NCERT Solutions · Class 11 Chemistry Equilibrium

pH measures how acidic a solution is through the hydrogen-ion concentration, using a negative log scale — smaller \(\displaystyle [\mathrm{H^+}]\) means a bigger, less negative log, so a larger pH.The defining formula is \[\mathrm{pH} = -\log_{10}[\mathrm{H^+}] \] where \(\displaystyle [\mathrm{H^+}]\) is the molar concentration of hydrogen ions, in mol L\(\displaystyle ^{-1}\) (equivalently mol/dm\(\displaystyle ^3\), written M).You are given \[[\mathrm{H^+}] = 3.8 \times 10^{-3}\ \mathrm{M} \]Substitute directly — do not simplify the concentration first, since the log of a product splits cleanly: \[\mathrm{pH} = -\log_{10}\left(3.8 \times 10^{-3}\right) \]The step people rush: split the log of a product into a sum, don't try to take \(\displaystyle \log(3.8\times10^{-3})\) as one lump. \[\log_{10}\left(3.8 \times 10^{-3}\right) = \log_{10}(3.8) + \log_{10}\left(10^{-3}\right) \]Now evaluate each piece: \[\log_{10}(3.8) = 0.5798 \] \[\log_{10}\left(10^{-3}\right) = -3 \]So \[\log_{10}\left(3.8 \times 10^{-3}\right) = 0.5798 + (-3) = -2.4202 \]Then \[\mathrm{pH} = -(-2.4202) = 2.4202 \]The given concentration, \(\displaystyle 3.8 \times 10^{-3}\), 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 \(\displaystyle [\mathrm{H^+}]\), since the digits before the decimal point just track the power of ten): \[\mathrm{pH} = 2.42 \]This is comfortably on the acidic side of pH $\displaystyle 7$, which fits a soft drink — most are mildly acidic from dissolved carbon dioxide and added acids.Answer: pH = $\displaystyle 2.42$

pH tells you \(\displaystyle -\log_{10}[\mathrm{H^+}]\), so to get the concentration back you undo the log by raising $\displaystyle 10$ to the power of \(\displaystyle -\text{pH}\).The defining formula is \[\mathrm{pH} = -\log_{10}[\mathrm{H^+}] \] where \(\displaystyle [\mathrm{H^+}]\) is the hydrogen-ion concentration in mol L\(\displaystyle ^{-1}\) (mol/L), the quantity you're asked to find.Step $\displaystyle 1$ — Rearrange for \(\displaystyle [\mathrm{H^+}]\).Starting from \(\displaystyle \mathrm{pH} = -\log_{10}[\mathrm{H^+}]\), multiply both sides by \(\displaystyle -1\): \[\log_{10}[\mathrm{H^+}] = -\mathrm{pH} \] Then remove the log by raising $\displaystyle 10$ to the power of each side: \[[\mathrm{H^+}] = 10^{-\mathrm{pH}} \]Step $\displaystyle 2$ — Substitute the given pH.The pH of the vinegar sample is \(\displaystyle 3.76\), so \[[\mathrm{H^+}] = 10^{-3.76}\ \text{mol L}^{-1} \]Step $\displaystyle 3$ — 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 \(\displaystyle 10^{-3.76}\) "about \(\displaystyle 10^{-4}\)"; the leading digit matters, and it comes from the decimal part of the exponent. Write \[-3.76 = -4 + 0.24 \] so that \[[\mathrm{H^+}] = 10^{-4}\times 10^{0.24}\ \text{mol L}^{-1} \]Step $\displaystyle 4$ — Evaluate \(\displaystyle 10^{0.24}\).\[10^{0.24} = 1.738\ (\text{to 4 significant figures}) \]Step $\displaystyle 5$ — Combine.\[[\mathrm{H^+}] = 1.738 \times 10^{-4}\ \text{mol L}^{-1} \]Since the pH was given to two decimal places ($\displaystyle 3.76$), the mantissa of \(\displaystyle [\mathrm{H^+}]\) is only reliable to about three significant figures, so round to \[[\mathrm{H^+}] \approx 1.74 \times 10^{-4}\ \text{mol L}^{-1} \]Answer: \(\displaystyle [\mathrm{H^+}] = 1.74 \times 10^{-4}\ \text{mol L}^{-1}\) (i.e., \(\displaystyle 10^{-3.76}\) mol L\(\displaystyle ^{-1}\)).

