NCERT Solutions · Class 11 Chemistry Equilibrium

The unknown \(\displaystyle x\) here is how much pressure reacts away, and \(\displaystyle K_p\) is built from what is LEFT at equilibrium — so the last step is to report \(\displaystyle x\) itself, not the leftover \(\displaystyle 4.0-x\). That single slip is the whole trap in this question.The reaction and what it does to pressuresThe water-gas shift reaction is\[\mathrm{CO}(g) + \mathrm{H_2O}(g) \rightleftharpoons \mathrm{CO_2}(g) + \mathrm{H_2}(g) \]Every mole of CO that reacts consumes one mole of H\(\displaystyle _2\)O and makes one mole of CO\(\displaystyle _2\) and one mole of H\(\displaystyle _2\). At constant temperature and volume, partial pressure is directly proportional to moles (\(\displaystyle p_i = n_iRT/V\)), so the same $\displaystyle 1$ : $\displaystyle 1$ : $\displaystyle 1$ : $\displaystyle 1$ 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 ($\displaystyle 2$ on the left, $\displaystyle 2$ on the right). That is why nothing extra happens to the total pressure, and why \(\displaystyle K_p\) here is dimensionless in the usual sense: the bar units cancel top and bottom.Set up the ICE table in barLet \(\displaystyle x\) = 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\(\displaystyle _2\) and H\(\displaystyle _2\) are zero.

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

The question asks for the partial pressure of H\(\displaystyle _2\) at equilibrium, and that is \(\displaystyle x\) — the row of the table headed "Equilibrium", column H\(\displaystyle _2\). Keep hold of that; it is exactly the thing that gets mixed up at the end.Write \(\displaystyle K_p\) and substituteFor this reaction the equilibrium constant in terms of partial pressures is\[K_p = \frac{p_{\mathrm{CO_2}}\, p_{\mathrm{H_2}}}{p_{\mathrm{CO}}\, p_{\mathrm{H_2O}}} \]where each \(\displaystyle p_i\) 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 $\displaystyle 1$ here).Substituting the equilibrium row, with \(\displaystyle K_p = 10.1\) at \(\displaystyle 400\,^\circ\mathrm{C}\):\[10.1 = \frac{(x)(x)}{(4.0-x)(4.0-x)} = \frac{x^{2}}{(4.0-x)^{2}} \]Solve — take the square root, don't expandBoth sides are perfect squares, so instead of multiplying out into a quadratic, take the positive square root of both sides (pressures are positive, and \(\displaystyle x < 4.0\) because you cannot consume more CO than you started with, so \(\displaystyle 4.0-x > 0\)):\[\frac{x}{4.0-x} = \sqrt{10.1} = 3.17805 \]Cross-multiplying:\[x = 3.17805\,(4.0 - x) = 12.71220 - 3.17805\,x \]\[x + 3.17805\,x = 12.71220 \]\[4.17805\,x = 12.71220 \]\[x = \frac{12.71220}{4.17805} = 3.04262\ \text{bar} \]Carrying the extra digits through and rounding only now: the data are \(\displaystyle K_p = 10.1\) (three significant figures) and \(\displaystyle 4.0\) bar, so quoting to three significant figures — two decimal places in bar — gives\[p_{\mathrm{H_2}} = 3.04\ \text{bar} \]Check by putting it back inEquilibrium pressures: \(\displaystyle p_{\mathrm{H_2}} = p_{\mathrm{CO_2}} = 3.0426\) bar, and \(\displaystyle p_{\mathrm{CO}} = p_{\mathrm{H_2O}} = 4.0 - 3.0426 = 0.9574\) bar. Then\[K_p = \frac{(3.0426)(3.0426)}{(0.9574)(0.9574)} = \frac{9.2574}{0.9166} = 10.10 \]which reproduces the given \(\displaystyle K_p = 10.1\). The answer is consistent.A sanity check on the size, too: \(\displaystyle K_p = 10.1\) is greater than $\displaystyle 1$, so at equilibrium the products are favoured over the reactants. The H\(\displaystyle _2\) pressure must therefore come out larger than the leftover CO pressure. \(\displaystyle 3.04 > 0.96\), as required. Had you obtained \(\displaystyle 0.96\) bar