupy equal gas volume, so compressing the vessel raises every gas's concentration by the same factor on both sides and the equilibrium quotient is unchanged. (iii) CO 2(g) + C (s) 2 CO (g) The carbon is solid, so it is left out of the count: n g( reactants ) = 1 (from CO 2 alone), n g( products ) = 2. Δ n g = 2 - 1 = +1 ≠ 0 Affected. Fewer gas moles sit on the reactant side, so increasing pressure pushes the equilibrium backward, toward CO 2(g) + C (s). (iv) 2 H 2(g) + CO (g) CH 3 OH (g) n g( reactants ) = 2 + 1 = 3, n g( products ) = 1. Δ n g = 1 - 3 = -2 ≠ 0 Affected. Here it is the product side that is smaller, so increasing pressure pushes the equilibrium forward, toward CH 3 OH (g). (v) CaCO 3(s) CaO (s) + CO 2(g) Both calcium compounds are solids; only CO 2 is gas: n g( reactants ) = 0, n g( products ) = 1. Δ n g = 1 - 0 = +1 ≠ 0 Affected. The reactant side has zero gas moles — as small as a side can get — so increasing pressure pushes the equilibrium backward, suppressing the decomposition of CaCO 3. (vi) 4 NH 3(g) + 5 O 2(g) 4 NO (g) + 6 H 2 O (g) n g( reactants ) = 4 + 5 = 9, n g( products ) = 4 + 6 = 10. Δ n g = 10 - 9 = +1 ≠ 0 Affected. The reactant side (9 mol gas) is smaller than the product side (10 mol gas), so increasing pressure pushes the equilibrium backward. Answer: Reactions (i), (iii), (iv), (v) and (vi) are all affected by increasing pressure, because each has Δ n g ≠ 0; reaction (ii) is unaffected, because Δ n g = 0. Direction of shift: (i) backward, (ii) no shift, (iii) backward, (iv) forward, (v) backward, (vi) backward — in every affected case the equilibrium moves toward the side with the smaller number of gas moles.

Question 27

The equilibrium constant for the following reaction is 1.6 ×105 at 1024K H 2(g) + Br 2(g) ⇌ 2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

Accepted Answer

[PH2]eq = [PBr2]eq = 2.5 × 10–2bar, [PHBr] = 10.0 bar. Answer: p(H₂) = p(Br₂) ≈ 0.0249 bar, p(HBr) ≈ 9.95 bar A huge equilibrium constant does not mean "no reverse reaction" — it means only a tiny amount of HBr breaks back down before equilibrium is reached. The reaction is H 2(g) + Br 2(g) 2HBr(g), K p = 1.6×10^5 at 1024 K Here the container starts with only HBr — 10.0 bar of it, and zero H₂, zero Br₂. With no H₂ or Br₂ present, the forward reaction has nothing to consume; the system can only move by running in reverse, decomposing a little HBr into H₂ and Br₂ until the ratio of pressures matches K p. Setting up the ICE table (in terms of the reverse reaction) Let 2x bar of HBr decompose: 2HBr(g) H 2(g) + Br 2(g) HBr H₂ Br₂ --- --- --- --- Initial (bar) 10.0 0 0 Change (bar) -2x +x +x Equilibrium (bar) 10.0-2x x x The stoichiometry ties H₂ and Br₂ to the same x because the balanced equation consumes 2 mol HBr for every 1 mol H₂ and 1 mol Br₂ formed — that 2:1:1 ratio is the step people drop, writing all three changes as the same x. Writing K p for the reaction as originally given K p here uses partial pressures (in bar) exactly like K c uses concentrations, one factor per gas raised to its stoichiometric coefficient: K p = (p HBr )^2 p H 2 · p Br 2 Substituting the equilibrium row of the table: 1.6×10^5 = ((10.0-2x)^2)/(x· x) = ((10.0-2x)^2)/x^2 Solving exactly — no approximation needed Both sides are squares of positive quantities, so take the square root of each side directly: √ 1.6×10^5 = 10.0-2x/x An aside on why this is convenient: 1.6×10^5 = 160000 = 400^2, a perfect square, so the square root comes out exact — √ 1.6×10^5 = 400 — with no rounding of K p itself. 400 = 10.0-2x/x 400x = 10.0 - 2x 402x = 10.0 x = 10.0/402 = 0.024876… bar Rounding once, at the end, to three significant figures (matching the 10.0 bar of the given data): x ≈ 0.0249 bar Reading off the equilibrium pressures p H 2 = p Br 2 = x ≈ 0.0249 bar p HBr = 10.0 - 2x = 10.0 - 2(0.024876) = 10.0 - 0.04975 ≈ 9.95 bar Checking the numbers actually satisfy K p K p = ((9.95)^2)/((0.0249)(0.0249)) = 99.0/0.000620 ≈ 1.6×10^5 This matches the given K p, confirming the equilibrium pressures. Only about 0.5% of the original HBr decomposed — consistent with a very large K p meaning the products (here, HBr) are strongly favored, so the container still holds mostly HBr with just a trace of H₂ and Br₂. Answer: p(H₂) = p(Br₂) ≈ 0.0249 bar and p(HBr) ≈ 9.95 bar

Question 28

Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH 4 (g) + H 2O (g) ⇌ CO (g) + 3H 2 (g) (a) Write as expression for Kp for the above reaction. (b) How will the values of Kp and composition of equilibrium mixture be affected by (i) increasing the pressure (ii) increasing the temperature (iii) using a catalyst?

Accepted Answer

K p is read straight off the balanced equation: partial pressures of products (raised to their coefficients) over partial pressures of reactants (raised to theirs). The reaction is CH 4(g) + H 2 O (g) CO (g) + 3 H 2(g) (a) Expression for K p For a gas-phase equilibrium, the law of mass action written in terms of partial pressures gives K p = p CO (p H 2 )^3 p CH 4 p H 2 O Here p CH 4, p H 2 O, p CO, p H 2 are the partial pressures of the respective gases at equilibrium, and each exponent is just the stoichiometric coefficient of that species in the balanced equation — H 2 appears cubed because its coefficient is 3. (b) Effect of changing conditions A step people get wrong: K p is a function of temperature only. Pressure and catalysts can shift where the equilibrium sits, but neither one is allowed to change the value of K p itself. (i) Increasing the pressure First count moles of gas on each side: - Reactant side: 1 ( CH 4) + 1 ( H 2 O ) = 2 mol of gas - Product side: 1 ( CO ) + 3 ( H 2) = 4 mol of gas So Δ n g = 4 - 2 = +2: the forward reaction increases the number of gas molecules. By Le Chatelier's principle, raising the pressure pushes the equilibrium toward the side with fewer moles of gas, since that is the side that lets the system reduce the pressure increase. Here that is the reverse direction (2 mol side). So: - The equilibrium mixture shifts backward — the amounts of CO and H 2 at the new equilibrium go down, and CH 4 and H 2 O go up, compared to before the pressure increase. - K p itself does not change, because temperature has not changed. (ii) Increasing the temperature The reaction is given as endothermic, i.e. it absorbs heat as it goes forward ( Δ H 0 ). Le Chatelier's principle says that raising the temperature drives the equilibrium in whichever direction absorbs the added heat — here, the forward direction. - The equilibrium mixture shifts forward: more CO and H 2 are formed at equilibrium, and CH 4, H 2 O are consumed further. - Because the forward (product-favoring) direction is now favoured, K p increases with rising temperature. This matches the van't Hoff relation (d( ln K p))/dT = (Δ H)/RT^2: with Δ H 0, ln K p rises as T rises. (iii) Using a catalyst A catalyst lowers the activation energy of both the forward and the reverse step by exactly the same amount, so it speeds up both rates equally. It gets the system to equilibrium faster, but it does not favour one direction over the other. - The composition of the equilibrium mixture is unchanged. - K p is unchanged. Answer: K p = p CO (p H 2 )^3 p CH 4 p H 2 O. (i) Increasing pressure shifts equilibrium backward (toward fewer gas moles), lowering CO and H 2 in the mixture, but K p is unchanged. (ii) Increasing temperature shifts equilibrium forward (endothermic direction), raising CO and H 2, and K p increases. (iii) A catalyst changes neither the equilibrium composition nor K p — it only makes equilibrium reach faster.

Question 29

Describe the effect of: a) addition of H 2 b) addition of CH 3OH c) removal of CO d) removal of CH 3OH on the equilibrium of the reaction: 2H 2(g) + CO (g) ⇌ CH 3OH (g)

Accepted Answer

Le Chatelier's principle: a system at equilibrium responds to a stress by shifting in whichever direction partly cancels that stress. There is no calculation here — this question asks for the direction of shift, and getting that direction right means always asking "which way undoes what I just did?", not "which way continues what I just did?" The equilibrium is 2 H 2(g) + CO (g) CH 3 OH (g) The forward reaction eats H 2 and CO and makes CH 3 OH. The reverse reaction does the opposite: it eats CH 3 OH and regenerates H 2 and CO. Every part below is just: find which species was disturbed, then ask which direction consumes the excess (if you added something) or replaces the shortfall (if you removed something). (a) Addition of H 2. H 2 is a reactant, and there is now more of it than the equilibrium had. The system shifts in the direction that uses up the extra H 2 — the forward direction. More CO is consumed along with it, and more CH 3 OH is produced. (b) Addition of CH 3 OH. CH 3 OH is the product, and there is now more of it than equilibrium allows. The system shifts to consume the extra product — the reverse direction. Some of the added CH 3 OH decomposes back into H 2 and CO. (c) Removal of CO. This is the step where the direction is easy to get backwards. CO is a reactant, and there is now less of it than equilibrium wants — the system does not "keep consuming" a reactant that is disappearing; it shifts to replace what was taken away. Replacing CO means running the reverse reaction, which regenerates H 2 and CO at the expense of CH 3 OH. So removing a reactant shifts the equilibrium backward, and the amount of CH 3 OH present falls. (d) Removal of CH 3 OH. CH 3 OH is the product, and now there is less of it than equilibrium wants. The system shifts to replace the missing product — the forward direction. More H 2 and CO combine to form fresh CH 3 OH. Notice the pattern that makes (a)–(d) consistent rather than four separate rules: adding a species always pushes the reaction away from that species (to consume the surplus), and removing a species always pushes the reaction toward that species (to restore it) — regardless of whether that species happens to be a reactant or a product. Answer: (a) shifts forward (more CH 3 OH forms); (b) shifts backward ( CH 3 OH decomposes); (c) shifts backward ( CH 3 OH decreases); (d) shifts forward (more CH 3 OH forms).

Question 30

At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl 5 is 8.3 ×10-3. If decomposition is depicted as, PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g) ∆rH0 = 124.0 kJ mol–1 a) write an expression for Kc for the reaction. b) what is the value of Kc for the reverse reaction at the same temperature? c) what would be the effect on Kc if (i) more PCl 5 is added (ii) pressure is increased (iii) the temperature is increased?

Accepted Answer

b) 120.48. K c only changes when the temperature changes — adding a reactant or squeezing the container moves the position of equilibrium, but not the value of the equilibrium constant. Only a change in T can change K c, and whether it goes up or down is decided by the sign of Δ rH^0. (a) Expression for K c For PCl 5(g) PCl 3(g) + Cl 2(g) the law of mass action gives K c = ([ PCl 3][ Cl 2])/([ PCl 5]) where each bracket is the equilibrium molar concentration (mol L^-1 ) of that species. (b) K c for the reverse reaction The reverse reaction is PCl 3(g) + Cl 2(g) PCl 5(g) so its equilibrium constant is just the reciprocal expression: K c' = ([ PCl 5])/([ PCl 3][ Cl 2]) = 1/K c This reciprocal relationship holds only because both reactions are at the same temperature (473 K) — you cannot take the reciprocal of a K c measured at one temperature and use it for a different temperature. Substituting K c = 8.3 × 10^-3: K c' = 1/(8.3 × 10^-3 mol L^-1) = 120.48 mol^-1 L 8.3 × 10^-3 carries 2 significant figures, so the reciprocal is rounded to 2 significant figures: K c' ≈ 1.2 × 10^2 mol^-1 L ( i.e., 1.2 × 10^2 dm^3 mol^-1) (c) Effect on K c (i) More PCl 5 is added. K c is a function of temperature alone, so at constant T = 473 K it stays exactly 8.3 × 10^-3. Adding PCl 5 raises [ PCl 5], which makes the reaction quotient Q c momentarily smaller than K c. The system responds by decomposing more PCl 5 — the equilibrium shifts to the right — until Q c climbs back up to equal K c again. It is the concentrations that move, not the constant. (ii) Pressure is increased. This is done by shrinking the volume, which raises all three concentrations at once — again K c itself is unaffected because T has not changed; it remains 8.3 × 10^-3. But the forward reaction turns 1 mole of gas into 2 moles ( PCl 5 → PCl 3 + Cl 2), so squeezing the system favors the side with fewer gas molecules by Le Chatelier's principle. The equilibrium shifts to the left, back toward PCl 5, until Q c again equals the unchanged K c. (iii) Temperature is increased. Here K c itself changes. Δ rH^0 = +124.0 kJ mol^-1 is positive, so the forward decomposition of PCl 5 is endothermic — it absorbs heat. Raising T supplies more of that heat and pushes the equilibrium further toward products, so the numerator [ PCl 3][ Cl 2] grows relative to [ PCl 5] and K c increases above 8.3×10^-3. Answer: (a) K c = ([ PCl 3][ Cl 2])/([ PCl 5]); (b) K c( reverse ) = 1/K c ≈ 1.2 × 10^2 mol^-1 L; (c)(i) K c unchanged at 8.3×10^-3, equilibrium shifts forward (more dissociation); (ii) K c unchanged at 8.3×10^-3, equilibrium shifts backward toward PCl 5 (the side with fewer gas moles); (iii) K c increases above 8.3×10^-3 because the reaction is endothermic (Δ rH^0 = +124.0 kJ mol^-1), so raising T favors the forward direction.