A conjugate acid–base pair is always linked by \(\displaystyle K_a \times K_b = K_w \). 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 $\displaystyle 298$ K,\[K_a \times K_b = K_w = 1.0 \times 10^{-14} \]Here \(\displaystyle K_a\) is the ionization constant of the acid, \(\displaystyle K_b\) is the ionization constant of its conjugate base, and \(\displaystyle K_w\) is the ionic product of water at $\displaystyle 298$ K. Rearranging,\[K_b = \frac{K_w}{K_a} \]The step people skip is checking which species is the conjugate base before plugging in — it is the acid with one \(\displaystyle \text{H}^+\) removed, not the acid itself: Conjugate base of HF (i.e. F⁻):\[K_b(\text{F}^-) = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.4706 \times 10^{-11} \]Rounded to two significant figures (matching the two sig. figs. given for \(\displaystyle K_a\)):\[K_b(\text{F}^-) = 1.5 \times 10^{-11} \]Conjugate base of HCOOH (i.e. HCOO⁻):\[K_b(\text{HCOO}^-) = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} = 5.5556 \times 10^{-11} \]Rounded to two significant figures:\[K_b(\text{HCOO}^-) = 5.6 \times 10^{-11} \]Conjugate base of HCN (i.e. CN⁻):\[K_b(\text{CN}^-) = \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-9}} = 2.0833 \times 10^{-6} \]Rounded to two significant figures:\[K_b(\text{CN}^-) = 2.1 \times 10^{-6} \]Notice the pattern: HCN has the smallest \(\displaystyle K_a\) (it is the weakest acid of the three), so its conjugate base \(\displaystyle \text{CN}^-\) has the largest \(\displaystyle K_b\) — 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: \(\displaystyle K_b(\text{F}^-) = 1.5 \times 10^{-11}\), \(\displaystyle K_b(\text{HCOO}^-) = 5.6 \times 10^{-11}\), \(\displaystyle K_b(\text{CN}^-) = 2.1 \times 10^{-6}\)

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: \[\mathrm{C_6H_5OH} \rightleftharpoons \mathrm{C_6H_5O^-} + \mathrm{H^+} \]The ionization constant is \[K_a = \frac{[\mathrm{C_6H_5O^-}][\mathrm{H^+}]}{[\mathrm{C_6H_5OH}]} \] where \(\displaystyle [\mathrm{C_6H_5O^-}]\), \(\displaystyle [\mathrm{H^+}]\), and \(\displaystyle [\mathrm{C_6H_5OH}]\) are equilibrium molar concentrations of phenolate ion, hydrogen ion, and un-ionized phenol.Part $\displaystyle 1$ — plain $\displaystyle 0.05$ M phenol, no added saltLet \(\displaystyle x\) be the concentration of phenol that ionizes. Since the only source of \(\displaystyle \mathrm{C_6H_5O^-}\) and \(\displaystyle \mathrm{H^+}\) is this ionization, they are produced in a $\displaystyle 1$:$\displaystyle 1$ ratio, so \(\displaystyle [\mathrm{C_6H_5O^-}] = [\mathrm{H^+}] = x\), and \(\displaystyle [\mathrm{C_6H_5OH}] = 0.05 - x\).\[K_a = \frac{x \cdot x}{0.05 - x} = \frac{x^2}{0.05-x} \]Because \(\displaystyle K_a = 1.0\times10^{-10}\) is minuscule compared with \(\displaystyle 0.05\), the amount that ionizes, \(\displaystyle x\), will turn out to be far smaller than \(\displaystyle 0.05\) — so \(\displaystyle 0.05 - x \approx 0.05\). (This is the step people skip and then get stuck solving a quadratic that never needed solving.) That gives\[x^2 \approx K_a \times 0.05 = (1.0\times10^{-10})(0.05) = 5.0\times10^{-12} \] \[x = \sqrt{5.0\times10^{-12}} = 2.2\times10^{-6}\ \text{M} \]Check the approximation: \(\displaystyle 2.2\times10^{-6}\) is about \(\displaystyle 0.004\%\) of \(\displaystyle 0.05\), completely negligible next to it — the shortcut was valid.So in plain $\displaystyle 0.05$ M phenol, \(\displaystyle [\mathrm{C_6H_5O^-}] = 2.2\times10^{-6}\ \text{M}\).Part $\displaystyle 2$ — the same phenol, now also $\displaystyle 0.01$ M in sodium phenolateSodium phenolate, \(\displaystyle \mathrm{C_6H_5ONa}\), is an ionic salt and dissociates completely: \[\mathrm{C_6H_5ONa} \rightarrow \mathrm{C_6H_5O^-} + \mathrm{Na^+} \] This dumps \(\displaystyle 0.01\) M of \(\displaystyle \mathrm{C_6H_5O^-}\) 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 $\displaystyle 1$'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 \(\displaystyle x\) be the (now much smaller) concentration of phenol that still manages to ionize. Track each species starting from these initial concentrations:\[[\mathrm{C_6H_5OH}] = 0.05 - x \approx 0.05, \qquad [\mathrm{C_6H_5O^-}] = 0.01 + x \approx 0.01, \qquad [\mathrm{H^+}] = x \]Both approximations drop \(\displaystyle x\) next to \(\displaystyle 0.05\) and \(\displaystyle 0.01\) — reasonable to try first, and worth checking once \(\displaystyle x\) is known. Substituting into \(\displaystyle K_a\):\[K_a = \frac{[\mathrm{H^+}][\mathrm{C_6H_5O^-}]}{[\mathrm{C_6H_5OH}]} = \frac{x(0.01)}{0.05} \]\[1.0\times10^{-10} = \frac{0.01\,x}{0.05} = 0.2\,x \]\[x = \frac{1.0\times10^{-10}}{0.2} = 5.0\times10^{-10}\ \text{M} \]Check: \(\displaystyle 5.0\times10^{-10} \ll 0.01\) and \(\displaystyle \ll 0.05\) — both approximations hold comfortably, even better than in Part $\displaystyle 1$, 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 \[\alpha = \frac{x}{C} \] where \(\displaystyle C = 0.05\) M is the starting phenol concentration:\[\alpha = \frac{5.0\times10^{-10}}{0.05} = 1.0\times10^{-8} \]Comparing the two parts: ionization dropped from \(\displaystyle x = 2.2\times10^{-6}\) M down to \(\displaystyle x = 5.0\times10^{-10}\) M — over four orders of magnitude smaller — purely because \(\displaystyle 0.01\) M of phenolate ion was already present. That is the common-ion effect in numbers, not just in words.Answer: In $\displaystyle 0.05$ M phenol alone, \(\displaystyle [\mathrm{C_6H_5O^-}] = 2.2\times10^{-6}\ \text{M}\). With the solution also $\displaystyle 0.01$ M in sodium phenolate, the degree of ionization of phenol falls to \(\displaystyle \alpha = 1.0\times10^{-8}\) (i.e. \(\displaystyle 1.0\times10^{-6}\%\)).

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 \(\displaystyle K_{a_1} = 9.1\times10^{-8}\) is about half a million times bigger than \(\displaystyle K_{a_2}=1.2\times10^{-13}\), so essentially all the \(\displaystyle \text{H}^+\) and \(\displaystyle \text{HS}^-\) in the solution come from the first step; the second step barely disturbs those numbers.Step $\displaystyle 1$ — plain $\displaystyle 0.1$ M H₂S, first ionization\[\text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \qquad K_{a_1}=\dfrac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]}=9.1\times10^{-8} \]Let \(\displaystyle x\) be the concentration of \(\displaystyle \text{H}_2\text{S}\) that ionizes. Starting from $\displaystyle 0.1$ M and nothing else present, at equilibrium \(\displaystyle [\text{H}^+]=[\text{HS}^-]=x\) and \(\displaystyle [\text{H}_2\text{S}]=0.1-x\):\[K_{a_1}=\dfrac{x^2}{0.1-x}=9.1\times10^{-8} \]\(\displaystyle \mathrm{K_{a_1}}\) is tiny next to $\displaystyle 0.1$, so \(\displaystyle x\) should be tiny too — assume \(\displaystyle 0.1-x\approx 0.1\) (this substitution is the step to justify, not skip: it's only safe when \(\displaystyle K_{a_1}\ll\) the initial concentration, which it is here by a factor of a million).