for H\(\displaystyle _2\) with \(\displaystyle 3.04\) bar of CO left over, that would mean reactants dominating, i.e. \(\displaystyle K_p < 1\) — which contradicts the value the question hands you.A note on the printed key. NCERT's answer key gives "\(\displaystyle [\mathrm{H_2}]_{eq} = 0.96\) bar" for this question, and that is a slip in the key, not a different convention or a rounding difference. \(\displaystyle 0.96\) bar is precisely \(\displaystyle 4.0 - 3.04\), the unreacted CO (and steam) pressure — the \(\displaystyle 4.0-x\) entry of the table rather than the \(\displaystyle x\) entry the question asks for. You can see it must be wrong without redoing the algebra: substituting \(\displaystyle p_{\mathrm{H_2}} = p_{\mathrm{CO_2}} = 0.96\) bar and \(\displaystyle p_{\mathrm{CO}} = p_{\mathrm{H_2O}} = 3.04\) bar into the \(\displaystyle K_p\) expression gives \(\displaystyle (0.96)^2/(3.04)^2 = 0.099\), which is \(\displaystyle 1/10.1\), not \(\displaystyle 10.1\). The correct partial pressure of dihydrogen is \(\displaystyle 3.04\) bar.NCERT's answer key prints "[H₂]eq = $\displaystyle 0.96$ bar", and that is the wrong species. $\displaystyle 0.96$ bar is \(\displaystyle 4.0 - 3.04\), 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 \(\displaystyle p_{\mathrm{H_2}} = 3.04\), \(\displaystyle K_p = 3.04^2/0.96^2 = 10.1\) ✓, whereas \(\displaystyle p_{\mathrm{H_2}} = 0.96\) gives \(\displaystyle 0.96^2/3.04^2 = 0.099 = 1/10.1\) — the constant for the reverse reaction. And \(\displaystyle K_p = 10.1 > 1\) means products predominate, so \(\displaystyle p_{\mathrm{H_2}}\) has to be the larger value. Answer: \(\displaystyle p_{\mathrm{H_2}} = 3.04\) bar (with \(\displaystyle p_{\mathrm{CO_2}} = 3.04\) bar and \(\displaystyle p_{\mathrm{CO}} = p_{\mathrm{H_2O}} = 0.96\) bar left unreacted); NCERT's printed $\displaystyle 0.96$ bar is the leftover CO/steam pressure, not the H\(\displaystyle _2\) pressure.

A reaction's equilibrium constant \(\displaystyle K_c\) tells you where the balance point sits — how far the reaction runs before the forward and reverse rates match. A huge \(\displaystyle K_c\) means equilibrium lies almost entirely with products; a tiny \(\displaystyle K_c\) means it lies almost entirely with reactants; only when \(\displaystyle K_c\) is close to $\displaystyle 1$ do both sides hold measurable amounts at the same time.The rule of thumb, and why it works: at equilibrium, \[K_c = \frac{[\text{products}]^{\text{stoich. powers}}}{[\text{reactants}]^{\text{stoich. powers}}} \] 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 $\displaystyle 1$ 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\[K_c > 10^{3}: \text{reaction goes almost to completion (products only)} \] \[K_c < 10^{-3}: \text{reaction barely proceeds (reactants only)} \] \[10^{-3} < K_c < 10^{3}: \text{appreciable amounts of both} \]This is where people go wrong: a "large" or "small" \(\displaystyle K_c\) is judged against $\displaystyle 1$, not against zero — \(\displaystyle K_c = 1.8\) is not small just because it's a single-digit number; it is close to $\displaystyle 1$, which is exactly the regime where both reactants and products coexist.Now check each equilibrium (note the reversible arrow \(\displaystyle \rightleftharpoons\) in every case — this is an equilibrium, so both directions are always happening):(a) \(\displaystyle \mathrm{Cl_2(g) \rightleftharpoons 2Cl(g)}\), \(\displaystyle K_c = 5\times10^{-39}\)This is far below \(\displaystyle 10^{-3}\) — vanishingly small. The numerator \(\displaystyle [\mathrm{Cl}]^2\) must be minuscule compared to \(\displaystyle [\mathrm{Cl_2}]\) for the ratio to come out this tiny. Essentially none of the \(\displaystyle \mathrm{Cl_2}\) dissociates; at equilibrium there is reactant, but no appreciable product.