Question 31

Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, CO (g) + H 2O (g) ⇌ CO 2 (g) + H 2 (g) If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that pco = pH2O = 4.0 bar, what will be the partial pressure of H 2 at equilibrium? Kp= 10.1 at 400°C

Accepted Answer

[H2]eq = 0.96 bar. The unknown x here is how much pressure reacts away, and K p is built from what is LEFT at equilibrium — so the last step is to report x itself, not the leftover 4.0-x. That single slip is the whole trap in this question. The reaction and what it does to pressures The water-gas shift reaction is CO (g) + H 2O (g) CO 2 (g) + H 2 (g) Every mole of CO that reacts consumes one mole of H 2 O and makes one mole of CO 2 and one mole of H 2. At constant temperature and volume, partial pressure is directly proportional to moles ( p i = n iRT/V ), so the same 1: 1: 1: 1 bookkeeping can be done straight in bar — no need to convert to moles at all. Note also that the total number of gas molecules does not change (2 on the left, 2 on the right). That is why nothing extra happens to the total pressure, and why K p here is dimensionless in the usual sense: the bar units cancel top and bottom. Set up the ICE table in bar Let x = the partial pressure (in bar) of CO that is consumed by the time equilibrium is reached. The vessel starts with only CO and steam, so the initial pressures of CO 2 and H 2 are zero. CO H 2O CO 2 H 2 --- --- --- --- --- Initial / bar 4.0 4.0 0 0 Change / bar -x -x +x +x Equilibrium / bar 4.0-x 4.0-x x x The question asks for the partial pressure of H 2 at equilibrium, and that is x — the row of the table headed "Equilibrium", column H 2. Keep hold of that; it is exactly the thing that gets mixed up at the end. Write K p and substitute For this reaction the equilibrium constant in terms of partial pressures is K p = p CO 2 p H 2 p CO p H 2O where each p i is the equilibrium partial pressure of that gas in bar (products on top, reactants underneath, each raised to its stoichiometric coefficient — all four coefficients are 1 here). Substituting the equilibrium row, with K p = 10.1 at 400^° C: 10.1 = ((x)(x))/((4.0-x)(4.0-x)) = (x^2)/((4.0-x)^2) Solve — take the square root, don't expand Both sides are perfect squares, so instead of multiplying out into a quadratic, take the positive square root of both sides (pressures are positive, and x < 4.0 because you cannot consume more CO than you started with, so 4.0-x 0 ): x/4.0-x = √ 10.1 = 3.17805 Cross-multiplying: x = 3.17805 (4.0 - x) = 12.71220 - 3.17805 x x + 3.17805 x = 12.71220 4.17805 x = 12.71220 x = 12.71220/4.17805 = 3.04262 bar Carrying the extra digits through and rounding only now: the data are K p = 10.1 (three significant figures) and 4.0 bar, so quoting to three significant figures — two decimal places in bar — gives p H 2 = 3.04 bar Check by putting it back in Equilibrium pressures: p H 2 = p CO 2 = 3.0426 bar, and p CO = p H 2O = 4.0 - 3.0426 = 0.9574 bar. Then K p = ((3.0426)(3.0426))/((0.9574)(0.9574)) = 9.2574/0.9166 = 10.10 which reproduces the given K p = 10.1. The answer is consistent. A sanity check on the size, too: K p = 10.1 is greater than 1, so at equilibrium the products are favoured over the reactants. The H 2 pressure must therefore come out larger than the leftover CO pressure. 3.04 0.96, as required. Had you obtained 0.96 bar for H 2 with 3.04 bar of CO left over, that would mean reactants dominating, i.e. K p < 1 — which contradicts the value the question hands you. A note on the printed key. NCERT's answer key gives " [ H 2 ] eq = 0.96 bar" for this question, and that is a slip in the key, not a different convention or a rounding difference. 0.96 bar is precisely 4.0 - 3.04, the unreacted CO (and steam) pressure — the 4.0-x entry of the table rather than the x entry the question asks for. You can see it must be wrong without redoing the algebra: substituting p H 2 = p CO 2 = 0.96 bar and p CO = p H 2O = 3.04 bar into the K p expression gives (0.96)^2/(3.04)^2 = 0.099, which is 1/10.1, not 10.1. The correct partial pressure of dihydrogen is 3.04 bar. NCERT's answer key prints "[H₂]eq = 0.96 bar", and that is the wrong species. 0.96 bar is 4.0 - 3.04, the equilibrium pressure of the leftover CO and H₂O — a correct number from this same working, attached to the wrong row. Check it against the constant the question gives you: with p H 2 = 3.04, K p = 3.04^2/0.96^2 = 10.1 ✓, whereas p H 2 = 0.96 gives 0.96^2/3.04^2 = 0.099 = 1/10.1 — the constant for the reverse reaction. And K p = 10.1 1 means products predominate, so p H 2 has to be the larger value. Answer: p H 2 = 3.04 bar (with p CO 2 = 3.04 bar and p CO = p H 2O = 0.96 bar left unreacted); NCERT's printed 0.96 bar is the leftover CO/steam pressure, not the H 2 pressure.

Question 32

Predict which of the following reaction will have appreciable concentration of reactants and products: a) Cl 2 (g) ⇌ 2Cl (g) Kc = 5 ×10–39 b) Cl 2 (g) + 2NO (g) ⇌ 2NOCl (g) Kc = 3.7 × 108 c) Cl 2 (g) + 2NO 2 (g) ⇌ 2NO 2Cl (g) Kc = 1.8

Accepted Answer

A reaction's equilibrium constant K c tells you where the balance point sits — how far the reaction runs before the forward and reverse rates match. A huge K c means equilibrium lies almost entirely with products; a tiny K c means it lies almost entirely with reactants; only when K c is close to 1 do both sides hold measurable amounts at the same time. The rule of thumb, and why it works: at equilibrium, K c = [ products ]^ stoich. powers [ reactants ]^ stoich. powers This is just a ratio. If the ratio is enormous, the numerator (products) must dominate the denominator (reactants) — so reactant concentration is squeezed down to something negligible. If the ratio is minuscule, the reverse happens: reactants dominate and product concentration is negligible. Only when the ratio is of order 1 can neither side be forced to a vanishing amount — both must be present in comparable, appreciable quantities. In practice the cutoffs are taken as roughly K c 10^3: reaction goes almost to completion (products only) K c < 10^-3: reaction barely proceeds (reactants only) 10^-3 < K c < 10^3: appreciable amounts of both This is where people go wrong: a "large" or "small" K c is judged against 1, not against zero — K c = 1.8 is not small just because it's a single-digit number; it is close to 1, which is exactly the regime where both reactants and products coexist. Now check each equilibrium (note the reversible arrow in every case — this is an equilibrium, so both directions are always happening): (a) Cl 2(g) 2Cl(g), K c = 5×10^-39 This is far below 10^-3 — vanishingly small. The numerator [ Cl ]^2 must be minuscule compared to [ Cl 2 ] for the ratio to come out this tiny. Essentially none of the Cl 2 dissociates; at equilibrium there is reactant, but no appreciable product. (b) Cl 2(g) + 2NO(g) 2NOCl(g), K c = 3.7×10^8 This is far above 10^3 — enormous. Here the numerator [ NOCl ]^2 must swamp the denominator, so the reaction runs almost to completion. At equilibrium there is product, but no appreciable reactant left. (c) Cl 2(g) + 2NO 2(g) 2NO 2Cl(g), K c = 1.8 This sits squarely between 10^-3 and 10^3, close to 1. Neither the numerator nor the denominator is forced toward zero — the concentrations of Cl 2, NO 2, and NO 2Cl can all be comparable in size at equilibrium. This is the case with genuinely appreciable amounts of both reactants and products. Answer: Reaction (c), Cl 2(g) + 2NO 2(g) 2NO 2Cl(g) with K c = 1.8, is the one with appreciable concentrations of both reactants and products at equilibrium — reaction (a) lies almost entirely toward reactants ( K c = 5×10^-39 1 ) and reaction (b) lies almost entirely toward products ( K c = 3.7×10^8 1 ).

Question 33

The value of Kc for the reaction 3O 2 (g) ⇌ 2O 3 (g) is 2.0 ×10–50 at 25°C. If the equilibrium concentration of O 2 in air at 25°C is 1.6 ×10–2, what is the concentration of O 3?

Accepted Answer

2.86 × 10–28 M. The exponents in K c are the stoichiometric coefficients — O 3 is squared because it has coefficient 2, and O 2 is cubed because it has coefficient 3. Mixing those two up is the error to watch for here. The equilibrium is 3 O 2(g) 2 O 3(g) so the equilibrium constant expression is K c = ([ O 3]^2)/([ O 2]^3) where [ O 3] and [ O 2] are the equilibrium molar concentrations (mol L ^-1 ) of ozone and oxygen. Given values K c = 2.0 × 10^-50, [ O 2] = 1.6 × 10^-2 mol L^-1 Step 1 — cube the oxygen concentration. Cubing means cubing the whole number, the coefficient and the power of ten together — cube only the 10^-2 and forget the 1.6, and the answer comes out wrong by a factor of 4.096. [ O 2]^3 = (1.6 × 10^-2)^3 = (1.6)^3 × 10^-6 = 4.096 × 10^-6 mol^3 L^-3 Step 2 — solve the K c expression for [ O 3]^2. [ O 3]^2 = K c × [ O 2]^3 = (2.0 × 10^-50) × (4.096 × 10^-6) [ O 3]^2 = (2.0 × 4.096) × 10^-50-6 = 8.192 × 10^-56 mol^2 L^-2 Step 3 — take the square root to get [ O 3]. Split the square root across the decimal part and the power of ten separately, keeping the power of ten even so it comes out cleanly: [ O 3] = √ 8.192 × 10^-56 = √ 8.192 × 10^-28 mol L^-1 √ 8.192 = 2.862… [ O 3] = 2.862 × 10^-28 mol L^-1 Both given data points ( K c and [ O 2] ) carry two significant figures, so the final concentration is rounded to two significant figures: [ O 3] ≈ 2.9 × 10^-28 mol L^-1 This is an extraordinarily small number — it says that at ordinary atmospheric oxygen levels, essentially no ozone would exist at equilibrium under these conditions; the ozone actually present in the atmosphere is sustained by continuous photochemical formation, not by this thermal equilibrium. Answer: [ O 3] ≈ 2.9 × 10^-28 mol L^-1 (i.e. 2.86 × 10^-28 mol L^-1 before rounding).

Question 34

The reaction, CO(g) + 3H 2(g) ⇌ CH 4(g) + H 2O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H 2 and 0.02 mol of H 2O and an unknown amount of CH 4 in the flask. Determine the concentration of CH 4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.

Accepted Answer

5.85x10–2. At equilibrium, the law of mass action fixes one ratio of concentrations — K c^- and every species' concentration must satisfy it together. The reaction as it occurs at equilibrium is reversible, so it is written with a double arrow, not a plain one: CO(g) + 3 H 2 (g) CH 4 (g) + H 2 O(g) Because the flask has a volume of exactly 1 L, moles and molar concentration are numerically identical here — this is a convenience special to this problem, not a general rule, so still write every quantity with its unit of mol L^-1: [ CO ] = 0.30 mol L^-1, [ H 2] = 0.10 mol L^-1, [ H 2 O ] = 0.02 mol L^-1 and [ CH 4] is the unknown to find. Writing the equilibrium constant expression. For this reaction, K c = ([ CH 4][ H 2 O ])/([ CO ][ H 2]^3) Here K c is the equilibrium constant in terms of concentrations, and each bracketed term is that species' equilibrium concentration raised to its stoichiometric coefficient in the balanced equation — H 2 is cubed because its coefficient is 3. This is the step people rush: forgetting to cube [ H 2] (treating it as if its coefficient were 1) throws the answer off by a factor of (0.10)^2 = 0.01, a hundredfold error. Solving for [ CH 4]. Rearranging for the one unknown, [ CH 4] = (K c [ CO ][ H 2]^3)/([ H 2 O ]) Substituting the given values, with K c = 3.90: [ CH 4] = (3.90 × (0.30 mol L^-1) × (0.10 mol L^-1)^3)/(0.02 mol L^-1) Work the numerator first. (0.10 mol L^-1)^3 = 1.0×10^-3 mol^3 L^-3, so 3.90 × 0.30 mol L^-1 × 1.0×10^-3 mol^3 L^-3 = 1.17×10^-3 mol^4 L^-4 Now divide by [ H 2 O ] = 0.02 mol L^-1: [ CH 4] = (1.17×10^-3 mol^4 L^-4)/(0.02 mol L^-1) = 0.0585 mol L^-1 The units divide cleanly to leave mol^3 L^-3... but check the bookkeeping: K c itself carries whatever units make the overall expression consistent, and since every concentration here is expressed in mol L^-1, the arithmetic on the numbers alone correctly returns [ CH 4] in mol L^-1 — this is the usual convention for concentration equilibrium constants, so track the number and attach mol L^-1 to the final result. The data given (0.30, 0.10, 0.02, 3.90) each carry two or three significant figures, so the result is kept to three significant figures. Answer: [ CH 4] = 0.0585 mol L^-1

Question 35

What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO 2, CN^-, HClO 4, F^-, OH^-, CO 3^2-, and S^2-

Accepted Answer

NO2 –, HCN, ClO4, HF, H2O, HCO3 –, HS–. A conjugate acid–base pair is two species that differ from each other by exactly one proton, H^+ — nothing else changes. When an acid HA gives away a proton, HA H^+ + A^- what is left, A^-, is called the conjugate base of HA. Read the same equilibrium the other way and A^- is a base that, on accepting a proton, turns back into HA — its conjugate acid. So HA and A^- form one conjugate pair: the same skeleton, with or without one extra proton. The step people get wrong: a conjugate pair is not "the acid and the base that reacted with each other" in a neutralisation. It is a single species and the one species you get when that same species alone loses or gains one H^+ — the two never appear as separate reactants together in the defining equation. To answer for each species below, first decide which way it goes. A species that still has an ionisable proton to give away is acting as the acid, so you strip a proton and report its conjugate base. A species with no proton to spare, but a lone pair able to accept one, is acting as the base, so you add a proton and report its conjugate acid. Species with a proton to give — strip H^+ to get the conjugate base: - HNO 2 (nitrous acid) loses a proton to give NO 2^- — conjugate base of HNO 2 is NO 2^-. - HClO 4 (perchloric acid) loses a proton to give ClO 4^- — conjugate base of HClO 4 is ClO 4^-. Species with no proton to spare — add H^+ to get the conjugate acid: - CN^- (cyanide ion) accepts a proton to give HCN — conjugate acid of CN^- is HCN. - F^- (fluoride ion) accepts a proton to give HF — conjugate acid of F^- is HF. - OH^- (hydroxide ion) accepts a proton to give H 2O — conjugate acid of OH^- is H 2O. - CO 3^2- (carbonate ion) accepts a proton to give HCO 3^- — conjugate acid of CO 3^2- is HCO 3^-. It picks up only one proton at a time, so the conjugate acid is HCO 3^- (hydrogen carbonate), not H 2CO 3, which would need a second proton on top of that. - S^2- (sulfide ion) accepts a proton to give HS^- — conjugate acid of S^2- is HS^-. Answer: A conjugate acid–base pair is a pair of species differing by exactly one H^+. Conjugate base of HNO 2 is NO 2^-; of HClO 4 is ClO 4^-. Conjugate acid of CN^- is HCN; of F^- is HF; of OH^- is H 2O; of CO 3^2- is HCO 3^-; of S^2- is HS^-.