\[x^2 = (9.1\times10^{-8})(0.1) = 9.1\times10^{-9} \] \[x = \sqrt{9.1\times10^{-9}} = 9.5\times10^{-5}\ \text{M} \]Checking the assumption: \(\displaystyle x/0.1 = 9.5\times10^{-4}\), under $\displaystyle 0.1$% of the starting concentration, so dropping it from \(\displaystyle 0.1-x\) was valid.So in plain $\displaystyle 0.1$ M H₂S: \[[\text{HS}^-] = [\text{H}^+] = 9.5\times10^{-5}\ \text{M} \]Step $\displaystyle 2$ — using this to get [S²⁻], plain solution\[\text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \qquad K_{a_2}=\dfrac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]}=1.2\times10^{-13} \]Solve for \(\displaystyle [\text{S}^{2-}]\): \[[\text{S}^{2-}] = K_{a_2}\times\dfrac{[\text{HS}^-]}{[\text{H}^+]} \]The \(\displaystyle [\text{H}^+]\) and \(\displaystyle [\text{HS}^-]\) here are the ones just found in Step $\displaystyle 1$, and — because the second ionization removes only a negligible sliver of the \(\displaystyle \text{HS}^-\) that Step $\displaystyle 1$ produced — they are still equal to each other (\(\displaystyle 9.5\times10^{-5}\) M each). That equality is the trick: it makes the ratio \(\displaystyle [\text{HS}^-]/[\text{H}^+]\) equal to $\displaystyle 1$, so it cancels outright:\[[\text{S}^{2-}] = K_{a_2} = 1.2\times10^{-13}\ \text{M} \]This is a shortcut specific to this situation (no other source of \(\displaystyle \text{H}^+\) or \(\displaystyle \text{HS}^-\)), 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 $\displaystyle 0.1$ M in HCl — common-ion effectHCl is a strong acid: it dissociates completely and dumps $\displaystyle 0.1$ M of \(\displaystyle \text{H}^+\) into the solution, which is over a thousand times more \(\displaystyle \text{H}^+\) than the H₂S alone produced. By Le Chatelier's principle this extra \(\displaystyle \text{H}^+\) pushes the first equilibrium, \(\displaystyle \text{H}_2\text{S}\rightleftharpoons \text{H}^+ + \text{HS}^-\), back toward \(\displaystyle \text{H}_2\text{S}\) — this is the common-ion effect, and it means \(\displaystyle [\text{HS}^-]\) must fall from what it was before.Because HCl supplies almost all the \(\displaystyle \text{H}^+\) 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): \[[\text{H}^+] \approx 0.1\ \text{M} \]And because the equilibrium has been pushed back toward H₂S, almost none of it ionizes, so: \[[\text{H}_2\text{S}] \approx 0.1\ \text{M} \ \text{(essentially unchanged from the starting concentration)} \]Substitute both into \(\displaystyle \mathrm{K_{a_1}}\) and solve for \(\displaystyle [\text{HS}^-]\): \[[\text{HS}^-] = K_{a_1}\times\dfrac{[\text{H}_2\text{S}]}{[\text{H}^+]} = (9.1\times10^{-8})\times\dfrac{0.1}{0.1} = 9.1\times10^{-8}\ \text{M} \]Compare the two cases: \(\displaystyle [\text{HS}^-]\) drops from \(\displaystyle 9.5\times10^{-5}\) M to \(\displaystyle 9.1\times10^{-8}\) M — roughly a thousandfold suppression, purely from the common \(\displaystyle \text{H}^+\) ion added by the HCl.Now redo Step $\displaystyle 2$ with the new numbers. This time \(\displaystyle [\text{H}^+]=0.1\) M (from HCl) is NOT equal to \(\displaystyle [\text{HS}^-]=9.1\times10^{-8}\) M, so the earlier cancellation trick does not apply — the two must be substituted separately:\[[\text{S}^{2-}] = K_{a_2}\times\dfrac{[\text{HS}^-]}{[\text{H}^+]} = (1.2\times10^{-13})\times\dfrac{9.1\times10^{-8}}{0.1} \]\[[\text{S}^{2-}] = (1.2\times9.1)\times10^{-13-8+1} = 10.92\times10^{-20} = 1.1\times10^{-19}\ \text{M} \]So adding the strong acid suppresses \(\displaystyle \mathrm{[\text{S}^{2-}]}\) even more drastically than it suppresses \(\displaystyle [\text{HS}^-]\) from \(\displaystyle 1.2\times10^{-13}\) M down to \(\displaystyle 1.1\times10^{-19}\) M, a factor of about a million, because the second ionization depends on \(\displaystyle [\text{H}^+]\) being small in a way the first ionization does not need to.Answer: In plain $\displaystyle 0.1$ M H₂S, \(\displaystyle [\text{HS}^-] = 9.5\times10^{-5}\) M and \(\displaystyle [\text{S}^{2-}] = 1.2\times10^{-13}\) M. With the solution also $\displaystyle 0.1$ M in HCl, the common-ion effect lowers these to \(\displaystyle [\text{HS}^-] = 9.1\times10^{-8}\) M and \(\displaystyle [\text{S}^{2-}] = 1.1\times10^{-19}\) M.