(b) \(\displaystyle \mathrm{Cl_2(g) + 2NO(g) \rightleftharpoons 2NOCl(g)}\), \(\displaystyle K_c = 3.7\times10^{8}\)This is far above \(\displaystyle 10^{3}\) — enormous. Here the numerator \(\displaystyle [\mathrm{NOCl}]^2\) must swamp the denominator, so the reaction runs almost to completion. At equilibrium there is product, but no appreciable reactant left.(c) \(\displaystyle \mathrm{Cl_2(g) + 2NO_2(g) \rightleftharpoons 2NO_2Cl(g)}\), \(\displaystyle K_c = 1.8\)This sits squarely between \(\displaystyle 10^{-3}\) and \(\displaystyle 10^{3}\), close to 1. Neither the numerator nor the denominator is forced toward zero — the concentrations of \(\displaystyle \mathrm{Cl_2}\), \(\displaystyle \mathrm{NO_2}\), and \(\displaystyle \mathrm{NO_2Cl}\) can all be comparable in size at equilibrium. This is the case with genuinely appreciable amounts of both reactants and products.**Answer: Reaction (c), \(\displaystyle \mathrm{Cl_2(g) + 2NO_2(g) \rightleftharpoons 2NO_2Cl(g)}\) with \(\displaystyle K_c = 1.8\), is the one with appreciable concentrations of both reactants and products at equilibrium — reaction (a) lies almost entirely toward reactants (\(\displaystyle K_c = 5\times10^{-39}\ll1\)) and reaction (b) lies almost entirely toward products (\(\displaystyle K_c = 3.7\times10^{8}\gg1\)).

The exponents in \(\displaystyle K_c\) are the stoichiometric coefficients — \(\displaystyle \text{O}_3\) is squared because it has coefficient $\displaystyle 2$, and \(\displaystyle \text{O}_2\) is cubed because it has coefficient 3. Mixing those two up is the error to watch for here.The equilibrium is\[3\text{O}_2(g) \rightleftharpoons 2\text{O}_3(g) \]so the equilibrium constant expression is\[K_c = \frac{[\text{O}_3]^2}{[\text{O}_2]^3} \]where \(\displaystyle [\text{O}_3]\) and \(\displaystyle [\text{O}_2]\) are the equilibrium molar concentrations (mol L\(\displaystyle ^{-1}\)) of ozone and oxygen.Given values\[K_c = 2.0 \times 10^{-50}, \qquad [\text{O}_2] = 1.6 \times 10^{-2}\ \text{mol L}^{-1} \]Step $\displaystyle 1$ — cube the oxygen concentration.Cubing means cubing the whole number, the coefficient and the power of ten together — cube only the \(\displaystyle 10^{-2}\) and forget the \(\displaystyle 1.6\), and the answer comes out wrong by a factor of \(\displaystyle 4.096\).\[[\text{O}_2]^3 = (1.6 \times 10^{-2})^3 = (1.6)^3 \times 10^{-6} = 4.096 \times 10^{-6}\ \text{mol}^3\text{L}^{-3} \]Step $\displaystyle 2$ — solve the \(\displaystyle K_c\) expression for \(\displaystyle [\text{O}_3]^2\).\[[\text{O}_3]^2 = K_c \times [\text{O}_2]^3 = (2.0 \times 10^{-50}) \times (4.096 \times 10^{-6}) \]\[[\text{O}_3]^2 = (2.0 \times 4.096) \times 10^{-50-6} = 8.192 \times 10^{-56}\ \text{mol}^2\text{L}^{-2} \]Step $\displaystyle 3$ — take the square root to get \(\displaystyle [\text{O}_3]\).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:\[[\text{O}_3] = \sqrt{8.192 \times 10^{-56}} = \sqrt{8.192} \times 10^{-28}\ \text{mol L}^{-1} \]\[\sqrt{8.192} = 2.862\ldots \]\[[\text{O}_3] = 2.862 \times 10^{-28}\ \text{mol L}^{-1} \]Both given data points (\(\displaystyle K_c\) and \(\displaystyle [\text{O}_2]\)) carry two significant figures, so the final concentration is rounded to two significant figures:\[[\text{O}_3] \approx 2.9 \times 10^{-28}\ \text{mol L}^{-1} \]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: \(\displaystyle [\text{O}_3] \approx 2.9 \times 10^{-28}\ \text{mol L}^{-1}\) (i.e. \(\displaystyle 2.86 \times 10^{-28}\ \text{mol L}^{-1}\) before rounding).