Question 36

Which of the followings are Lewis acids? H 2O, BF 3, H^+, and NH 4^+

Accepted Answer

BF3, H+, NH4 +. A Lewis acid needs somewhere to PUT the incoming pair — an empty orbital. Carrying a positive charge is not the test. G.N. Lewis defined an acid as a species that accepts an electron pair, and a base as a species that donates an electron pair. So the working test has two parts, and you must apply both: - To be a Lewis base, the species must own a lone pair it can hand out. - To be a Lewis acid, the species must have a vacant (empty) orbital in its valence shell to receive that pair, so that a new coordinate bond can actually form. This is exactly the criterion the chapter itself uses: electron-deficient species such as AlCl 3, Co^3+, Mg^2+ act as Lewis acids, while H 2O, NH 3, OH^- act as Lewis bases. So for each species, count the electrons around the central atom and ask: is there a lone pair to give, and is there an empty orbital to fill? H 2O. Oxygen has 6 valence electrons; two go into the two O - H bonds, leaving two lone pairs. Count around O: 2 × 2 (bonding) + 2 × 2 (lone pairs) = 8 electrons — a complete octet, with every valence orbital occupied. There is nothing empty to accept a pair into, but there are two pairs to give away. Water is therefore a Lewis base (it is the pair-donor when it attaches to H^+ to make H 3O^+ ). Not a Lewis acid. BF 3. Boron has only 3 valence electrons, all used in three B - F bonds. Count around B: 3 × 2 = 6 electrons — two short of an octet — and the 2p z orbital is completely empty. That empty 2p orbital is the vacancy, so BF 3 takes a lone pair from ammonia: BF 3 +: NH 3 → F 3B ← NH 3 BF 3 is a Lewis acid. H^+. A bare proton has zero electrons and an entirely empty 1s orbital. Nothing is more electron-hungry than that; it is the textbook Lewis acid, accepting a pair from OH^-, F^-, H 2O and so on. A Lewis acid. NH 4^+. This is the one to be careful with. Nitrogen is sp^3 hybridised and all four hybrid orbitals are used in four N - H bonds. Count around N: 4 × 2 = 8 electrons — a complete octet — with no lone pair left and no empty orbital left. Nitrogen sits in Period 2, so it has no d orbitals and cannot expand its octet to ten electrons; there is no such thing as NH 5. With nothing to donate it cannot be a Lewis base, and with nowhere to receive a pair it cannot be a Lewis acid either. The step people get wrong here: seeing the + charge and concluding "cation, therefore Lewis acid". That rule of thumb works for Mg^2+, Co^3+, Ag^+ because those ions really do have empty valence orbitals. NH 4^+ is the standard exception: its positive charge is a book-keeping charge spread over the hydrogens, while nitrogen's own shell is full and closed. A second aside — Brønsted acid is not the same thing as Lewis acid. NH 4^+ is genuinely a Brønsted acid: it donates a proton, NH 4^+ + OH^- → NH 3 + H 2O But read that reaction in Lewis language. No new bond forms to nitrogen. What happens is that H^+ is handed from NH 3 over to OH^-: the electron-pair acceptor is the proton, and NH 4^+ is merely the vehicle carrying it. A proton-donor is not automatically a pair-acceptor. The cleanest way to see it: NH 4^+ is itself a Lewis adduct — the finished product of the Lewis acid H^+ and the Lewis base: NH 3, H^+ +: NH 3 → NH 4^+ which is the same shape of reaction as BF 3 +: NH 3 → F 3B ← NH 3. Nobody calls the adduct F 3B ← NH 3 a Lewis acid — the acid was BF 3, before it was satisfied. For exactly the same reason NH 4^+ is not a Lewis acid: the acid was H^+, and in NH 4^+ it has already been satisfied. So of the four species, two have a vacancy and two do not. A note on the printed answer. The answer key at the back of the book gives BF 3, H^+ and NH 4^+. The first two are right; the inclusion of NH 4^+ is an error, and it contradicts the book's own definition on the very page where Lewis acids are introduced. Notice that the printed set is not what either possible criterion gives you: - Judge by the vacant-orbital rule the chapter states ("electron deficient species … can act as Lewis acids") and you get BF 3, H^+ — NH 4^+ fails, because its nitrogen has a closed octet and no empty orbital. - Stretch the definition instead to "anything that can hand over a proton counts", and NH 4^+ would get in — but then H 2O must get in too, since water also donates a proton ( H 2O → H^+ + OH^-, the ionisation of water that gives K w ). That criterion yields BF 3, H^+, NH 4^+, H 2O. The printed set matches neither list: it admits NH 4^+ on a rule that would also have to admit H 2O, and then excludes H 2O. Since the chapter teaches the vacant-orbital rule, that is the rule this question must be answered by, and it gives two species, not three. Write BF 3 and H^+, and add the one-line reason for NH 4^+ — an examiner following the key will see that you know precisely why it does not belong. NCERT's answer key prints "BF₃, H⁺, NH₄⁺", and NH₄⁺ does not belong there. A Lewis acid must be able to ACCEPT an electron pair, which needs a vacant orbital. In NH₄⁺ the nitrogen has already used all four of its orbitals — three N–H bonds plus the one formed by donating its lone pair to the fourth proton — so it has nothing left to accept with. NH₄⁺ is a Brønsted acid: it donates a proton. That is a different definition, and the two do not coincide here. Answer: BF 3 and H^+ are the Lewis acids — BF 3 has a vacant 2p orbital on boron and H^+ has an empty 1s orbital. H 2O is a Lewis base (two lone pairs to donate), and NH 4^+ is neither a Lewis acid nor a Lewis base — nitrogen's octet is complete with no lone pair to give and no empty orbital to receive one; it is a Brønsted acid only.

Question 37

What will be the conjugate bases for the Brönsted acids: HF, H 2SO 4 and HCO^- 3?

Accepted Answer

F–, HSO4 –, CO3 2–. A Brønsted acid is a proton ( H^+ ) donor, so its conjugate base is exactly what is left behind once that one H^+ is removed. The rule is: Acid Conjugate base + H^+ To get the conjugate base, take away one H^+ from the acid and adjust the charge by -1 (removing a positively charged proton leaves the remaining species one unit more negative than the acid was). 1. HF HF is neutral. Removing H^+ leaves the fluoride ion: HF → H^+ + F^- Conjugate base: F^- 2. H 2SO 4 This acid has two ionisable protons, but the conjugate base is defined by removing only one H^+ — the species formed immediately after the first proton leaves, not after both leave. This is the step most people rush past: H 2SO 4 is neutral, so taking away one H^+ (charge +1 ) leaves a species of charge -1, not SO 4^2-. H 2SO 4 → H^+ + HSO 4^- Conjugate base: HSO 4^- (the hydrogen sulphate ion — it still has one more proton it can donate) 3. HCO 3^- Here the acid itself already carries a charge of -1. Removing one more H^+ (charge +1 ) makes the resulting species -1 + (-1) = -2: HCO 3^- → H^+ + CO 3^2- Conjugate base: CO 3^2- (the carbonate ion) Answer: The conjugate base of HF is F^-; of H 2SO 4 is HSO 4^-; of HCO 3^- is CO 3^2-.

Question 38

Write the conjugate acids for the following Brönsted bases: NH 2^-, NH 3 and HCOO–.

Accepted Answer

NH3, NH4 +, HCOOH. A conjugate acid is what you get when a base GAINS one H^+ ion. By the Brønsted–Lowry definition, a base is a species that accepts a proton ( H^+ ). When it does, the new species formed — base plus that one proton — is called its conjugate acid: Base + H^+ → Conjugate acid The step people skip is checking the charge bookkeeping: adding one positive H^+ to a species raises its overall charge by exactly one unit — a -1 ion becomes neutral, a neutral molecule becomes +1. Get the charge on the product wrong and the "conjugate acid" you write is really a different ion altogether. Apply this to each base in turn. 1. NH 2^- (amide ion) This ion carries a -1 charge. Adding H^+ brings the charge to -1 + 1 = 0: NH 2^- + H^+ → NH 3 The conjugate acid of NH 2^- is NH 3 (ammonia). 2. NH 3 (ammonia) This molecule is neutral. Adding H^+ brings the charge to 0 + 1 = +1: NH 3 + H^+ → NH 4^+ The conjugate acid of NH 3 is NH 4^+ (ammonium ion). Notice that NH 3 shows up twice in this problem — once as a conjugate acid (of NH 2^- ) and once as the base being converted (into NH 4^+ ). This is exactly what Brønsted–Lowry theory predicts: the same species can act as an acid in one pairing and a base in another, since "acid" and "conjugate base" (or "base" and "conjugate acid") are always relative to a specific partner, not fixed labels on a molecule. 3. HCOO^- (formate ion, the conjugate base of formic acid) This ion carries a -1 charge. Adding H^+ brings the charge to -1 + 1 = 0: HCOO^- + H^+ → HCOOH The conjugate acid of HCOO^- is HCOOH (formic acid, methanoic acid). Answer: The conjugate acid of NH 2^- is NH 3; the conjugate acid of NH 3 is NH 4^+; the conjugate acid of HCOO^- is HCOOH.

Question 39

The species: H 2O, HCO 3^-, HSO 4^- and NH 3 can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

Accepted Answer

A Brønsted acid donates a proton (H⁺); a Brønsted base accepts one. A species that can do both is called amphiprotic, and each direction gives it a different partner: the conjugate base is what's left after it gives away a proton, and the conjugate acid is what it becomes after it takes one on. The rule to keep straight: removing H^+ from a species gives its conjugate base (one less proton, one more negative charge); adding H^+ gives its conjugate acid (one more proton, one less negative charge, or one more positive charge). People mix these up because "conjugate acid of X" sounds like it should be X acting as an acid — it is the opposite: X acting as a base, having just accepted a proton. Work through each species both ways. H 2 O As an acid (donates H^+ ): H 2 O H^+ + OH^- conjugate base = OH^- As a base (accepts H^+ ): H 2 O + H^+ H 3 O^+ conjugate acid = H 3 O^+ HCO 3^- As an acid: HCO 3^- H^+ + CO 3^2- conjugate base = CO 3^2- As a base: HCO 3^- + H^+ H 2 CO 3 conjugate acid = H 2 CO 3 HSO 4^- As an acid: HSO 4^- H^+ + SO 4^2- conjugate base = SO 4^2- As a base: HSO 4^- + H^+ H 2 SO 4 conjugate acid = H 2 SO 4 NH 3 As an acid — this is the one that feels wrong at first, since ammonia is usually thought of only as a base, but it can still give up a proton from one of its N–H bonds: NH 3 H^+ + NH 2^- conjugate base = NH 2^- (the amide ion) As a base — its familiar role: NH 3 + H^+ NH 4^+ conjugate acid = NH 4^+ Notice the pattern across all four: as an acid, each species loses one H^+ to reach its conjugate base; as a base, each species gains one H^+ to reach its conjugate acid. Charge always shifts by exactly one unit toward more negative (losing H^+ ) or more positive (gaining H^+ ). Answer: - H 2 O: conjugate base OH^-, conjugate acid H 3 O^+ - HCO 3^-: conjugate base CO 3^2-, conjugate acid H 2 CO 3 - HSO 4^-: conjugate base SO 4^2-, conjugate acid H 2 SO 4 - NH 3: conjugate base NH 2^-, conjugate acid NH 4^+

Question 40

Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH^- (b) F^- (c) H^+ (d) BCl 3.

Accepted Answer

A Lewis acid has an empty orbital that can accept an electron pair; a Lewis base has a lone pair it can donate. This is a broader definition than the Arrhenius or Brønsted-Lowry one — it does not require H ^+ or OH ^- at all, just the ability to accept or donate an electron pair, so even species with no protons in sight (like BCl 3 ) can be classified. (a) OH^- — Lewis base. The oxygen atom carries three lone pairs and an overall negative charge, so it is electron-rich. It donates one of these lone pairs to an electron-deficient centre: OH^- + H^+ → H 2 O Here OH^- supplies the electron pair that forms the new O–H bond, so it is acting as a Lewis base. (b) F^- — Lewis base. Fluoride has four lone pairs and a negative charge, making it a good electron-pair donor: F^- + BF 3 → BF 4^- The lone pair on F^- is donated into the empty orbital on boron, forming the new B–F bond. F^- is the electron-pair donor, so it is the Lewis base (and BF 3 is the Lewis acid in this pairing). (c) H^+ — Lewis acid. A bare proton has no electrons at all — its 1s orbital is completely empty. It cannot donate anything; it can only accept a lone pair offered to it, as already seen in (a): H^+ + OH^- → H 2 O H^+ accepts the electron pair from OH^-, so it is a Lewis acid. This is the step people miss: under the Brønsted-Lowry picture H^+ is "the acid" because it is the proton donor in a proton-transfer reaction, but under the Lewis picture the same species is an acid for the opposite reason — because it is an electron-pair acceptor, not a donor of anything. (d) BCl 3 — Lewis acid. Boron has only 3 valence electrons and forms three bonds to Cl, using all 3 electrons but leaving boron's octet incomplete — it has just 6 electrons around it, with one orbital empty. That empty orbital can accept an incoming lone pair, for example from ammonia: BCl 3 +: NH 3 → Cl 3 B - NH 3 Nitrogen's lone pair fills boron's empty orbital, forming a new coordinate (dative) B–N bond. BCl 3 accepts the pair, so it is a Lewis acid. The short aside worth remembering here: it is boron's incomplete octet, not any charge, that makes a neutral molecule like BCl 3 behave as a Lewis acid — you don't need a formal positive charge to be an electron-pair acceptor. Answer: OH^- and F^- are Lewis bases (electron-pair donors, via their lone pairs); H^+ and BCl 3 are Lewis acids (electron-pair acceptors — H^+ via its empty 1s orbital, BCl 3 via boron's incomplete octet).

Question 41

The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. What is its pH?

Accepted Answer

2.42. pH measures how acidic a solution is through the hydrogen-ion concentration, using a negative log scale — smaller [ H^+ ] means a bigger, less negative log, so a larger pH. The defining formula is pH = - log 10[ H^+ ] where [ H^+ ] is the molar concentration of hydrogen ions, in mol L ^-1 (equivalently mol/dm ^3, written M). You are given [ H^+ ] = 3.8 × 10^-3 M Substitute directly — do not simplify the concentration first, since the log of a product splits cleanly: pH = - log 10(3.8 × 10^-3) The step people rush: split the log of a product into a sum, don't try to take log (3.8×10^-3) as one lump. log 10(3.8 × 10^-3) = log 10(3.8) + log 10(10^-3) Now evaluate each piece: log 10(3.8) = 0.5798 log 10(10^-3) = -3 So log 10(3.8 × 10^-3) = 0.5798 + (-3) = -2.4202 Then pH = -(-2.4202) = 2.4202 The given concentration, 3.8 × 10^-3, has two significant figures, so the pH should be reported to two decimal places (pH values by convention carry as many decimal places as the significant figures in [ H^+ ], since the digits before the decimal point just track the power of ten): pH = 2.42 This is comfortably on the acidic side of pH 7, which fits a soft drink — most are mildly acidic from dissolved carbon dioxide and added acids. Answer: pH = 2.42

Question 42

The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

Accepted Answer

1.7 x 10–4M. pH tells you - log 10[ H^+ ], so to get the concentration back you undo the log by raising 10 to the power of - pH. The defining formula is pH = - log 10[ H^+ ] where [ H^+ ] is the hydrogen-ion concentration in mol L ^-1 (mol/L), the quantity you're asked to find. Step 1 — Rearrange for [ H^+ ]. Starting from pH = - log 10[ H^+ ], multiply both sides by -1: log 10[ H^+ ] = - pH Then remove the log by raising 10 to the power of each side: [ H^+ ] = 10^ - pH Step 2 — Substitute the given pH. The pH of the vinegar sample is 3.76, so [ H^+ ] = 10^-3.76 mol L^-1 Step 3 — Split the exponent into a whole number and a fraction so it can be evaluated. This is the step people get wrong — you cannot just call 10^-3.76 "about 10^-4 "; the leading digit matters, and it comes from the decimal part of the exponent. Write -3.76 = -4 + 0.24 so that [ H^+ ] = 10^-4× 10^0.24 mol L^-1 Step 4 — Evaluate 10^0.24. 10^0.24 = 1.738 ( to 4 significant figures ) Step 5 — Combine. [ H^+ ] = 1.738 × 10^-4 mol L^-1 Since the pH was given to two decimal places (3.76), the mantissa of [ H^+ ] is only reliable to about three significant figures, so round to [ H^+ ] ≈ 1.74 × 10^-4 mol L^-1 Answer: [ H^+ ] = 1.74 × 10^-4 mol L^-1 (i.e., 10^-3.76 mol L ^-1 ).