The dissociation is small enough to use the shortcut \(\displaystyle \alpha = \sqrt{K_a/C} \) — but only after checking that assumption is actually valid.Step $\displaystyle 1$ — Set up the equilibrium.Acetic acid dissociates as \[\mathrm{CH_3COOH \rightleftharpoons CH_3COO^- + H^+} \]Let the initial concentration be \(\displaystyle C = 0.05\ \text{M}\) and let \(\displaystyle \alpha\) be the degree of dissociation (the fraction of acid molecules that have ionized). Then at equilibrium:

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

Step $\displaystyle 2$ — Write the equilibrium constant.\[K_a = \frac{[\mathrm{CH_3COO^-}][\mathrm{H^+}]}{[\mathrm{CH_3COOH}]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \]Here \(\displaystyle K_a\) 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 $\displaystyle 3$ — Simplify using \(\displaystyle 1-\alpha \approx 1\).Acetic acid is a weak acid, so \(\displaystyle \alpha\) is expected to be small (much less than $\displaystyle 1$). This is the step people skip without checking: assuming \(\displaystyle 1-\alpha\approx 1\) is only valid when \(\displaystyle \alpha\) turns out to be small, so it must be verified after the fact, which is done in Step 4.With that approximation: \[K_a \approx C\alpha^2 \quad\Rightarrow\quad \alpha = \sqrt{\frac{K_a}{C}} \]Step $\displaystyle 4$ — Substitute the numbers.\[\alpha = \sqrt{\frac{1.74\times10^{-5}}{0.05}} = \sqrt{3.48\times10^{-4}} \]\[\alpha = 1.866\times10^{-2} \approx 0.0187 \]Since \(\displaystyle \alpha \approx 0.019 \ll 1\), the approximation \(\displaystyle 1-\alpha\approx1\) in Step $\displaystyle 3$ was justified — check this every time, because for a stronger acid or a more dilute solution \(\displaystyle \alpha\) can come out too large for the shortcut to hold.Step $\displaystyle 5$ — Concentration of acetate ion.At equilibrium, \(\displaystyle [\mathrm{CH_3COO^-}] = [\mathrm{H^+}] = C\alpha\) (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):\[[\mathrm{CH_3COO^-}] = C\alpha = 0.05\ \text{M} \times 1.866\times10^{-2} \]\[[\mathrm{CH_3COO^-}] = 9.33\times10^{-4}\ \text{M} \]Step $\displaystyle 6$ — pH of the solution.pH is defined as \(\displaystyle \mathrm{pH} = -\log_{10}[\mathrm{H^+}]\), where \(\displaystyle [\mathrm{H^+}]\) is in mol/L. Since \(\displaystyle [\mathrm{H^+}] = [\mathrm{CH_3COO^-}] = 9.33\times10^{-4}\ \text{M}\):\[\mathrm{pH} = -\log(9.33\times10^{-4}) = -\left(\log 9.33 + \log 10^{-4}\right) \]\[\mathrm{pH} = -(0.970 - 4) = 4 - 0.970 \]\[\mathrm{pH} = 3.03 \]The data (\(\displaystyle K_a\) to $\displaystyle 3$ significant figures, \(\displaystyle C\) to $\displaystyle 1$ significant figure but treated as exact for this calculation) supports rounding the degree of dissociation to $\displaystyle 3$ significant figures and the pH to $\displaystyle 2$ decimal places — the conventional precision for pH.Answer: degree of dissociation \(\displaystyle \alpha \approx 0.0187\) (i.e., $\displaystyle 1.87$% ionized); \(\displaystyle [\mathrm{CH_3COO^-}] \approx 9.33\times10^{-4}\ \text{M}\); pH \(\displaystyle \approx 3.03\).