At equilibrium, the law of mass action fixes one ratio of concentrations — \(\displaystyle \mathrm{K_c^{-}}\) 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:\[\text{CO(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)} \]Because the flask has a volume of exactly $\displaystyle 1$ 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 \(\displaystyle \text{mol L}^{-1}\):\[[\text{CO}] = 0.30\ \text{mol L}^{-1}, \quad [\text{H}_2] = 0.10\ \text{mol L}^{-1}, \quad [\text{H}_2\text{O}] = 0.02\ \text{mol L}^{-1} \]and \(\displaystyle [\text{CH}_4]\) is the unknown to find.Writing the equilibrium constant expression. For this reaction,\[K_c = \dfrac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3} \]Here \(\displaystyle K_c\) 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 — \(\displaystyle \text{H}_2\) is cubed because its coefficient is 3. This is the step people rush: forgetting to cube \(\displaystyle [\text{H}_2]\) (treating it as if its coefficient were $\displaystyle 1$) throws the answer off by a factor of \(\displaystyle (0.10)^2 = 0.01\), a hundredfold error.Solving for \(\displaystyle [\text{CH}_4]\). Rearranging for the one unknown,\[[\text{CH}_4] = \dfrac{K_c \, [\text{CO}][\text{H}_2]^3}{[\text{H}_2\text{O}]} \]Substituting the given values, with \(\displaystyle K_c = 3.90\):\[[\text{CH}_4] = \dfrac{3.90 \times (0.30\ \text{mol L}^{-1}) \times (0.10\ \text{mol L}^{-1})^3}{0.02\ \text{mol L}^{-1}} \]Work the numerator first. \(\displaystyle (0.10\ \text{mol L}^{-1})^3 = 1.0\times10^{-3}\ \text{mol}^3\text{L}^{-3}\), so\[3.90 \times 0.30\ \text{mol L}^{-1} \times 1.0\times10^{-3}\ \text{mol}^3\text{L}^{-3} = 1.17\times10^{-3}\ \text{mol}^4\text{L}^{-4} \]Now divide by \(\displaystyle [\text{H}_2\text{O}] = 0.02\ \text{mol L}^{-1}\):\[[\text{CH}_4] = \dfrac{1.17\times10^{-3}\ \text{mol}^4\text{L}^{-4}}{0.02\ \text{mol L}^{-1}} = 0.0585\ \text{mol L}^{-1} \]The units divide cleanly to leave \(\displaystyle \text{mol}^3\text{L}^{-3}\)... but check the bookkeeping: \(\displaystyle K_c\) itself carries whatever units make the overall expression consistent, and since every concentration here is expressed in \(\displaystyle \text{mol L}^{-1}\), the arithmetic on the numbers alone correctly returns \(\displaystyle [\text{CH}_4]\) in \(\displaystyle \text{mol L}^{-1}\) — this is the usual convention for concentration equilibrium constants, so track the number and attach \(\displaystyle \text{mol L}^{-1}\) to the final result.The data given ($\displaystyle 0.30$, $\displaystyle 0.10$, $\displaystyle 0.02$, $\displaystyle 3.90$) each carry two or three significant figures, so the result is kept to three significant figures.Answer: \(\displaystyle [\text{CH}_4] = 0.0585\ \text{mol L}^{-1}\)

A conjugate acid–base pair is two species that differ from each other by exactly one proton, \(\displaystyle \mathrm{H^+}\) — nothing else changes.When an acid \(\displaystyle \mathrm{HA}\) gives away a proton, \[\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-} \] what is left, \(\displaystyle \mathrm{A^-}\), is called the conjugate base of \(\displaystyle \mathrm{HA}\). Read the same equilibrium the other way and \(\displaystyle \mathrm{A^-}\) is a base that, on accepting a proton, turns back into \(\displaystyle \mathrm{HA}\) — its conjugate acid. So \(\displaystyle \mathrm{HA}\) and \(\displaystyle \mathrm{A^-}\) 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 \(\displaystyle \mathrm{H^+}\) — 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 \(\displaystyle \mathrm{H^+}\) to get the conjugate base: Species with no proton to spare — add \(\displaystyle \mathrm{H^+}\) to get the conjugate acid: Answer: A conjugate acid–base pair is a pair of species differing by exactly one \(\displaystyle \mathrm{H^+}\). Conjugate base of \(\displaystyle \mathrm{HNO_2}\) is \(\displaystyle \mathrm{NO_2^-}\); of \(\displaystyle \mathrm{HClO_4}\) is \(\displaystyle \mathrm{ClO_4^-}\). Conjugate acid of \(\displaystyle \mathrm{CN^-}\) is \(\displaystyle \mathrm{HCN}\); of \(\displaystyle \mathrm{F^-}\) is \(\displaystyle \mathrm{HF}\); of \(\displaystyle \mathrm{OH^-}\) is \(\displaystyle \mathrm{H_2O}\); of \(\displaystyle \mathrm{CO_3^{2-}}\) is \(\displaystyle \mathrm{HCO_3^-}\); of \(\displaystyle \mathrm{S^{2-}}\) is \(\displaystyle \mathrm{HS^-}\).