Question 43

The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Accepted Answer

F–= 1.5 x 10–11, HCOO–= 5.6 × 10–11, CN–= 2.08 x 10–6. A conjugate acid–base pair is always linked by K a × K b = K w. When an acid HA loses a proton, it turns into its conjugate base A⁻. The strength of that conjugate base (how well it grabs a proton back from water) is fixed by how strong the acid was — a strong acid leaves behind a weak conjugate base, and this trade-off is captured by one constant: at 298 K, K a × K b = K w = 1.0 × 10^-14 Here K a is the ionization constant of the acid, K b is the ionization constant of its conjugate base, and K w is the ionic product of water at 298 K. Rearranging, K b = K w/K a The step people skip is checking which species is the conjugate base before plugging in — it is the acid with one H^+ removed, not the acid itself: - HF loses H^+ to give F^- - HCOOH loses H^+ to give HCOO^- - HCN loses H^+ to give CN^- Conjugate base of HF (i.e. F⁻): K b( F^-) = (1.0 × 10^-14)/(6.8 × 10^-4) = 1.4706 × 10^-11 Rounded to two significant figures (matching the two sig. figs. given for K a ): K b( F^-) = 1.5 × 10^-11 Conjugate base of HCOOH (i.e. HCOO⁻): K b( HCOO^-) = (1.0 × 10^-14)/(1.8 × 10^-4) = 5.5556 × 10^-11 Rounded to two significant figures: K b( HCOO^-) = 5.6 × 10^-11 Conjugate base of HCN (i.e. CN⁻): K b( CN^-) = (1.0 × 10^-14)/(4.8 × 10^-9) = 2.0833 × 10^-6 Rounded to two significant figures: K b( CN^-) = 2.1 × 10^-6 Notice the pattern: HCN has the smallest K a (it is the weakest acid of the three), so its conjugate base CN^- has the largest K b — it holds onto the proton it picks up from water most readily, of the three bases here. The stronger the parent acid, the weaker its conjugate base, and vice versa. Answer: K b( F^-) = 1.5 × 10^-11, K b( HCOO^-) = 5.6 × 10^-11, K b( CN^-) = 2.1 × 10^-6

Question 44

The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

Accepted Answer

[phenolate ion]= 2.2 × 10–6, α = 4.47 × 10–5, α in sodium phenolate = 10–8. The common-ion effect crushes an already-tiny ionization even further — track [H⁺] itself, and don't let a hydrolysis common ion trick you into a percentage that looks reasonable. Setting up the equilibrium. Phenol is a weak acid: C 6H 5OH C 6H 5O^- + H^+ The ionization constant is K a = ([ C 6H 5O^- ][ H^+ ])/([ C 6H 5OH ]) where [ C 6H 5O^- ], [ H^+ ], and [ C 6H 5OH ] are equilibrium molar concentrations of phenolate ion, hydrogen ion, and un-ionized phenol. Part 1 — plain 0.05 M phenol, no added salt Let x be the concentration of phenol that ionizes. Since the only source of C 6H 5O^- and H^+ is this ionization, they are produced in a 1:1 ratio, so [ C 6H 5O^- ] = [ H^+ ] = x, and [ C 6H 5OH ] = 0.05 - x. K a = (x · x)/(0.05 - x) = x^2/0.05-x Because K a = 1.0×10^-10 is minuscule compared with 0.05, the amount that ionizes, x, will turn out to be far smaller than 0.05 — so 0.05 - x ≈ 0.05. (This is the step people skip and then get stuck solving a quadratic that never needed solving.) That gives x^2 ≈ K a × 0.05 = (1.0×10^-10)(0.05) = 5.0×10^-12 x = √ 5.0×10^-12 = 2.2×10^-6 M Check the approximation: 2.2×10^-6 is about 0.004 % of 0.05, completely negligible next to it — the shortcut was valid. So in plain 0.05 M phenol, [ C 6H 5O^- ] = 2.2×10^-6 M. Part 2 — the same phenol, now also 0.01 M in sodium phenolate Sodium phenolate, C 6H 5ONa, is an ionic salt and dissociates completely: C 6H 5ONa → C 6H 5O^- + Na^+ This dumps 0.01 M of C 6H 5O^- into the solution before phenol ionizes at all. That phenolate ion is already a product of the phenol equilibrium — adding it from outside pushes the equilibrium in Part 1's reaction backward. This is the common-ion effect: phenol's own ionization is suppressed because one of its products is now sitting there in bulk. Let x be the (now much smaller) concentration of phenol that still manages to ionize. Track each species starting from these initial concentrations: [ C 6H 5OH ] = 0.05 - x ≈ 0.05, [ C 6H 5O^- ] = 0.01 + x ≈ 0.01, [ H^+ ] = x Both approximations drop x next to 0.05 and 0.01 — reasonable to try first, and worth checking once x is known. Substituting into K a: K a = ([ H^+ ][ C 6H 5O^- ])/([ C 6H 5OH ]) = (x(0.01))/0.05 1.0×10^-10 = (0.01 x)/0.05 = 0.2 x x = (1.0×10^-10)/0.2 = 5.0×10^-10 M Check: 5.0×10^-10 0.01 and 0.05 — both approximations hold comfortably, even better than in Part 1, because the common ion has pinned the equilibrium down harder. Degree of ionization is the fraction of the original phenol that ionized — do not divide by the total phenolate concentration (that would mix up "how much of the acid reacted" with "how much product exists"). Using α = x/C where C = 0.05 M is the starting phenol concentration: α = (5.0×10^-10)/0.05 = 1.0×10^-8 Comparing the two parts: ionization dropped from x = 2.2×10^-6 M down to x = 5.0×10^-10 M — over four orders of magnitude smaller — purely because 0.01 M of phenolate ion was already present. That is the common-ion effect in numbers, not just in words. Answer: In 0.05 M phenol alone, [ C 6H 5O^- ] = 2.2×10^-6 M. With the solution also 0.01 M in sodium phenolate, the degree of ionization of phenol falls to α = 1.0×10^-8 (i.e. 1.0×10^-6 % ).

Question 45

The first ionization constant of H 2S is 9.1 × 10–8. Calculate the concentration of HS^- ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H 2S is 1.2 × 10–13, calculate the concentration of S 2 – under both conditions.

Accepted Answer

[HS–]= 9.54 x 10–5, in 0.1M HCl [HS–] = 9.1 × 10–8M, [S2–] = 1.2 × 10–13M, in 0.1M HCl [S2–]= 1.09 × 10–19M. A weak diprotic acid ionizes in two steps whose equilibrium constants are wildly different — solve the first step completely, then feed its numbers into the second. Here K a 1 = 9.1×10^-8 is about half a million times bigger than K a 2=1.2×10^-13, so essentially all the H^+ and HS^- in the solution come from the first step; the second step barely disturbs those numbers. Step 1 — plain 0.1 M H₂S, first ionization H 2 S H^+ + HS^- K a 1=([ H^+][ HS^-])/([ H 2 S ])=9.1×10^-8 Let x be the concentration of H 2 S that ionizes. Starting from 0.1 M and nothing else present, at equilibrium [ H^+]=[ HS^-]=x and [ H 2 S ]=0.1-x: K a 1=x^2/0.1-x=9.1×10^-8 K a 1 is tiny next to 0.1, so x should be tiny too — assume 0.1-x≈ 0.1 (this substitution is the step to justify, not skip: it's only safe when K a 1 the initial concentration, which it is here by a factor of a million). x^2 = (9.1×10^-8)(0.1) = 9.1×10^-9 x = √ 9.1×10^-9 = 9.5×10^-5 M Checking the assumption: x/0.1 = 9.5×10^-4, under 0.1% of the starting concentration, so dropping it from 0.1-x was valid. So in plain 0.1 M H₂S: [ HS^-] = [ H^+] = 9.5×10^-5 M Step 2 — using this to get [S²⁻], plain solution HS^- H^+ + S^2- K a 2=([ H^+][ S^2-])/([ HS^-])=1.2×10^-13 Solve for [ S^2-]: [ S^2-] = K a 2×([ HS^-])/([ H^+]) The [ H^+] and [ HS^-] here are the ones just found in Step 1, and — because the second ionization removes only a negligible sliver of the HS^- that Step 1 produced — they are still equal to each other ( 9.5×10^-5 M each). That equality is the trick: it makes the ratio [ HS^-]/[ H^+] equal to 1, so it cancels outright: [ S^2-] = K a 2 = 1.2×10^-13 M This is a shortcut specific to this situation (no other source of H^+ or HS^- ), not a general rule — it breaks the moment another acid is added, which is exactly the next part of the question. Now the solution is also 0.1 M in HCl — common-ion effect HCl is a strong acid: it dissociates completely and dumps 0.1 M of H^+ into the solution, which is over a thousand times more H^+ than the H₂S alone produced. By Le Chatelier's principle this extra H^+ pushes the first equilibrium, H 2 S H^+ + HS^-, back toward H 2 S — this is the common-ion effect, and it means [ HS^-] must fall from what it was before. Because HCl supplies almost all the H^+ now (the tiny amount H₂S itself contributes is swamped and can be ignored — this shortcut only works because HCl is the much stronger acid, not for two comparably weak acids): [ H^+] ≈ 0.1 M And because the equilibrium has been pushed back toward H₂S, almost none of it ionizes, so: [ H 2 S ] ≈ 0.1 M (essentially unchanged from the starting concentration) Substitute both into K a 1 and solve for [ HS^-]: [ HS^-] = K a 1×([ H 2 S ])/([ H^+]) = (9.1×10^-8)×0.1/0.1 = 9.1×10^-8 M Compare the two cases: [ HS^-] drops from 9.5×10^-5 M to 9.1×10^-8 M — roughly a thousandfold suppression, purely from the common H^+ ion added by the HCl. Now redo Step 2 with the new numbers. This time [ H^+]=0.1 M (from HCl) is NOT equal to [ HS^-]=9.1×10^-8 M, so the earlier cancellation trick does not apply — the two must be substituted separately: [ S^2-] = K a 2×([ HS^-])/([ H^+]) = (1.2×10^-13)×(9.1×10^-8)/0.1 [ S^2-] = (1.2×9.1)×10^-13-8+1 = 10.92×10^-20 = 1.1×10^-19 M So adding the strong acid suppresses [ S^2-] even more drastically than it suppresses [ HS^-]^- from 1.2×10^-13 M down to 1.1×10^-19 M, a factor of about a million, because the second ionization depends on [ H^+] being small in a way the first ionization does not need to. Answer: In plain 0.1 M H₂S, [ HS^-] = 9.5×10^-5 M and [ S^2-] = 1.2×10^-13 M. With the solution also 0.1 M in HCl, the common-ion effect lowers these to [ HS^-] = 9.1×10^-8 M and [ S^2-] = 1.1×10^-19 M.

Question 46

The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Accepted Answer

[Ac–]= 0.00093, pH= 3.03. The dissociation is small enough to use the shortcut α = √ K a/C — but only after checking that assumption is actually valid. Step 1 — Set up the equilibrium. Acetic acid dissociates as CH 3COOH CH 3COO^- + H^+ Let the initial concentration be C = 0.05 M and let α be the degree of dissociation (the fraction of acid molecules that have ionized). Then at equilibrium: CH₃COOH CH₃COO⁻ H⁺ --- --- --- --- Initial C 0 0 Equilibrium C(1-α) Cα Cα Step 2 — Write the equilibrium constant. K a = ([ CH 3COO^- ][ H^+ ])/([ CH 3COOH ]) = ((Cα)(Cα))/(C(1-α)) = (Cα^2)/(1-α) Here K a is the acid ionization constant — a fixed number for acetic acid at a given temperature — and each bracket is an equilibrium concentration, not the starting concentration. Step 3 — Simplify using 1-α ≈ 1. Acetic acid is a weak acid, so α is expected to be small (much less than 1). This is the step people skip without checking: assuming 1-α≈ 1 is only valid when α turns out to be small, so it must be verified after the fact, which is done in Step 4. With that approximation: K a ≈ Cα^2 ⇒ α = √ K a/C Step 4 — Substitute the numbers. α = √ (1.74×10^-5)/0.05 = √ 3.48×10^-4 α = 1.866×10^-2 ≈ 0.0187 Since α ≈ 0.019 1, the approximation 1-α≈1 in Step 3 was justified — check this every time, because for a stronger acid or a more dilute solution α can come out too large for the shortcut to hold. Step 5 — Concentration of acetate ion. At equilibrium, [ CH 3COO^- ] = [ H^+ ] = Cα (one acetate ion and one hydrogen ion are produced together for every molecule that dissociates — this is the step people get wrong by forgetting the two concentrations are equal here): [ CH 3COO^- ] = Cα = 0.05 M × 1.866×10^-2 [ CH 3COO^- ] = 9.33×10^-4 M Step 6 — pH of the solution. pH is defined as pH = - log 10[ H^+ ], where [ H^+ ] is in mol/L. Since [ H^+ ] = [ CH 3COO^- ] = 9.33×10^-4 M: pH = - log (9.33×10^-4) = -( log 9.33 + log 10^-4) pH = -(0.970 - 4) = 4 - 0.970 pH = 3.03 The data ( K a to 3 significant figures, C to 1 significant figure but treated as exact for this calculation) supports rounding the degree of dissociation to 3 significant figures and the pH to 2 decimal places — the conventional precision for pH. Answer: degree of dissociation α ≈ 0.0187 (i.e., 1.87% ionized); [ CH 3COO^- ] ≈ 9.33×10^-4 M; pH ≈ 3.03.

Question 47

It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.