pH tells you \(\displaystyle [\text{H}^+]\) directly, and for an acid that dissociates one-to-one, every \(\displaystyle \text{H}^+\) ion released is matched by exactly one \(\displaystyle \text{A}^-\) ion — so at equilibrium \(\displaystyle [\text{A}^-] = [\text{H}^+]\).Step $\displaystyle 1$ — get \(\displaystyle [\text{H}^+]\) from the pH.By definition \(\displaystyle \text{pH} = -\log_{10}[\text{H}^+]\), so\[[\text{H}^+] = 10^{-\text{pH}} = 10^{-4.15} \]Split the exponent: \(\displaystyle 10^{-4.15} = 10^{0.85}\times 10^{-5} = 7.079\times10^{-5}\ \text{M}\).Step $\displaystyle 2$ — set up the equilibrium.Let the acid be HA, with initial concentration \(\displaystyle C = 0.01\ \text{M}\):\[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]

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

Here \(\displaystyle x\) is the amount of acid that has ionized per litre. Since essentially all the \(\displaystyle \text{H}^+\) in this solution comes from the acid (water's own contribution, \(\displaystyle 10^{-7}\ \text{M}\), is negligible next to \(\displaystyle 7.08\times10^{-5}\ \text{M}\)),\[[\text{A}^-] = [\text{H}^+] = x = 7.079\times10^{-5}\ \text{M} \approx 7.08\times10^{-5}\ \text{M} \]This is the concentration of the anion asked for in the question.Step $\displaystyle 3$ — find the equilibrium concentration of unreacted HA.The mistake to avoid: plugging the initial concentration \(\displaystyle C\) into the equilibrium expression instead of \(\displaystyle C - x\). Only a tiny fraction of the acid has ionized, but the correct equilibrium value is still \(\displaystyle C-x\), not \(\displaystyle C\):\[[\text{HA}]_{eq} = C - x = 0.01 - 0.00007079 = 9.929\times10^{-3}\ \text{M} \]Step $\displaystyle 4$ — write the ionization constant \(\displaystyle K_a\) and substitute.\(\displaystyle K_a\) (the acid ionization/dissociation constant) is the equilibrium constant for the dissociation, written as\[K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]with all three concentrations taken at equilibrium (not the initial value). Substituting:\[K_a = \frac{(7.079\times10^{-5})(7.079\times10^{-5})}{9.929\times10^{-3}} = \frac{5.011\times10^{-9}}{9.929\times10^{-3}} = 5.048\times10^{-7} \]Rounding to three significant figures (matching the precision of the given pH, $\displaystyle 4.15$):\[K_a \approx 5.05\times10^{-7} \]Step $\displaystyle 5$ — convert \(\displaystyle K_a\) to \(\displaystyle pK_a\).By the same log definition used for pH, \(\displaystyle pK_a = -\log_{10}K_a\). This is the step people get backwards — \(\displaystyle pK_a\) is the negative log, so a smaller \(\displaystyle K_a\) gives a larger \(\displaystyle pK_a\):\[pK_a = -\log_{10}(5.048\times10^{-7}) = 7 - \log_{10}(5.048) = 7 - 0.703 = 6.297 \]Rounded to two decimal places (consistent with the pH data given):\[pK_a \approx 6.30 \]Answer: \(\displaystyle [\text{A}^-] \approx 7.08\times10^{-5}\ \text{M}\), \(\displaystyle K_a \approx 5.05\times10^{-7}\), \(\displaystyle pK_a \approx 6.30\).

Complete dissociation means the molarity of the acid or base IS the molarity of \(\displaystyle \mathrm{H^+} \) or \(\displaystyle \mathrm{OH^-} \) it releases — for a strong monoprotic acid or monoacidic base, no equilibrium calculation is needed.The defining formula throughout is \[\text{pH} = -\log_{10}[\mathrm{H^+}] \] where \(\displaystyle [\mathrm{H^+}]\) is the hydrogen-ion concentration in \(\displaystyle \text{mol L}^{-1}\). For a base, it is easier to find \(\displaystyle \text{pOH} = -\log_{10}[\mathrm{OH^-}]\) first and then use \[\text{pH} + \text{pOH} = 14 \quad \text{(at 298 K)} \](a) $\displaystyle 0.003$ M HClHCl is a strong acid, so it dissociates completely: \[\mathrm{HCl \rightarrow H^+ + Cl^-} \] Each mole of HCl gives one mole of \(\displaystyle \mathrm{H^+}\), so \(\displaystyle [\mathrm{H^+}] = 0.003\ \text{mol L}^{-1} = 3 \times 10^{-3}\ \text{mol L}^{-1}\).