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: This is exactly the criterion the chapter itself uses: electron-deficient species such as \(\displaystyle \mathrm{AlCl_3}\), \(\displaystyle \mathrm{Co^{3+}}\), \(\displaystyle \mathrm{Mg^{2+}}\) act as Lewis acids, while \(\displaystyle \mathrm{H_2O}\), \(\displaystyle \mathrm{NH_3}\), \(\displaystyle \mathrm{OH^-}\) 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?\(\displaystyle \mathrm{H_2O}\). Oxygen has $\displaystyle 6$ valence electrons; two go into the two \(\displaystyle \mathrm{O\!-\!H}\) bonds, leaving two lone pairs. Count around O: \(\displaystyle 2 \times 2\) (bonding) \(\displaystyle +\;2 \times 2\) (lone pairs) \(\displaystyle =\) $\displaystyle 8$ 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 \(\displaystyle \mathrm{H^+}\) to make \(\displaystyle \mathrm{H_3O^+}\)). Not a Lewis acid.\(\displaystyle \mathrm{BF_3}\). Boron has only $\displaystyle 3$ valence electrons, all used in three \(\displaystyle \mathrm{B\!-\!F}\) bonds. Count around B: \(\displaystyle 3 \times 2 = 6\) electrons — two short of an octet — and the \(\displaystyle 2p_z\) orbital is completely empty. That empty \(\displaystyle 2p\) orbital is the vacancy, so \(\displaystyle \mathrm{BF_3}\) takes a lone pair from ammonia: \[\mathrm{BF_3} + \mathrm{:\!NH_3} \longrightarrow \mathrm{F_3B\!\leftarrow\!NH_3}\] \(\displaystyle \mathrm{BF_3}\) is a Lewis acid.\(\displaystyle \mathrm{H^+}\). A bare proton has zero electrons and an entirely empty \(\displaystyle 1s\) orbital. Nothing is more electron-hungry than that; it is the textbook Lewis acid, accepting a pair from \(\displaystyle \mathrm{OH^-}\), \(\displaystyle \mathrm{F^-}\), \(\displaystyle \mathrm{H_2O}\) and so on. A Lewis acid.\(\displaystyle \mathrm{NH_4^+}\). This is the one to be careful with. Nitrogen is \(\displaystyle sp^3\) hybridised and all four hybrid orbitals are used in four \(\displaystyle \mathrm{N\!-\!H}\) bonds. Count around N: \(\displaystyle 4 \times 2 = 8\) electrons — a complete octet — with no lone pair left and no empty orbital left. Nitrogen sits in Period $\displaystyle 2$, so it has no \(\displaystyle d\) orbitals and cannot expand its octet to ten electrons; there is no such thing as \(\displaystyle \mathrm{NH_5}\). 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 \(\displaystyle +\) charge and concluding "cation, therefore Lewis acid". That rule of thumb works for \(\displaystyle \mathrm{Mg^{2+}}\), \(\displaystyle \mathrm{Co^{3+}}\), \(\displaystyle \mathrm{Ag^+}\) because those ions really do have empty valence orbitals. \(\displaystyle \mathrm{NH_4^+}\) 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. \(\displaystyle \mathrm{NH_4^+}\) is genuinely a Brønsted acid: it donates a proton, \[\mathrm{NH_4^+} + \mathrm{OH^-} \longrightarrow \mathrm{NH_3} + \mathrm{H_2O}\] But read that reaction in Lewis language. No new bond forms to nitrogen. What happens is that \(\displaystyle \mathrm{H^+}\) is handed from \(\displaystyle \mathrm{NH_3}\) over to \(\displaystyle \mathrm{OH^-}\): the electron-pair acceptor is the proton, and \(\displaystyle \mathrm{NH_4^+}\) is merely the vehicle carrying it. A proton-donor is not automatically a pair-acceptor.The cleanest way to see it: \(\displaystyle \mathrm{NH_4^+}\) is itself a Lewis adduct — the finished product of the Lewis acid \(\displaystyle \mathrm{H^+}\) and the Lewis base \(\displaystyle \mathrm{:\!NH_3}\), \[\mathrm{H^+} + \mathrm{:\!NH_3} \longrightarrow \mathrm{NH_4^+}\] which is the same shape of reaction as \(\displaystyle \mathrm{BF_3} + \mathrm{:\!NH_3} \to \mathrm{F_3B\!