Accepted Answer

[A–] = 7.08 x10–5M, Ka= 5.08 × 10–7, pKa= 6.29. pH tells you [ H^+] directly, and for an acid that dissociates one-to-one, every H^+ ion released is matched by exactly one A^- ion — so at equilibrium [ A^-] = [ H^+]. Step 1 — get [ H^+] from the pH. By definition pH = - log 10[ H^+], so [ H^+] = 10^ - pH = 10^-4.15 Split the exponent: 10^-4.15 = 10^0.85× 10^-5 = 7.079×10^-5 M. Step 2 — set up the equilibrium. Let the acid be HA, with initial concentration C = 0.01 M: HA H^+ + A^- HA H^+ A^- --- --- --- --- Initial C 0 0 Change -x +x +x Equilibrium C-x x x Here x is the amount of acid that has ionized per litre. Since essentially all the H^+ in this solution comes from the acid (water's own contribution, 10^-7 M, is negligible next to 7.08×10^-5 M ), [ A^-] = [ H^+] = x = 7.079×10^-5 M ≈ 7.08×10^-5 M This is the concentration of the anion asked for in the question. Step 3 — find the equilibrium concentration of unreacted HA. The mistake to avoid: plugging the initial concentration C into the equilibrium expression instead of C - x. Only a tiny fraction of the acid has ionized, but the correct equilibrium value is still C-x, not C: [ HA ] eq = C - x = 0.01 - 0.00007079 = 9.929×10^-3 M Step 4 — write the ionization constant K a and substitute. K a (the acid ionization/dissociation constant) is the equilibrium constant for the dissociation, written as K a = ([ H^+][ A^-])/([ HA ]) with all three concentrations taken at equilibrium (not the initial value). Substituting: K a = ((7.079×10^-5)(7.079×10^-5))/(9.929×10^-3) = (5.011×10^-9)/(9.929×10^-3) = 5.048×10^-7 Rounding to three significant figures (matching the precision of the given pH, 4.15): K a ≈ 5.05×10^-7 Step 5 — convert K a to pK a. By the same log definition used for pH, pK a = - log 10K a. This is the step people get backwards — pK a is the negative log, so a smaller K a gives a larger pK a: pK a = - log 10(5.048×10^-7) = 7 - log 10(5.048) = 7 - 0.703 = 6.297 Rounded to two decimal places (consistent with the pH data given): pK a ≈ 6.30 Answer: [ A^-] ≈ 7.08×10^-5 M, K a ≈ 5.05×10^-7, pK a ≈ 6.30.

Question 48

Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

Accepted Answer

a) 2.52 b) 11.70 c) 2.70 d) 11.30. Complete dissociation means the molarity of the acid or base IS the molarity of H^+ or OH^- it releases — for a strong monoprotic acid or monoacidic base, no equilibrium calculation is needed. The defining formula throughout is pH = - log 10[ H^+ ] where [ H^+ ] is the hydrogen-ion concentration in mol L^-1. For a base, it is easier to find pOH = - log 10[ OH^- ] first and then use pH + pOH = 14 (at 298 K) (a) 0.003 M HCl HCl is a strong acid, so it dissociates completely: HCl → H^+ + Cl^- Each mole of HCl gives one mole of H^+, so [ H^+ ] = 0.003 mol L^-1 = 3 × 10^-3 mol L^-1. pH = - log 10(3 × 10^-3) = -( log 103 + log 1010^-3) = -(0.4771 - 3) = 3 - 0.4771 = 2.5229 Rounding to two decimal places (the data is given to one significant figure, but pH is conventionally reported to two decimals since the digits before the decimal point only fix the power of ten): pH ≈ 2.52 (b) 0.005 M NaOH NaOH is a strong base: NaOH → Na^+ + OH^- so [ OH^- ] = 0.005 mol L^-1 = 5 × 10^-3 mol L^-1. This is where it is easy to go wrong: you find pOH from [ OH^- ] directly — you do not take - log of the concentration and call it pH. The concentration you were given is of OH^-, not H^+. pOH = - log 10(5 × 10^-3) = -( log 105 - 3) = 3 - 0.6990 = 2.301 pH = 14 - pOH = 14 - 2.301 = 11.699 pH ≈ 11.70 (c) 0.002 M HBr HBr is a strong acid, fully dissociated: HBr → H^+ + Br^- [ H^+ ] = 0.002 mol L^-1 = 2 × 10^-3 mol L^-1 pH = - log 10(2 × 10^-3) = -( log 102 - 3) = 3 - 0.3010 = 2.699 pH ≈ 2.70 (d) 0.002 M KOH KOH is a strong base: KOH → K^+ + OH^- [ OH^- ] = 0.002 mol L^-1 = 2 × 10^-3 mol L^-1 pOH = - log 10(2 × 10^-3) = 3 - 0.3010 = 2.699 pH = 14 - 2.699 = 11.301 pH ≈ 11.30 Answer: (a) pH ≈ 2.52, (b) pH ≈ 11.70, (c) pH ≈ 2.70, (d) pH ≈ 11.30

Question 49

Calculate the pH of the following solutions: a) 2 g of TlOH dissolved in water to give 2 litre of solution. b) 0.3 g of Ca(OH) 2 dissolved in water to give 500 mL of solution. c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

Accepted Answer

a) 11.65 b) 12.21 c) 12.57 c) 1.87. A strong acid or base dissociates completely, so the first job in every part is turning "so many grams in so many litres" into a molar concentration — then the dissociation equation tells you [ H^+] or [ OH^-] directly, with no equilibrium to solve. a) TlOH Name the formula: molar mass M is the mass of one mole, found by adding atomic masses. Taking the atomic mass of Tl as 204 u, M( TlOH ) = 204 + 16 + 1 = 221 g/mol Moles of solute, n = ( mass )/M: n = (2 g )/(221 g/mol ) = 9.05×10^-3 mol Molarity C = n/V, where V is the volume of solution in litres — not the water you poured in to make it. That distinction is exactly why the question tells you the final volume ("2 litre of solution") rather than how much water was added: C = (9.05×10^-3 mol )/(2 L ) = 4.525×10^-3 mol/L TlOH is a strong, monoacidic base — it dissociates completely as TlOH → Tl^+ + OH^-, one OH^- per formula unit, so [ OH^-] = C: [ OH^-] = 4.525×10^-3 M Using pOH = - log [ OH^-]: pOH = - log (4.525×10^-3) = 3 - log (4.525) = 3 - 0.656 = 2.34 Using pH + pOH = 14 (from K w = 10^-14 at 298 K): pH = 14 - 2.34 = 11.66 b) Ca(OH)₂ M( Ca(OH) 2) = 40 + 2(16+1) = 74 g/mol n = (0.3 g )/(74 g/mol ) = 4.054×10^-3 mol, C = (4.054×10^-3 mol )/(0.5 L ) = 8.108×10^-3 mol/L The step people skip here: Ca(OH) 2 → Ca^2+ + 2 OH^- releases two hydroxide ions per formula unit, so [ OH^-] is twice the molarity of the base, not equal to it: [ OH^-] = 2C = 2 × 8.108×10^-3 = 1.6216×10^-2 M pOH = - log (1.6216×10^-2) = 2 - log (1.6216) = 2 - 0.210 = 1.79 pH = 14 - 1.79 = 12.21 c) NaOH M( NaOH ) = 23 + 16 + 1 = 40 g/mol n = (0.3 g )/(40 g/mol ) = 7.5×10^-3 mol, C = (7.5×10^-3 mol )/(0.2 L ) = 3.75×10^-2 mol/L NaOH is a strong, monoacidic base, so [ OH^-] = C = 3.75×10^-2 M: pOH = - log (3.75×10^-2) = 2 - log (3.75) = 2 - 0.574 = 1.43 pH = 14 - 1.43 = 12.57 d) HCl diluted This is a dilution, not a dissolution — the moles of solute don't change, only the volume they sit in. The dilution law M 1V 1 = M 2V 2 is just that conservation statement, where M is molarity and V is volume (any consistent unit, since it cancels): M 1 = 13.6 M, V 1 = 1 mL, V 2 = 1000 mL (=1 L ) M 2 = M 1V 1/V 2 = (13.6 M × 1 mL )/(1000 mL ) = 1.36×10^-2 M HCl is a strong, monoprotic acid, so it dissociates fully and [ H^+] equals this diluted molarity directly — no equilibrium constant needed: [ H^+] = 1.36×10^-2 M pH = - log (1.36×10^-2) = 2 - log (1.36) = 2 - 0.133 = 1.87 Answer: pH = 11.66 (TlOH solution), 12.21 (Ca(OH)₂ solution), 12.57 (NaOH solution), 1.87 (diluted HCl)

Question 50

The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.

Accepted Answer

pH = 1.88, pKa = 2.70. Degree of ionization tells you what fraction of the acid molecules have split into ions — multiply it by the starting concentration to get [ H^+], don't guess at it from the concentration alone. Bromoacetic acid ionizes as BrCH 2 COOH BrCH 2 COO^- + H^+ Let C = initial concentration = 0.1 M and α = degree of ionization = 0.132 (a pure number — the fraction of molecules ionized, not a concentration itself). At equilibrium, the concentration of H^+ produced is [ H^+] = Cα Step 1: Find [ H^+] [ H^+] = (0.1 M )(0.132) = 0.0132 M = 1.32×10^-2 M Step 2: Find pH The formula for pH is pH = - log 10[ H^+] pH = - log 10(1.32×10^-2) = - ( log 101.32 + log 1010^-2 ) = -(0.1206 - 2) = 1.8794 Rounded to three significant figures (matching the three figures in α = 0.132 ): pH = 1.88 Step 3: Find K a For a weak acid of initial concentration C and degree of ionization α, the equilibrium concentrations are: undissociated acid = C(1-α), H^+ = Cα, and BrCH 2 COO^- = Cα. The acid dissociation constant is K a = ([ H^+][ BrCH 2 COO^-])/([ BrCH 2 COOH ]) = ((Cα)(Cα))/(C(1-α)) = (Cα^2)/(1-α) This is the step most people skip past — the denominator uses (1-α) because only that fraction of the acid is left un-ionized; using C instead of C(1-α) there is the classic mistake for an acid this weakly ionized ( α is small but not negligible). Substituting C = 0.1 M and α = 0.132: K a = ((0.1 M )(0.132)^2)/(1 - 0.132) = ((0.1 M )(0.017424))/0.868 = (0.0017424 M )/0.868 K a = 2.007×10^-3 M Rounded to three significant figures: K a = 2.01×10^-3 ( K a is dimensionless in the thermodynamic sense, but on the molarity scale it is often quoted with units of mol L ^-1; either convention is accepted.) Step 4: Find pKa The formula for pKa is p K a = - log 10K a p K a = - log 10(2.01×10^-3) = -( log 102.01 - 3) = -(0.3032 - 3) = 2.6968 Rounded to three significant figures: p K a = 2.70 Answer: pH = 1.88 and pKa = 2.70 (with K a = 2.01×10^-3 ).

Question 51

The pH of 0.005M codeine (C 18H 21NO 3) solution is 9.95. Calculate its ionization constant and pKb.

Accepted Answer

Kb = 1.6 × 10–6, pKb = 5.8. A weak base's ionization constant comes from its [ OH^-] at equilibrium, and that means going through pOH first — not pH directly. Codeine (call it B, since C 18 H 21 NO 3 is unwieldy to keep rewriting) is a weak base. In water it partly ionizes: B + H 2 O BH^+ + OH^- Step 1 — get pOH from pH. Use pH + pOH = 14 (this holds for water at 298 K, where K w = 10^-14 ). pOH = 14 - pH = 14 - 9.95 = 4.05 A base gives a high pH, so it's easy to accidentally plug 9.95 straight into [ OH^-] = 10^ - pH — that would silently compute [ H^+] instead. Convert to pOH first. Step 2 — get [ OH^-] from pOH. [ OH^-] = 10^ - pOH = 10^-4.05 = 10^-4 × 10^-0.05 mol L^-1 Since 10^-0.05 = 0.891, [ OH^-] = 8.91 × 10^-5 mol L^-1 Step 3 — set up the equilibrium table. Let C = 0.005 mol L^-1 be the starting concentration of codeine, and let x be how much of it ionizes. B BH^+ OH^- --- --- --- --- initial C 0 0 at equilibrium C-x x x Every OH^- ion produced comes with one BH^+, and both equal x. We already found [ OH^-] at equilibrium, so x = 8.91 × 10^-5 mol L^-1 Step 4 — check whether C - x ≈ C is safe. x/C = (8.91× 10^-5)/(5× 10^-3) = 0.0178 = 1.78 % That's under 5%, so approximating C - x ≈ C in the denominator is justified — skipping this check is the step people get wrong, because it silently turns an approximation into an assumption. Step 5 — write K b and substitute. The base ionization constant is K b = ([ BH^+][ OH^-])/([ B ]) = x^2/(C - x) ≈ x^2/C K b = ((8.91× 10^-5 mol L^-1)^2)/(5× 10^-3 mol L^-1) = (7.94× 10^-9 mol^2 L^-2)/(5× 10^-3 mol L^-1) K b = 1.59 × 10^-6 mol L^-1 Step 6 — convert to pK b. pK b = - log K b = - log (1.59× 10^-6) = 6 - log (1.59) = 6 - 0.20 = 5.80 pK b has no units — it's a log of a number, not a physical quantity — but K b itself is quoted in mol L ^-1, since the ionization is a 1:1:1 reaction and the mol ^2 L ^-2 from the numerator divided by mol L ^-1 leaves one power of concentration. Answer: K b ≈ 1.59 × 10^-6 mol L^-1, pK b ≈ 5.80

Question 52

What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Accepted Answer

α = 6.53 × 10–4, Ka = 2.35 × 10–5. Aniline is a weak base — set up its ionization equilibrium, use the fact that K b C to simplify, and go via [OH⁻] and pOH, since it's K b you were given, not K a. In water, aniline accepts a proton from water to form the anilinium ion: C 6H 5NH 2 + H 2O C 6H 5NH 3^+ + OH^- From Table 6.7, the base ionization constant of aniline is K b = 4.27 × 10^-10 Let C = 0.001 M = 1×10^-3 M be the initial concentration of aniline, and let α be its degree of ionization — the fraction of the dissolved aniline that has actually reacted with water. At equilibrium the concentrations are: aniline = C(1-α), anilinium ion = Cα, hydroxide ion = Cα. K b = ([C 6H 5NH 3^+][OH^-])/([C 6H 5NH 2]) = ((Cα)(Cα))/(C(1-α)) = (Cα^2)/(1-α) Here is the step people skip: you cannot drop (1-α) just because it looks convenient — you drop it only after checking that α will turn out to be tiny. Since K b 10^-10 is far smaller than C 10^-3, α will indeed be small, so 1-α ≈ 1: K b ≈ Cα^2 ⇒ α = √ K b/C Substituting the numbers: α = √ (4.27×10^-10)/(1×10^-3) = √ 4.27×10^-7 = 6.53×10^-4 Since α = 6.53×10^-4 1, the approximation 1-α ≈ 1 was justified — no need to solve the quadratic. Now find the hydroxide concentration: [OH^-] = Cα = (1×10^-3 mol/L )(6.53×10^-4) = 6.53×10^-7 mol/L Because you were given K b, this calculation naturally lands you on [OH^-], not [H^+]^- so find pOH first, then convert. pOH = - log [OH^-] = - log (6.53×10^-7) = 7 - log (6.53) = 7 - 0.815 = 6.19 At 298 K, K w = 1.0×10^-14, so pH + pOH = 14: pH = 14 - pOH = 14 - 6.19 = 7.82 Finding K a of the conjugate acid. The conjugate acid of aniline is the anilinium ion, which itself can donate a proton back to water: C 6H 5NH 3^+ + H 2O C 6H 5NH 2 + H 3O^+ For any conjugate acid–base pair, their ionization constants are tied together through the ionic product of water: K a × K b = K w = 1.0×10^-14 ( at 298 K ) This relation only holds for a genuine conjugate pair (aniline and anilinium ion here) — it is not a formula for combining two unrelated acids and bases. Solving for K a: K a = K w/K b = (1.0×10^-14)/(4.27×10^-10) = 2.34×10^-5 Rounding each result to three significant figures, matching the precision of the given K b: Answer: pH ≈ 7.82; degree of ionization α ≈ 6.53 × 10⁻⁴; ionization constant of the conjugate acid (anilinium ion) K a ≈ 2.34 × 10⁻⁵

Question 53

Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl?