\[\text{pH} = -\log_{10}(3 \times 10^{-3}) = -\left(\log_{10}3 + \log_{10}10^{-3}\right) = -(0.4771 - 3) = 3 - 0.4771 = 2.5229 \]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): \[\text{pH} \approx 2.52 \](b) $\displaystyle 0.005$ M NaOHNaOH is a strong base: \[\mathrm{NaOH \rightarrow Na^+ + OH^-} \] so \(\displaystyle [\mathrm{OH^-}] = 0.005\ \text{mol L}^{-1} = 5 \times 10^{-3}\ \text{mol L}^{-1}\).This is where it is easy to go wrong: you find pOH from \(\displaystyle [\mathrm{OH^-}]\) directly — you do not take \(\displaystyle -\log\) of the concentration and call it pH. The concentration you were given is of \(\displaystyle \mathrm{OH^-}\), not \(\displaystyle \mathrm{H^+}\).\[\text{pOH} = -\log_{10}(5 \times 10^{-3}) = -(\log_{10}5 - 3) = 3 - 0.6990 = 2.301 \] \[\text{pH} = 14 - \text{pOH} = 14 - 2.301 = 11.699 \]\[\text{pH} \approx 11.70 \](c) $\displaystyle 0.002$ M HBrHBr is a strong acid, fully dissociated: \[\mathrm{HBr \rightarrow H^+ + Br^-} \] \[[\mathrm{H^+}] = 0.002\ \text{mol L}^{-1} = 2 \times 10^{-3}\ \text{mol L}^{-1} \] \[\text{pH} = -\log_{10}(2 \times 10^{-3}) = -(\log_{10}2 - 3) = 3 - 0.3010 = 2.699 \]\[\text{pH} \approx 2.70 \](d) $\displaystyle 0.002$ M KOHKOH is a strong base: \[\mathrm{KOH \rightarrow K^+ + OH^-} \] \[[\mathrm{OH^-}] = 0.002\ \text{mol L}^{-1} = 2 \times 10^{-3}\ \text{mol L}^{-1} \] \[\text{pOH} = -\log_{10}(2 \times 10^{-3}) = 3 - 0.3010 = 2.699 \] \[\text{pH} = 14 - 2.699 = 11.301 \]\[\text{pH} \approx 11.30 \]Answer: (a) pH ≈ $\displaystyle 2.52$, (b) pH ≈ $\displaystyle 11.70$, (c) pH ≈ $\displaystyle 2.70$, (d) pH ≈ $\displaystyle 11.30$

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 \(\displaystyle [\text{H}^+]\) or \(\displaystyle [\text{OH}^-]\) directly, with no equilibrium to solve.a) TlOHName the formula: molar mass \(\displaystyle M\) is the mass of one mole, found by adding atomic masses. Taking the atomic mass of Tl as $\displaystyle 204$ u, \[M(\text{TlOH}) = 204 + 16 + 1 = 221\ \text{g/mol} \]Moles of solute, \(\displaystyle n = \dfrac{\text{mass}}{M}\): \[n = \frac{2\ \text{g}}{221\ \text{g/mol}} = 9.05\times10^{-3}\ \text{mol} \]Molarity \(\displaystyle C = \dfrac{n}{V}\), where \(\displaystyle V\) 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 ("$\displaystyle 2$ litre of solution") rather than how much water was added: \[C = \frac{9.05\times10^{-3}\ \text{mol}}{2\ \text{L}} = 4.525\times10^{-3}\ \text{mol/L} \]TlOH is a strong, monoacidic base — it dissociates completely as \(\displaystyle \text{TlOH} \to \text{Tl}^+ + \text{OH}^-\), one \(\displaystyle \text{OH}^-\) per formula unit, so \(\displaystyle [\text{OH}^-] = C\): \[[\text{OH}^-] = 4.525\times10^{-3}\ \text{M} \]Using \(\displaystyle \text{pOH} = -\log[\text{OH}^-]\): \[\text{pOH} = -\log(4.525\times10^{-3}) = 3 - \log(4.525) = 3 - 0.656 = 2.34 \]Using \(\displaystyle \text{pH} + \text{pOH} = 14\) (from \(\displaystyle K_w = 10^{-14}\) at $\displaystyle 298$ K): \[\text{pH} = 14 - 2.34 = 11.66 \]b) Ca(OH)₂\[M(\text{Ca(OH)}_2) = 40 + 2(16+1) = 74\ \text{g/mol} \] \[n = \frac{0.3\ \text{g}}{74\ \text{g/mol}} = 4.054\times10^{-3}\ \text{mol}, \qquad C = \frac{4.054\times10^{-3}\ \text{mol}}{0.5\ \text{L}} = 8.108\times10^{-3}\ \text{mol/L} \]The step people skip here: \(\displaystyle \text{Ca(OH)}_2 \to \text{Ca}^{2+} + 2\text{OH}^-\) releases two hydroxide ions per formula unit, so \(\displaystyle [\text{OH}^-]\) is twice the molarity of the base, not equal to it: \[[\text{OH}^-] = 2C = 2 \times 8.108\times10^{-3} = 1.6216\times10^{-2}\ \text{M} \] \[\text{pOH} = -\log(1.6216\times10^{-2}) = 2 - \log(1.6216) = 2 - 0.210 = 