\leftarrow\!NH_3}\). Nobody calls the adduct \(\displaystyle \mathrm{F_3B\!\leftarrow\!NH_3}\) a Lewis acid — the acid was \(\displaystyle \mathrm{BF_3}\), before it was satisfied. For exactly the same reason \(\displaystyle \mathrm{NH_4^+}\) is not a Lewis acid: the acid was \(\displaystyle \mathrm{H^+}\), and in \(\displaystyle \mathrm{NH_4^+}\) 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 \(\displaystyle \mathrm{BF_3}\), \(\displaystyle \mathrm{H^+}\) and \(\displaystyle \mathrm{NH_4^+}\). The first two are right; the inclusion of \(\displaystyle \mathrm{NH_4^+}\) 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: The printed set matches neither list: it admits \(\displaystyle \mathrm{NH_4^+}\) on a rule that would also have to admit \(\displaystyle \mathrm{H_2O}\), and then excludes \(\displaystyle \mathrm{H_2O}\). 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 \(\displaystyle \mathrm{BF_3}\) and \(\displaystyle \mathrm{H^+}\), and add the one-line reason for \(\displaystyle \mathrm{NH_4^+}\) — 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: \(\displaystyle \mathrm{BF_3}\) and \(\displaystyle \mathrm{H^+}\) are the Lewis acids — \(\displaystyle \mathrm{BF_3}\) has a vacant \(\displaystyle 2p\) orbital on boron and \(\displaystyle \mathrm{H^+}\) has an empty \(\displaystyle 1s\) orbital. \(\displaystyle \mathrm{H_2O}\) is a Lewis base (two lone pairs to donate), and \(\displaystyle \mathrm{NH_4^+}\) 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.

A Brønsted acid is a proton \(\displaystyle (\mathrm{H^+})\) donor, so its conjugate base is exactly what is left behind once that one \(\displaystyle \mathrm{H^+}\) is removed.The rule is: \[\text{Acid} \;\rightleftharpoons\; \text{Conjugate base} + \mathrm{H^+} \]To get the conjugate base, take away one \(\displaystyle \mathrm{H^+}\) from the acid and adjust the charge by \(\displaystyle -1\) (removing a positively charged proton leaves the remaining species one unit more negative than the acid was).1. \(\displaystyle \mathrm{HF}\)\(\displaystyle \mathrm{HF}\) is neutral. Removing \(\displaystyle \mathrm{H^+}\) leaves the fluoride ion:\[\mathrm{HF} \rightarrow \mathrm{H^+} + \mathrm{F^-} \]Conjugate base: \(\displaystyle \mathrm{F^-}\)2. \(\displaystyle \mathrm{H_2SO_4}\)This acid has two ionisable protons, but the conjugate base is defined by removing only one \(\displaystyle \mathrm{H^+}\) — the species formed immediately after the first proton leaves, not after both leave. This is the step most people rush past: \(\displaystyle \mathrm{H_2SO_4}\) is neutral, so taking away one \(\displaystyle \mathrm{H^+}\) (charge \(\displaystyle +1\)) leaves a species of charge \(\displaystyle -1\), not \(\displaystyle \mathrm{SO_4^{2-}}\).\[\mathrm{H_2SO_4} \rightarrow \mathrm{H^+} + \mathrm{HSO_4^-} \]Conjugate base: \(\displaystyle \mathrm{HSO_4^-}\) (the hydrogen sulphate ion — it still has one more proton it can donate)3. \(\displaystyle \mathrm{HCO_3^-}\)Here the acid itself already carries a charge of \(\displaystyle -1\). Removing one more \(\displaystyle \mathrm{H^+}\) (charge \(\displaystyle +1\)) makes the resulting species \(\displaystyle -1 + (-1) = -2\):\[\mathrm{HCO_3^-} \rightarrow \mathrm{H^+} + \mathrm{CO_3^{2-}} \]Conjugate base: \(\displaystyle \mathrm{CO_3^{2-}}\) (the carbonate ion)Answer: The conjugate base of \(\displaystyle \mathrm{HF}\) is \(\displaystyle \mathrm{F^-}\); of \(\displaystyle \mathrm{H_2SO_4}\) is \(\displaystyle \mathrm{HSO_4^-}\); of \(\displaystyle \mathrm{HCO_3^-}\) is \(\displaystyle \mathrm{CO_3^{2-}}\).