Accepted Answer

a) 0.0018 b) 0.00018. The degree of ionization of a weak acid is fixed by its dissociation constant K a^- and adding a strong acid dumps extra H^+ into the solution, which pushes the acid's own equilibrium back toward the undissociated form. This is the common-ion effect. Step 1 — get K a out of the pKa. K a is the acid dissociation constant of acetic acid, and by definition pK a = - log 10K a, so K a = 10^-pK a = 10^-4.74 = 10^-5×10^0.26 Since 10^0.26≈ 1.82, K a ≈ 1.8×10^-5 mol L^-1 (pKa is a log scale — you must exponentiate it before using it in ordinary algebra. Never subtract pKa values and treat the result as a concentration.) Step 2 — degree of ionization of the acetic acid on its own. Let c = 0.05 mol L^-1 be the starting concentration of CH 3 COOH, and α the fraction of it that ionizes: CH 3 COOH CH 3 COO^- + H^+ At equilibrium, [ CH 3 COOH ] = c(1-α), [ CH 3 COO^-] = cα, and [ H^+] = cα, so K a = ([ CH 3 COO^-][ H^+])/([ CH 3 COOH ]) = ((cα)(cα))/(c(1-α)) = (cα^2)/(1-α) K a is tiny compared with c, which means α will come out small, so 1-α≈1: K a ≈ cα^2 ⇒ α = √ K a/c α = √ (1.8×10^-5)/0.05 = √ 3.6×10^-4 = 1.9×10^-2 (Check the approximation you just made: α≈0.019, so 1-α≈0.98 — within 2% of 1, well inside the precision that a two-significant-figure K a can justify.) So on its own, about 1.9% of the acetic acid ionizes. Step 3 — adding HCl: the common-ion effect. HCl is a strong acid, so it dissociates completely: a "0.01 M HCl" solution already has [ H^+] = 0.01 mol L^-1 in it before the acetic acid ionizes at all. That H^+ is the same species the acetic acid equilibrium produces, so by Le Chatelier's principle this extra H^+ shifts the equilibrium back toward CH 3 COOH, shrinking the degree of ionization from α to a smaller value α'. Let C stand for the HCl concentration. Now [ CH 3 COOH ] = c(1-α'), [ CH 3 COO^-] = cα', and — this is the step that trips people up — [ H^+] is the total from both sources, C + cα', not just the acid's own contribution: K a = (cα' (C+cα'))/(c(1-α')) Because K a is so small, α' here will be far smaller than the α of Step 2, so cα' C and also 1-α'≈1. Both approximations collapse the expression to K a ≈ α' C ⇒ α' = K a/C (a) C = 0.01 mol L^-1: α' = (1.8×10^-5)/0.01 = 1.8×10^-3 Check: cα' = 0.05 × 1.8×10^-3 = 9×10^-5 mol L^-1 — under 1% of C = 0.01 mol L^-1, so [ H^+]≈ C was a safe approximation. (b) C = 0.1 mol L^-1: α' = (1.8×10^-5)/0.1 = 1.8×10^-4 Check: cα' = 0.05 × 1.8×10^-4 = 9×10^-6 mol L^-1 — negligible next to C = 0.1 mol L^-1, so the approximation holds even better here. Lay the three results side by side: 1.9×10^-2 → 1.8×10^-3 → 1.8×10^-4. Each ten-fold increase in the H^+ supplied by HCl cuts the acetic acid's own ionization by almost exactly a factor of ten, matching α' = K a/C — the signature of common-ion suppression, not a coincidence. Answer: In 0.05 M acetic acid alone, α ≈ 1.9×10^-2 (about 1.9%). Adding 0.01 M HCl suppresses this to α' ≈ 1.8×10^-3 (about 0.18%), and adding 0.1 M HCl suppresses it further to α' ≈ 1.8×10^-4 (about 0.018%) — the degree of dissociation falls sharply as more common H^+ ion is introduced.

Question 54

The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?

Accepted Answer

α = 0.0054. A base's ionization constant K b sets how far it splits into ions at a given concentration — and loading the solution with a strong base that supplies the very same product ion ( OH^- ) pushes that split back down. That's the common-ion effect, and it's the part of this problem that catches people off guard. Dimethylamine ionizes in water as ( CH 3 ) 2 NH + H 2O ( CH 3 ) 2 NH 2^+ + OH^- Part 1 — degree of ionization in plain 0.02 M solution Let α be the degree of ionization (the fraction of dimethylamine molecules that ionize) and C = 0.02 M the starting concentration. At equilibrium: [( CH 3 ) 2 NH ] = C(1-α), [( CH 3 ) 2 NH 2^+ ] = Cα, [ OH^- ] = Cα The base ionization constant is defined as K b = ([( CH 3 ) 2 NH 2^+ ][ OH^- ])/([( CH 3 ) 2 NH ]) = ((Cα)(Cα))/(C(1-α)) = (Cα^2)/(1-α) Aside — this is the step where a weak-electrolyte problem usually gets simplified: when α turns out to be small, 1-α ≈ 1, and the messy quotient collapses to K b ≈ Cα^2. Try that first; it's the standard move for this kind of ionization constant. K b ≈ Cα^2 ⇒ α = √ K b/C Substituting K b = 5.4×10^-4 and C = 0.02 M: α = √ (5.4×10^-4)/(0.02 M ) = √ 2.7×10^-2 = 0.164 So roughly 16.4% of the dimethylamine ionizes when it's on its own in solution. Part 2 — ionization once the solution is also 0.1 M in NaOH NaOH is a strong base: every formula unit dissociates completely, so it hands the solution 0.1 M OH^- before the dimethylamine ionizes at all. That OH^- is a product of the dimethylamine equilibrium too, so by Le Chatelier's principle the extra OH^- already present pushes the equilibrium back toward the un-ionized amine — suppressing ionization rather than leaving it unchanged. This is the common-ion effect, and skipping it (treating the two bases as independent) is the error to watch for here. Let x be the concentration of dimethylamine that now ionizes. At equilibrium: [( CH 3 ) 2 NH ] = 0.02 M -x, [( CH 3 ) 2 NH 2^+ ] = x, [ OH^- ] = 0.1 M +x K b = (x(0.1 M +x))/(0.02 M -x) Aside — because the common-ion effect suppresses ionization even further than in Part 1, x will be much smaller than either 0.02 M or 0.1 M (smaller, in fact, than the α found in Part 1). That makes 0.1 M +x ≈ 0.1 M and 0.02 M -x ≈ 0.02 M safe approximations, even though the same kind of approximation was borderline in Part 1. K b ≈ (x(0.1 M ))/(0.02 M ) Solving for x: x = (K b × 0.02 M )/(0.1 M ) = ((5.4×10^-4)(0.02 M ))/0.1 = 1.08×10^-4 M The percentage of dimethylamine ionized is this ionized concentration divided by the original concentration C = 0.02 M — not by 0.1 M, which is the NaOH concentration, not the amine's: % ionized = x/C× 100 = (1.08×10^-4 M )/(0.02 M )× 100 = 0.54 % Aside — that's about 30 times smaller than the 16.4% found with no NaOH present. K b itself hasn't changed (it's fixed at a given temperature); only the equilibrium position has shifted, which is exactly what the common-ion effect predicts. Rounding to match the precision K b = 5.4×10^-4 (two significant figures) supports: the degree of ionization in plain solution is α ≈ 0.164, and the percentage ionized with NaOH present is 0.54 %. Answer: In 0.02 M dimethylamine alone, the degree of ionization is α ≈ 0.164 (about 16.4%). With the solution also 0.1 M in NaOH, the common-ion effect suppresses ionization to about 0.54%.

Question 55

Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below: (a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2 (c) Human blood, 7.38 (d) Human saliva, 6.4.

Accepted Answer

a) 1.48 × 10–7M, b) 0.063 c) 4.17 × 10–8M d) 3.98 × 10–7. pH tells you the power of ten that [ H^+ ] sits at — you just have to undo the logarithm. The defining relation is pH = - log 10[ H^+ ] where [ H^+ ] is the hydrogen-ion concentration in mol/L (mol L ^-1 ). Rearranging for the concentration gives [ H^+ ] = 10^ - pH The step people fumble here is the algebra of the rearrangement itself: from pH = - log [ H^+ ], multiplying by -1 gives - pH = log [ H^+ ], and taking the antilog of both sides gives [ H^+ ] = 10^ - pH — not 10^ pH. A larger pH must give a smaller [ H^+ ], so the exponent has to stay negative. For each fluid you split - pH into a negative integer plus a positive decimal, so the answer comes out as a clean number times a power of ten: 10^-6.83 = 10^0.17× 10^-7, for instance, using 10^0.17 as the "clean number." (a) Human muscle fluid, pH = 6.83 [ H^+ ] = 10^-6.83 = 10^0.17× 10^-7 mol L^-1 10^0.17 = 1.479, so [ H^+ ] = 1.479× 10^-7 mol L^-1 ≈ 1.48× 10^-7 mol L^-1 (b) Human stomach fluid, pH = 1.2 [ H^+ ] = 10^-1.2 = 10^0.8× 10^-2 mol L^-1 10^0.8 = 6.310, so [ H^+ ] = 6.310× 10^-2 mol L^-1 ≈ 6.31× 10^-2 mol L^-1 The very low pH of stomach fluid is exactly why this concentration is the largest of the four — nearly a million times more concentrated in H^+ than the blood in part (c). (c) Human blood, pH = 7.38 [ H^+ ] = 10^-7.38 = 10^0.62× 10^-8 mol L^-1 10^0.62 = 4.169, so [ H^+ ] = 4.169× 10^-8 mol L^-1 ≈ 4.17× 10^-8 mol L^-1 (d) Human saliva, pH = 6.4 [ H^+ ] = 10^-6.4 = 10^0.6× 10^-7 mol L^-1 10^0.6 = 3.981, so [ H^+ ] = 3.981× 10^-7 mol L^-1 ≈ 3.98× 10^-7 mol L^-1 Each pH value was given to two decimal places, so each [ H^+ ] is rounded to three significant figures — carrying more digits would claim precision the data doesn't have. Answer: (a) [ H^+ ] ≈ 1.48× 10^-7 mol L ^-1; (b) [ H^+ ] ≈ 6.31× 10^-2 mol L ^-1; (c) [ H^+ ] ≈ 4.17× 10^-8 mol L ^-1; (d) [ H^+ ] ≈ 3.98× 10^-7 mol L ^-1

Question 56

The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

Accepted Answer

a) 1.5 × 10–7M, b) 10–5M, c) 6.31 × 10–5M d) 6.31 × 10–3M. pH is defined as pH = - log 10[ H^+], so reversing it gives [ H^+] = 10^ - pH — the hydrogen ion concentration is ten raised to the negative of the pH value. Do this for each liquid. To evaluate 10^ - pH by hand, split the exponent into a whole-number part and a decimal part, because 10^-n (n a whole number) is easy, and 10^0.x is a small factor you can read off (or compute) separately. Milk, pH = 6.8 Write -6.8 = 0.2 - 7, so [ H^+] = 10^-6.8 = 10^0.2 × 10^-7 = 1.585 × 10^-7 mol L^-1 Rounded to two significant figures (matching the one-decimal precision of the pH data): 1.6 × 10^-7 mol L^-1. Black coffee, pH = 5.0 Here the exponent is already a whole number, so no splitting is needed: [ H^+] = 10^-5.0 = 1.0 × 10^-5 mol L^-1 Tomato juice, pH = 4.2 Write -4.2 = 0.8 - 5, so [ H^+] = 10^-4.2 = 10^0.8 × 10^-5 = 6.310 × 10^-5 mol L^-1 Rounded: 6.3 × 10^-5 mol L^-1. Lemon juice, pH = 2.2 Write -2.2 = 0.8 - 3, so [ H^+] = 10^-2.2 = 10^0.8 × 10^-3 = 6.310 × 10^-3 mol L^-1 Rounded: 6.3 × 10^-3 mol L^-1. Egg white, pH = 7.8 Write -7.8 = 0.2 - 8, so [ H^+] = 10^-7.8 = 10^0.2 × 10^-8 = 1.585 × 10^-8 mol L^-1 Rounded: 1.6 × 10^-8 mol L^-1. A step people get wrong here: the sign in [ H^+] = 10^ - pH is easy to drop, giving a huge concentration instead of a tiny one — a bigger pH number must always turn into a smaller [ H^+], since pH and [ H^+] move in opposite directions (milk, the least acidic of the five, correctly comes out with the smallest [ H^+], and lemon juice, the most acidic, with the largest). Answer: [H⁺] = 1.6 × 10⁻⁷ mol L⁻¹ (milk), 1.0 × 10⁻⁵ mol L⁻¹ (black coffee), 6.3 × 10⁻⁵ mol L⁻¹ (tomato juice), 6.3 × 10⁻³ mol L⁻¹ (lemon juice), 1.6 × 10⁻⁸ mol L⁻¹ (egg white)

Question 57

If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Accepted Answer

[K+] = [OH–] = 0.05M, [H+] = 2.0 × 10–13M. KOH is a strong base — it dissociates completely, so the concentration of K^+ and OH^- equals the concentration of KOH itself, not some equilibrium fraction of it. Step 1 — moles of KOH. Molar mass of KOH: M = 39 + 16 + 1 = 56 g mol^-1 (K = 39, O = 16, H = 1). Moles, n = ( mass )/( molar mass ), where mass is the mass of solute (0.561 g) and molar mass is 56 g mol⁻¹: n( KOH ) = (0.561 g )/(56 g mol^-1) = 1.002 × 10^-2 mol Step 2 — molar concentration of the KOH solution. Molarity c = n/V, where n is moles of solute and V is the volume of the solution in litres. Aside: the volume that goes in the denominator is the 200 mL of final solution stated in the problem — not the volume of water used to make it. Mixing up "volume of solvent" with "volume of solution" is the error that most often derails this kind of question. V = 200 mL = 0.200 L c( KOH ) = (1.002 × 10^-2 mol )/(0.200 L ) = 5.01 × 10^-2 mol L^-1 Step 3 — concentrations of K^+ and OH^-. KOH dissociates completely, one mole of solid giving one mole each of the ions: KOH → K^+ + OH^- Because the stoichiometry is 1: 1: 1 and dissociation is complete (strong base — no equilibrium arrow needed here), each ion's concentration equals the concentration of KOH found above: [ K^+] = [ OH^-] = 5.01 × 10^-2 mol L^-1 Step 4 — [ H^+] from the ionic product of water. Ionic product of water: K w = [ H^+][ OH^-], with K w = 1.0 × 10^-14 mol^2 L^-2 at 298 K — this holds in any aqueous solution, acidic or basic, at this temperature. [ H^+] = K w/([ OH^-]) = (1.0 × 10^-14)/(5.01 × 10^-2) = 2.00 × 10^-13 mol L^-1 Step 5 — pH. Definition: pH = - log 10[ H^+], where [ H^+] is in mol L⁻¹. pH = - log (2.00 × 10^-13) = 13 - log (2.00) = 13 - 0.30 = 12.70 Aside — a quick check on this route: it is easy to forget the minus sign and to trip over subtracting a negative power of ten; going the other way through pOH catches such slips. pOH = - log [ OH^-] = - log (5.01×10^-2) = 1.30, and since pH + pOH = 14.00 at 298 K, pH = 14.00 - 1.30 = 12.70 — the same value, confirming the result. Data are given to 3 significant figures (0.561 g), so the concentrations are reported to 3 significant figures and pH, being a logarithmic quantity, to 2 decimal places. Answer: [ K^+] = [ OH^-] = 5.01 × 10^-2 mol L^-1 (≈ 0.05 M); [ H^+] = 2.00 × 10^-13 mol L^-1; pH = 12.70

Question 58

The solubility of Sr(OH) 2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Accepted Answer

[Sr2+] = 0.1581M, [OH–] = 0.3162M, pH = 13.50. The solubility is given as grams of compound per litre of solution — divide by molar mass to get molarity, the same way you always convert g/L to mol/L. Step 1: Molar mass of Sr(OH) 2 Using atomic masses M Sr = 87.6 g/mol, M O = 16.0 g/mol, M H = 1.0 g/mol: M ( Sr(OH) 2 ) = 87.6 + 2(16.0 + 1.0) = 87.6 + 34.0 = 121.6 g/mol Step 2: Convert solubility to molarity Molarity c — moles of solute per litre of solution (the 19.23 g is already given per litre of solution, so no further correction for solvent volume is needed) — is c = ( mass of solute per litre of solution )/( molar mass ) = (19.23 g/L )/(121.6 g/mol ) = 0.1581 mol/L This c is the concentration of dissolved Sr(OH) 2 as a whole, before it splits into ions. Step 3: Dissociation gives the ion concentrations Sr(OH) 2 is a strong base, so it dissociates completely in water: Sr(OH) 2(aq) → Sr^2+(aq) + 2 OH^-(aq) One formula unit releases one Sr^2+ but two OH^- — missing that factor of 2 is the step most people get wrong here. [ Sr^2+ ] = c = 0.1581 mol/L ≈ 0.158 M [ OH^- ] = 2c = 2 × 0.1581 mol/L = 0.3163 mol/L ≈ 0.316 M Step 4: pH from [ OH^- ] The formula pOH = - log 10[ OH^- ] must use the hydroxide-ion concentration, not the concentration of dissolved Sr(OH) 2 — plugging in c instead of 2c is the other common slip. pOH = - log 10(0.3163) = 0.500 At 298 K, pH + pOH = 14, so pH = 14 - 0.500 = 13.50 Answer: [ Sr^2+ ] ≈ 0.158 M, [ OH^- ] ≈ 0.316 M, and pH = 13.5

Question 59

The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Accepted Answer

α = 1.63 × 10–2, pH = 3.09. In presence of 0.01M HCl, α = 1.32 × 10–3. The acid barely ionizes on its own, but adding H⁺ from a strong acid pushes the equilibrium back — that's Le Chatelier acting through a common ion. Step 1: Set up the ionization equilibrium for the pure acid. Propanoic acid ionizes as C 2 H 5 COOH C 2 H 5 COO^- + H^+ Let C = 0.05 M be the initial concentration and α the degree of ionization (the fraction of acid molecules that have ionized). At equilibrium: C 2 H 5 COOH C 2 H 5 COO^- H^+ --- --- --- --- Initial C 0 0 Equilibrium C(1-α) Cα Cα The ionization constant is K a = ([ C 2 H 5 COO^-][ H^+])/([ C 2 H 5 COOH ]) = (Cα · Cα)/(C(1-α)) = (Cα^2)/(1-α) Since K a is small (of order 10^-5 ), α 1, so 1-α ≈ 1. This is the approximation everyone reaches for — but it is only valid when the resulting α turns out to be small, so check it at the end. With that simplification, K a ≈ Cα^2 ⇒ α = √ K a/C Step 2: Substitute the numbers. α = √ (1.32×10^-5)/0.05 = √ 2.64×10^-4 = 1.625×10^-2 Check the approximation: α = 0.01625, which is far below 0.05 (the usual 5% cutoff), so treating 1-α ≈ 1 was justified. Step 3: Find [ H^+] and then pH. [ H^+] = Cα = 0.05 M × 1.625×10^-2 = 8.12×10^-4 M pH = - log [ H^+] = - log (8.12×10^-4) Splitting the log: - log (8.12×10^-4) = 4 - log (8.12) = 4 - 0.910 = 3.09 Step 4: Now add 0.01 M HCl — the common-ion effect. HCl is a strong acid, so it dissociates completely and dumps 0.01 M H^+ into the solution before the propanoic acid equilibrium is even considered. This extra H^+ is a common ion (propanoic acid also produces H^+ ), and by Le Chatelier's principle it suppresses the acid's own ionization — the acid's degree of ionization must now be smaller than 1.625×10^-2. Let α' be the new degree of ionization of propanoic acid (still starting from C = 0.05 M ). The equilibrium concentrations are now: C 2 H 5 COOH C 2 H 5 COO^- H^+ --- --- --- --- Initial 0.05 0 0.01 (from HCl) Equilibrium 0.05(1-α') 0.05 α' 0.01 + 0.05 α' K a = ((0.05 α')(0.01 + 0.05 α'))/(0.05(1-α')) The step people skip: you cannot drop the 0.05α' sitting next to the 0.01 until you've confirmed it really is negligible next to it — not just small in absolute terms. Since the common-ion effect makes α' very small here, expect 0.05 α' 0.01 and α' 1, and verify after solving. With 1-α' ≈ 1 and 0.01+0.05α' ≈ 0.01: K a ≈ ((0.05 α')(0.01))/0.05 = 0.01 α' α' = K a/0.01 = (1.32×10^-5)/(1×10^-2) = 1.32×10^-3 Step 5: Verify the approximation. 0.05 α' = 0.05 × 1.32×10^-3 = 6.6×10^-5, which is indeed negligible next to 0.01 (less than 1% of it), and α' = 1.32×10^-3 1. Both simplifications hold. Comparing the two results: α fell from 1.625×10^-2 (about 1.6%) to 1.32×10^-3 (about 0.13%) once the common H^+ from HCl was added — roughly a 12-fold suppression, exactly the direction Le Chatelier's principle predicts. Answer: Degree of ionization of 0.05 M propanoic acid α = 1.62×10^-2 (1.6%), with pH = 3.09. In the presence of 0.01 M HCl, the degree of ionization drops to α' = 1.32×10^-3 (0.13%).

Question 60

The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Accepted Answer

Ka = 2.09 × 10–4 and degree of ionization = 0.0457. A weak acid's K a is built from the concentrations left at equilibrium — and the acid that ionized is no longer acid. That single bookkeeping step is the whole question here. Cyanic acid is a weak monoprotic acid, so it ionizes only partly: HCNO H^+ + CNO^- Step 1 — get [ H^+ ] from the pH. The definition of pH is pH = - log 10[ H^+ ], where [ H^+ ] is the hydrogen-ion concentration in mol L ^-1. Inverting it: [ H^+ ] = 10^ - pH = 10^-2.34 = 4.571 × 10^-3 mol L^-1 Aside: pH is a logarithm, so you invert it with a power of ten, never by dividing. And 10^-2.34 is not 2.34 × 10^-3 — write it as 10^0.66 × 10^-3 = 4.571 × 10^-3. Step 2 — build the equilibrium table. Let C = 0.1 mol L ^-1 be the concentration of acid taken initially, and let x be the amount per litre that ionizes. HCNO H^+ CNO^- --- --- --- --- Initial C = 0.1 0 0 Change -x +x +x Equilibrium C - x x x Each molecule that ionizes gives one H^+ and one CNO^-, so x = [ H^+ ] = 4.571 × 10^-3 mol L ^-1. (The H^+ from water, about 10^-7 mol L ^-1, is smaller by a factor of 4×10^4 and is safely ignored.) Equilibrium concentrations: [ H^+ ] = [ CNO^- ] = 4.571 × 10^-3 mol L^-1 [ HCNO ] = C - x = 0.1 - 0.004571 = 0.09543 mol L^-1 Aside — this is the step people lose the mark on: the denominator is the acid remaining, C - x, not the 0.1 you started with. What ionized has left the HCNO column and moved into the ion columns. Step 3 — the ionization constant. K a = ([ H^+ ][ CNO^- ])/([ HCNO ]) = x^2/C-x Substituting: K a = ((4.571 × 10^-3)^2)/0.09543 = (2.0893 × 10^-5)/0.09543 = 2.189 × 10^-4 The pH is given to two decimal places, which fixes [ H^+ ] to about three significant figures; K a depends on the square of it, so quote three figures at most: K a ≈ 2.19 × 10^-4 Equivalently p K a = - log K a = 3.66. Aside: K a is dimensionless as it is written here, because the concentrations are ratios to the standard state of 1 mol L ^-1. You will also see it quoted with mol L ^-1; either convention is accepted, but do not mix them. Step 4 — the degree of ionization. The degree of ionization α is the fraction of the acid taken that has ionized: α = ( amount ionized per litre )/( amount taken per litre ) = x/C α = (4.571 × 10^-3)/0.1 = 4.57 × 10^-2 = 0.0457 That is 4.57 % of the cyanic acid ionized — about one molecule in twenty-two. Aside: α is a fraction, so it has no unit and can never exceed 1. If you get α 1, you have divided by x instead of by C. A note on the two values of K a you will see. NCERT prints K a = 2.09 × 10^-4 for this question. That comes from the standard weak-acid shortcut, in which x is dropped from the denominator because it is small compared with C: K a ≈ x^2/C = (2.0893 × 10^-5)/0.1 = 2.09 × 10^-4 This is a deliberate approximation, not a misprint, and it is the one usually allowed when α < 5 % — here α = 4.57 %, just inside that limit. But notice how thin the margin is: dropping x shifts K a by 4.8 %, which is larger than the uncertainty the given pH carries (about 2 % in K a ). Since the exact denominator costs one subtraction, use C - x and report 2.19 × 10^-4; if your answer key or examiner works with the approximate form, 2.09 × 10^-4 is the number they expect, and both rest on the same [ H^+ ]. The degree of ionization, 0.0457, is identical either way — it never depended on that choice. Answer: K a = 2.19 × 10^-4 (three significant figures; 2.09 × 10^-4 if the denominator is approximated as 0.1 M, as NCERT does), and the degree of ionization α = 4.57 × 10^-2, i.e. 4.57 %.

Question 61

The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Accepted Answer

pH = 7.97. Degree of hydrolysis = 2.36 × 10–5. Sodium nitrite is the salt of a weak acid ( HNO 2) and a strong base ( NaOH ), so the nitrite ion hydrolyses -- it pulls a proton from water and leaves OH^- behind, making the solution basic even though no base was added directly. The hydrolysis reaction is NO 2^- + H 2 O HNO 2 + OH^- Step 1 -- get the hydrolysis constant K b from the given K a. NO 2^- is the conjugate base of HNO 2, so its equilibrium constant is related to K a of the acid through K b = K w/K a where K w = 1.0×10^-14 is the ion product of water and K a = 4.5×10^-4 is the given ionization constant of nitrous acid. This is the step people skip -- you are handed K a of the acid, but the species actually reacting with water is the conjugate base, so you must convert to K b before setting up the equilibrium table. K b = (1.0×10^-14)/(4.5×10^-4) = 2.22×10^-11 Step 2 -- set up the equilibrium in terms of the degree of hydrolysis h. Let C = 0.04 M be the initial concentration of NO 2^-, and let h be the fraction of it that hydrolyses. At equilibrium: [ NO 2^-] = C(1-h), [ HNO 2] = Ch, [ OH^-] = Ch so K b = ([ HNO 2][ OH^-])/([ NO 2^-]) = (Ch· Ch)/(C(1-h)) = Ch^2/1-h Because K b is so small compared with C, h 1, and 1-h ≈ 1. The expression simplifies to K b ≈ Ch^2 ⇒ h = √ K b/C Step 3 -- substitute and solve for h. h = √ (2.22×10^-11)/0.04 = √ 5.56×10^-10 = 2.36×10^-5 This is the degree of hydrolysis: only about 2.36×10^-5 (i.e. 2.36×10^-3 % ) of the nitrite ions actually hydrolyse -- most of the NO 2^- stays untouched, which is exactly why the assumption 1-h≈1 was safe to make. Step 4 -- get [ OH^-], then pOH, then pH. [ OH^-] = Ch = 0.04 M × 2.36×10^-5 = 9.43×10^-7 M pOH = - log [ OH^-] = - log (9.43×10^-7) = 6.03 The last step to pH uses pH + pOH = 14 at 25^° C -- the mistake to avoid here is reporting the pOH as if it were the pH, which would wrongly suggest the solution is acidic when a nitrite solution must be basic. pH = 14 - pOH = 14 - 6.03 = 7.97 The value above 7 confirms the chemistry: a salt of a weak acid and a strong base gives a basic solution, and pH = 7.97 is exactly that -- mildly basic, as expected. Answer: pH ≈ 7.97; degree of hydrolysis h ≈ 2.36×10^-5 (about 2.36×10^-3 % ).

Question 62

A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Accepted Answer

Kb = 1.5 × 10–9. Pyridinium hydrochloride is a salt of a strong acid (HCl) and a weak base (pyridine), so in water it hands you the pyridinium ion, C 5H 5NH^+ — the conjugate acid of pyridine — as the species that is actually ionizing. The pH lets you find that acid's ionization constant first; getting to pyridine's own constant, K b, takes one more step. Step 1 — What's dissolved, and how much of it. C 5H 5NHCl is an ionic salt: it dissociates completely in water. C 5H 5NHCl(aq) → C 5H 5NH^+(aq) + Cl^-(aq) So a 0.02 M solution of the salt starts with [ C 5H 5NH^+ ] 0 = 0.02 mol L^-1. Chloride is just a spectator ion here. Step 2 — Turn the pH into [ H^+ ]. By definition pH = - log 10[ H^+ ], where [ H^+ ] is the equilibrium hydrogen-ion concentration in mol L ^-1. So [ H^+ ] = 10^ - pH = 10^-3.44 = 3.63×10^-4 mol L^-1 This is the easy step to fumble: the data hands you pH, not a concentration — convert before touching the equilibrium expression. Step 3 — Set up the equilibrium that is actually ionizing. The pyridinium ion is a weak acid; it loses a proton to give back pyridine: C 5H 5NH^+(aq) C 5H 5N(aq) + H^+(aq) Essentially every H^+ in solution comes from this ionization (water's own contribution is negligible next to 10^-4 mol L^-1 ), so at equilibrium [ H^+ ] = [ C 5H 5N ] = x = 3.63×10^-4 mol L^-1, [ C 5H 5NH^+ ] = 0.02 - x Step 4 — Write K a for the pyridinium ion and substitute. Naming the formula: K a = ([ C 5H 5N ][ H^+ ])/([ C 5H 5NH^+ ]) — the ratio of the ionized products to the un-ionized acid, all at equilibrium. K a = ((3.63×10^-4)(3.63×10^-4))/(0.02 - 3.63×10^-4) = (1.318×10^-7)/0.01964 mol L^-1 = 6.71×10^-6 mol L^-1 Notice the denominator keeps the " -x " instead of rounding it away: x is about 1.8% of 0.02, small but not free to drop, since the extra arithmetic costs nothing here. Step 5 — Get from the acid's K a to the base's K b. C 5H 5NH^+ is the conjugate acid of pyridine, and any conjugate acid–base pair is tied together at a given temperature by K a × K b = K w = 1.0×10^-14 mol^2 L^-2 ( at 298 K ) This is the step the question is really testing: the pH could only ever measure the acid form's ionization, but the question asks for the base's constant — so you must divide by K a, not report it as if it were already K b. K b = K w/K a = (1.0×10^-14 mol^2 L^-2)/(6.71×10^-6 mol L^-1) = 1.49×10^-9 mol L^-1 Rounding once, to the two significant figures that the pH (given to two decimal places) and the 0.02 M concentration support: K b ≈ 1.5×10^-9 mol L^-1 Answer: K a( C 5H 5NH^+ ) ≈ 6.7×10^-6 mol L^-1, so the ionization constant of pyridine is K b ≈ 1.5×10^-9 mol L^-1.