A conjugate acid is what you get when a base GAINS one \(\displaystyle \text{H}^+ \) ion.By the Brønsted–Lowry definition, a base is a species that accepts a proton (\(\displaystyle \text{H}^+ \)). When it does, the new species formed — base plus that one proton — is called its conjugate acid:\[\text{Base} + \text{H}^+ \rightarrow \text{Conjugate acid} \]The step people skip is checking the charge bookkeeping: adding one positive \(\displaystyle \text{H}^+ \) to a species raises its overall charge by exactly one unit — a \(\displaystyle -1 \) ion becomes neutral, a neutral molecule becomes \(\displaystyle +1 \). 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. \(\displaystyle \text{NH}_2^- \) (amide ion)This ion carries a \(\displaystyle -1 \) charge. Adding \(\displaystyle \text{H}^+ \) brings the charge to \(\displaystyle -1 + 1 = 0\):\[\text{NH}_2^- + \text{H}^+ \rightarrow \text{NH}_3 \]The conjugate acid of \(\displaystyle \text{NH}_2^- \) is \(\displaystyle \text{NH}_3 \) (ammonia).2. \(\displaystyle \text{NH}_3 \) (ammonia)This molecule is neutral. Adding \(\displaystyle \text{H}^+ \) brings the charge to \(\displaystyle 0 + 1 = +1\):\[\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+ \]The conjugate acid of \(\displaystyle \text{NH}_3 \) is \(\displaystyle \text{NH}_4^+ \) (ammonium ion).Notice that \(\displaystyle \text{NH}_3 \) shows up twice in this problem — once as a conjugate acid (of \(\displaystyle \text{NH}_2^- \)) and once as the base being converted (into \(\displaystyle \text{NH}_4^+ \)). 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. \(\displaystyle \text{HCOO}^- \) (formate ion, the conjugate base of formic acid)This ion carries a \(\displaystyle -1 \) charge. Adding \(\displaystyle \text{H}^+ \) brings the charge to \(\displaystyle -1 + 1 = 0\):\[\text{HCOO}^- + \text{H}^+ \rightarrow \text{HCOOH} \]The conjugate acid of \(\displaystyle \text{HCOO}^- \) is \(\displaystyle \text{HCOOH} \) (formic acid, methanoic acid).Answer: The conjugate acid of \(\displaystyle \text{NH}_2^- \) is \(\displaystyle \text{NH}_3 \); the conjugate acid of \(\displaystyle \text{NH}_3 \) is \(\displaystyle \text{NH}_4^+ \); the conjugate acid of \(\displaystyle \text{HCOO}^- \) is \(\displaystyle \text{HCOOH} \).