Question 63

Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH 4NO 3, NaNO 2 and KF

Accepted Answer

NaCl, KBr solutions are neutral, NaCN, NaNO2 and KF solutions are basic and NH4NO3 solution is acidic.. A salt is only as "neutral" as the acid and base that made it — an ion born from a WEAK parent acid or base reacts with water (hydrolyses) and pulls the solution's pH away from 7; an ion born from a STRONG parent does not react with water at all. Every one of these salts breaks apart completely in water into its cation and anion. The question is whether either ion then reacts with water. The rule that decides this: - An anion is the conjugate base of some acid HA. If that acid was strong (fully ionised, like HCl, HBr, HNO 3 ), its conjugate base is too weak to take a proton back from water — no reaction, solution stays neutral from that ion. - If the parent acid was weak (like HCN, HNO 2, HF ), the conjugate base is strong enough to pull a proton off water, releasing OH^- and making the solution basic. - Symmetrically, a cation that is the conjugate acid of a weak base (like NH 4^+, conjugate acid of NH 3/ NH 4 OH ) donates a proton to water, releasing H 3 O^+ and making the solution acidic. A cation from a strong base (like Na^+, K^+, from NaOH, KOH ) does not hydrolyse. This is the same equilibrium chapter's relation K w = K a × K b at work: for a conjugate acid–base pair, the weaker one is fixed by how strong the other is, and it is the hydrolysis constant of the surviving weak partner ( K b = K w/K a for a weak acid's conjugate base, or K a = K w/K b for a weak base's conjugate acid) that measures how far the ion pushes the pH off 7. The trap: it is easy to mis-slot NaNO 2 with NaNO 3 -type salts because the formulas look alike. HNO 3 (nitric acid) is strong, but HNO 2 (nitrous acid) is weak — one extra oxygen is the whole difference between "neutral salt" and "basic salt." Now take each salt by identifying its parent acid and parent base: NaCl: parent base NaOH (strong), parent acid HCl (strong). Neither Na^+ nor Cl^- hydrolyses. → neutral KBr: parent base KOH (strong), parent acid HBr (strong). Neither K^+ nor Br^- hydrolyses. → neutral NaCN: parent base NaOH (strong), parent acid HCN (weak). Na^+ is inert, but CN^- hydrolyses: CN^-(aq) + H 2 O (l) HCN (aq) + OH^-(aq) This releases OH^-. → basic NH 4 NO 3: parent base NH 4 OH (weak), parent acid HNO 3 (strong). NO 3^- is inert, but NH 4^+ hydrolyses: NH 4^+(aq) + H 2 O (l) NH 4 OH (aq) + H^+(aq) This releases H^+. → acidic NaNO 2: parent base NaOH (strong), parent acid HNO 2 (weak — this is the one that looks like it should pair with strong-acid behaviour but does not). Na^+ is inert, but NO 2^- hydrolyses: NO 2^-(aq) + H 2 O (l) HNO 2(aq) + OH^-(aq) → basic KF: parent base KOH (strong), parent acid HF (weak). K^+ is inert, but F^- hydrolyses: F^-(aq) + H 2 O (l) HF (aq) + OH^-(aq) → basic Answer: NaCl and KBr are neutral (strong acid + strong base); NH₄NO₃ is acidic (NH₄⁺ hydrolyses, weak base + strong acid); NaCN, NaNO₂, and KF are all basic (CN⁻, NO₂⁻, F⁻ each hydrolyse — weak acid + strong base).

Question 64

The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

Accepted Answer

(a) pH of acid solution= 1.9 (b) pH of its salt solution= 7.9. A weak acid's own ionization can be too large to ignore -- when that happens, the shortcut [ H^+ ] = √ K a C is wrong, and you must solve the exact quadratic instead. Part 1 -- pH of 0.1 M chloroacetic acid Chloroacetic acid ( ClCH 2COOH, write it as HA ) ionizes as HA H^+ + A^- Let x = [ H^+ ] = [ A^- ] at equilibrium, starting from C = 0.1 M of HA, so [ HA ] = C - x. The ionization constant K a (the equilibrium constant for this proton-loss step) is K a = ([ H^+ ][ A^- ])/([ HA ]) = x^2/C-x = 1.35×10^-3 Check before approximating: the usual shortcut drops the x in C-x, valid only when x C. Here C/K a = 0.1/(1.35×10^-3) ≈ 74, well below the 400 benchmark that makes the shortcut safe -- so x must be kept, and the quadratic solved in full. Skipping this check is exactly where this problem goes wrong. Rearranging: x^2 + K a x - K a C = 0 x^2 + (1.35×10^-3)x - (1.35×10^-4) = 0 By the quadratic formula, x = (-K a + √ K a^2 + 4K aC )/2: x = -1.35×10^-3 + √ (1.35×10^-3)^2 + 4(1.35×10^-3)(0.1) 2 x = -1.35×10^-3 + √ 1.8225×10^-6 + 5.4×10^-4 2 = -1.35×10^-3 + √ 5.4182×10^-4 2 x = (-1.35×10^-3 + 2.3277×10^-2)/2 = 1.096×10^-2 M So [ H^+ ] = 1.096×10^-2 M (a check: x^2/(C-x) = 1.35×10^-3, which reproduces K a — the fraction ionized here is nearly 11%, confirming the shortcut would have understated it). pH = - log [ H^+ ] = - log (1.096×10^-2) = 1.96 Part 2 -- pH of 0.1 M sodium chloroacetate solution Sodium chloroacetate is the salt of a weak acid ( HA ) and a strong base ( NaOH ). In solution it is fully dissociated into Na^+ and A^-; Na^+ is a spectator, but A^- is the conjugate base of a weak acid and is itself weakly basic -- it pulls a proton back off water. This is the step people skip, assuming a "salt" must give pH 7: A^- + H 2O HA + OH^- The equilibrium constant for this hydrolysis is the base constant K b of A^-, linked to K a of the parent acid through the water constant K w = [ H^+ ][ OH^- ] = 1×10^-14 (at 298 K): K b = K w/K a = (1×10^-14)/(1.35×10^-3) = 7.407×10^-12 Let y = [ OH^- ] = [ HA ] formed, with initial [ A^- ] = C = 0.1 M: K b = ([ HA ][ OH^- ])/([ A^- ]) = y^2/C-y Because K b is tiny here (unlike K a in Part 1), y will be minute next to C, so C - y ≈ C is safe this time -- the same shortcut that failed above is fine here because the size of the equilibrium constant relative to C is what decides it, not a blanket rule: y^2 ≈ K b C = (7.407×10^-12)(0.1) = 7.407×10^-13 y = √ 7.407×10^-13 = 8.607×10^-7 M (Indeed y/C ≈ 8.6×10^-6, utterly negligible -- the approximation is justified after the fact.) pOH = - log [ OH^- ] = - log (8.607×10^-7) = 6.07 Using pH + pOH = 14 (from K w at 298 K): pH = 14 - 6.07 = 7.93 The salt solution is mildly basic (pH above 7), exactly because the anion of a weak acid is a weak base -- not neutral, even though it is "just a salt." Answer: pH of 0.1 M chloroacetic acid ≈ 1.96 (acidic); pH of 0.1 M sodium chloroacetate solution ≈ 7.93 (mildly basic).

	\(\displaystyle \mathrm{N_2}\)	\(\displaystyle \mathrm{O_2}\)	\(\displaystyle \mathrm{N_2O}\)
Initial	\(\displaystyle 0.0482\)	\(\displaystyle 0.0933\)	\(\displaystyle 0\)
Change	\(\displaystyle -2x\)	\(\displaystyle -x\)	\(\displaystyle +2x\)
Equilibrium	\(\displaystyle 0.0482-2x\)	\(\displaystyle 0.0933-x\)	\(\displaystyle 2x\)

	\(\displaystyle \text{HI(g)} \)	\(\displaystyle \text{H}_2\text{(g)} \)	\(\displaystyle \text{I}_2\text{(g)} \)
Initial	\(\displaystyle 0.2\ \text{atm} \)	\(\displaystyle 0 \)	\(\displaystyle 0 \)
Change	\(\displaystyle -p \)	\(\displaystyle +p/2 \)	\(\displaystyle +p/2 \)
Equilibrium	\(\displaystyle 0.2-p \)	\(\displaystyle p/2 \)	\(\displaystyle p/2 \)

	\(\displaystyle \mathrm{H_{2}O}\)	CO	\(\displaystyle \mathrm{H_{2}}\)	\(\displaystyle \mathrm{CO_{2}}\)
Initial	$\displaystyle 1$	$\displaystyle 1$	$\displaystyle 0$	$\displaystyle 0$
Change	-$\displaystyle 0.4$	-$\displaystyle 0.4$	+$\displaystyle 0.4$	+$\displaystyle 0.4$
Equilibrium	$\displaystyle 0.6$	$\displaystyle 0.6$	$\displaystyle 0.4$	$\displaystyle 0.4$

	\(\displaystyle \text{H}_2\)	\(\displaystyle \text{I}_2\)
At equilibrium	\(\displaystyle x\)	\(\displaystyle x\)

	\(\displaystyle 2\text{ICl}(g)\)	\(\displaystyle \text{I}_2(g)\)	\(\displaystyle \text{Cl}_2(g)\)
Initial (M)	$\displaystyle 0.78$	$\displaystyle 0$	$\displaystyle 0$
Change (M)	\(\displaystyle -2x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium (M)	\(\displaystyle 0.78-2x\)	\(\displaystyle x\)	\(\displaystyle x\)

	\(\displaystyle \mathrm{C_2H_6}\)	\(\displaystyle \mathrm{C_2H_4}\)	\(\displaystyle \mathrm{H_2}\)
Initial (atm)	\(\displaystyle 4.0\)	\(\displaystyle 0\)	\(\displaystyle 0\)
Change (atm)	\(\displaystyle -x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium (atm)	\(\displaystyle 4.0-x\)	\(\displaystyle x\)	\(\displaystyle x\)

	\(\displaystyle \text{PCl}_5\)	\(\displaystyle \text{PCl}_3\)	\(\displaystyle \text{Cl}_2\)
Initial	\(\displaystyle c_0\)	$\displaystyle 0$	$\displaystyle 0$
Change	\(\displaystyle -x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium	\(\displaystyle c_0-x\)	\(\displaystyle x\)	\(\displaystyle x\)

	\(\displaystyle p_{\text{CO}}\)	\(\displaystyle p_{\text{CO}_2}\)
Initial	\(\displaystyle 1.4\)	\(\displaystyle 0.80\)
Change	\(\displaystyle +x\)	\(\displaystyle -x\)
Equilibrium	\(\displaystyle 1.4+x\)	\(\displaystyle 0.80-x\)

	HBr	H₂	Br₂
Initial (bar)	$\displaystyle 10.0$	$\displaystyle 0$	$\displaystyle 0$
Change (bar)	\(\displaystyle -2x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium (bar)	\(\displaystyle 10.0-2x\)	\(\displaystyle x\)	\(\displaystyle x\)

	\(\displaystyle \mathrm{CO}\)	\(\displaystyle \mathrm{H_2O}\)	\(\displaystyle \mathrm{CO_2}\)	\(\displaystyle \mathrm{H_2}\)
Initial / bar	\(\displaystyle 4.0\)	\(\displaystyle 4.0\)	\(\displaystyle 0\)	\(\displaystyle 0\)
Change / bar	\(\displaystyle -x\)	\(\displaystyle -x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium / bar	\(\displaystyle 4.0-x\)	\(\displaystyle 4.0-x\)	\(\displaystyle x\)	\(\displaystyle x\)

	CH₃COOH	CH₃COO⁻	H⁺
Initial	\(\displaystyle C\)	$\displaystyle 0$	$\displaystyle 0$
Equilibrium	\(\displaystyle C(1-\alpha)\)	\(\displaystyle C\alpha\)	\(\displaystyle C\alpha\)

	HA	\(\displaystyle \text{H}^+\)	\(\displaystyle \text{A}^-\)
Initial	\(\displaystyle C\)	$\displaystyle 0$	$\displaystyle 0$
Change	\(\displaystyle -x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium	\(\displaystyle C-x\)	\(\displaystyle x\)	\(\displaystyle x\)

	B	\(\displaystyle \text{BH}^+\)	\(\displaystyle \text{OH}^-\)
initial	\(\displaystyle C\)	$\displaystyle 0$	$\displaystyle 0$
at equilibrium	\(\displaystyle C-x\)	\(\displaystyle x\)	\(\displaystyle x\)

	\(\displaystyle \text{C}_2\text{H}_5\text{COOH}\)	\(\displaystyle \text{C}_2\text{H}_5\text{COO}^-\)	\(\displaystyle \text{H}^+\)
Initial	\(\displaystyle C\)	$\displaystyle 0$	$\displaystyle 0$
Equilibrium	\(\displaystyle C(1-\alpha)\)	\(\displaystyle C\alpha\)	\(\displaystyle C\alpha\)

	\(\displaystyle \mathrm{HCNO}\)	\(\displaystyle \mathrm{H^+}\)	\(\displaystyle \mathrm{CNO^-}\)
Initial	\(\displaystyle C = 0.1\)	\(\displaystyle 0\)	\(\displaystyle 0\)
Change	\(\displaystyle -x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium	\(\displaystyle C - x\)	\(\displaystyle x\)	\(\displaystyle x\)

Exercises 6.1–6.10 (part 1 of 7)

Exercise 6.1

Exercise 6.2

Exercise 6.3

Exercise 6.4

Exercise 6.5

Exercise 6.6

Exercise 6.7

Exercise 6.8

Exercise 6.9

Exercise 6.10