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 \(\displaystyle \text{H}^+ \) from a species gives its conjugate base (one less proton, one more negative charge); adding \(\displaystyle \text{H}^+ \) 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.\(\displaystyle \text{H}_2\text{O} \)As an acid (donates \(\displaystyle \text{H}^+ \)): \[\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \] conjugate base = \(\displaystyle \text{OH}^- \)As a base (accepts \(\displaystyle \text{H}^+ \)): \[\text{H}_2\text{O} + \text{H}^+ \rightleftharpoons \text{H}_3\text{O}^+ \] conjugate acid = \(\displaystyle \text{H}_3\text{O}^+ \)\(\displaystyle \text{HCO}_3^- \)As an acid: \[\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-} \] conjugate base = \(\displaystyle \text{CO}_3^{2-} \)As a base: \[\text{HCO}_3^- + \text{H}^+ \rightleftharpoons \text{H}_2\text{CO}_3 \] conjugate acid = \(\displaystyle \text{H}_2\text{CO}_3 \)\(\displaystyle \text{HSO}_4^- \)As an acid: \[\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-} \] conjugate base = \(\displaystyle \text{SO}_4^{2-} \)As a base: \[\text{HSO}_4^- + \text{H}^+ \rightleftharpoons \text{H}_2\text{SO}_4 \] conjugate acid = \(\displaystyle \text{H}_2\text{SO}_4 \)\(\displaystyle \text{NH}_3 \)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: \[\text{NH}_3 \rightleftharpoons \text{H}^+ + \text{NH}_2^- \] conjugate base = \(\displaystyle \text{NH}_2^- \) (the amide ion)As a base — its familiar role: \[\text{NH}_3 + \text{H}^+ \rightleftharpoons \text{NH}_4^+ \] conjugate acid = \(\displaystyle \text{NH}_4^+ \)Notice the pattern across all four: as an acid, each species loses one \(\displaystyle \text{H}^+ \) to reach its conjugate base; as a base, each species gains one \(\displaystyle \text{H}^+ \) to reach its conjugate acid. Charge always shifts by exactly one unit toward more negative (losing \(\displaystyle \text{H}^+ \)) or more positive (gaining \(\displaystyle \text{H}^+ \)).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\(\displaystyle ^+\) or OH\(\displaystyle ^-\) at all, just the ability to accept or donate an electron pair, so even species with no protons in sight (like \(\displaystyle \text{BCl}_3\)) can be classified.(a) \(\displaystyle \text{OH}^-\) — 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: \[\text{OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O} \] Here \(\displaystyle \text{OH}^-\) supplies the electron pair that forms the new O–H bond, so it is acting as a Lewis base.(b) \(\displaystyle \text{F}^-\) — Lewis base. Fluoride has four lone pairs and a negative charge, making it a good electron-pair donor: \[\text{F}^- + \text{BF}_3 \rightarrow \text{BF}_4^- \] The lone pair on \(\displaystyle \text{F}^-\) is donated into the empty orbital on boron, forming the new B–F bond. \(\displaystyle \text{F}^-\) is the electron-pair donor, so it is the Lewis base (and \(\displaystyle \text{BF}_3\) is the Lewis acid in this pairing).(c) \(\displaystyle \text{H}^+\) — 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): \[\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \] \(\displaystyle \text{H}^+\) accepts the electron pair from \(\displaystyle \text{OH}^-\), so it is a Lewis acid. This is the step people miss: under the Brønsted-Lowry picture \(\displaystyle \text{H}^+\) 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) \(\displaystyle \text{BCl}_3\) — Lewis acid. Boron has only $\displaystyle 3$ valence electrons and forms three bonds to Cl, using all $\displaystyle 3$ electrons but leaving boron's octet incomplete — it has just $\displaystyle 6$ electrons around it, with one orbital empty. That empty orbital can accept an incoming lone pair, for example from ammonia: \[\text{BCl}_3 + :\text{NH}_3 \rightarrow \text{Cl}_3\text{B}\!-\!\text{NH}_3 \] Nitrogen's lone pair fills boron's empty orbital, forming a new coordinate (dative) B–N bond. \(\displaystyle \text{BCl}_3\) 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 \(\displaystyle \text{BCl}_3\) behave as a Lewis acid — you don't need a formal positive charge to be an electron-pair acceptor.Answer: \(\displaystyle \text{OH}^-\) and \(\displaystyle \text{F}^-\) are Lewis bases (electron-pair donors, via their lone pairs); \(\displaystyle \text{H}^+\) and \(\displaystyle \text{BCl}_3\) are Lewis acids (electron-pair acceptors — \(\displaystyle \text{H}^+\) via its empty 1s orbital, \(\displaystyle \text{BCl}_3\) via boron's incomplete octet).