1.79 \] \[\text{pH} = 14 - 1.79 = 12.21 \]c) NaOH\[M(\text{NaOH}) = 23 + 16 + 1 = 40\ \text{g/mol} \] \[n = \frac{0.3\ \text{g}}{40\ \text{g/mol}} = 7.5\times10^{-3}\ \text{mol}, \qquad C = \frac{7.5\times10^{-3}\ \text{mol}}{0.2\ \text{L}} = 3.75\times10^{-2}\ \text{mol/L} \]NaOH is a strong, monoacidic base, so \(\displaystyle [\text{OH}^-] = C = 3.75\times10^{-2}\ \text{M}\): \[\text{pOH} = -\log(3.75\times10^{-2}) = 2 - \log(3.75) = 2 - 0.574 = 1.43 \] \[\text{pH} = 14 - 1.43 = 12.57 \]d) HCl dilutedThis is a dilution, not a dissolution — the moles of solute don't change, only the volume they sit in. The dilution law \(\displaystyle M_1V_1 = M_2V_2\) is just that conservation statement, where \(\displaystyle M\) is molarity and \(\displaystyle V\) is volume (any consistent unit, since it cancels): \[M_1 = 13.6\ \text{M}, \quad V_1 = 1\ \text{mL}, \quad V_2 = 1000\ \text{mL} \ (=1\ \text{L}) \] \[M_2 = \frac{M_1V_1}{V_2} = \frac{13.6\ \text{M} \times 1\ \text{mL}}{1000\ \text{mL}} = 1.36\times10^{-2}\ \text{M} \]HCl is a strong, monoprotic acid, so it dissociates fully and \(\displaystyle [\text{H}^+]\) equals this diluted molarity directly — no equilibrium constant needed: \[[\text{H}^+] = 1.36\times10^{-2}\ \text{M} \] \[\text{pH} = -\log(1.36\times10^{-2}) = 2 - \log(1.36) = 2 - 0.133 = 1.87 \]Answer: pH = $\displaystyle 11.66$ (TlOH solution), $\displaystyle 12.21$ (Ca(OH)₂ solution), $\displaystyle 12.57$ (NaOH solution), $\displaystyle 1.87$ (diluted HCl)

Degree of ionization tells you what fraction of the acid molecules have split into ions — multiply it by the starting concentration to get \(\displaystyle [\text{H}^+]\), don't guess at it from the concentration alone.Bromoacetic acid ionizes as\[\text{BrCH}_2\text{COOH} \rightleftharpoons \text{BrCH}_2\text{COO}^- + \text{H}^+ \]Let \(\displaystyle C\) = initial concentration = \(\displaystyle 0.1\ \text{M}\) and \(\displaystyle \alpha\) = degree of ionization = \(\displaystyle 0.132\) (a pure number — the fraction of molecules ionized, not a concentration itself).At equilibrium, the concentration of \(\displaystyle \text{H}^+\) produced is\[[\text{H}^+] = C\alpha \]Step $\displaystyle 1$: Find \(\displaystyle [\text{H}^+]\)\[[\text{H}^+] = (0.1\ \text{M})(0.132) = 0.0132\ \text{M} = 1.32\times10^{-2}\ \text{M} \]Step $\displaystyle 2$: Find pHThe formula for pH is\[\text{pH} = -\log_{10}[\text{H}^+] \]\[\text{pH} = -\log_{10}(1.32\times10^{-2}) = -\big(\log_{10}1.32 + \log_{10}10^{-2}\big) = -(0.1206 - 2) = 1.8794 \]Rounded to three significant figures (matching the three figures in \(\displaystyle \alpha = 0.132\)):\[\text{pH} = 1.88 \]Step $\displaystyle 3$: Find \(\displaystyle K_a\)For a weak acid of initial concentration \(\displaystyle C\) and degree of ionization \(\displaystyle \alpha\), the equilibrium concentrations are: undissociated acid \(\displaystyle = C(1-\alpha)\), \(\displaystyle \text{H}^+\) \(\displaystyle = C\alpha\), and \(\displaystyle \text{BrCH}_2\text{COO}^-\) \(\displaystyle = C\alpha\). The acid dissociation constant is\[K_a = \frac{[\text{H}^+][\text{BrCH}_2\text{COO}^-]}{[\text{BrCH}_2\text{COOH}]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \]This is the step most people skip past — the denominator uses \(\displaystyle (1-\alpha)\) because only that fraction of the acid is left un-ionized; using \(\displaystyle C\) instead of \(\displaystyle C(1-\alpha)\) there is the classic mistake for an acid this weakly ionized (\(\displaystyle \alpha\) is small but not negligible).Substituting \(\displaystyle C = 0.1\ \text{M}\) and \(\displaystyle \alpha = 0.132\):\[K_a = \frac{(0.1\ \text{M})(0.132)^2}{1 - 0.132} = \frac{(0.1\ \text{M})(0.017424)}{0.868} = \frac{0.0017424\ \text{M}}{0.868} \]\[K_a = 2.007\times10^{-3}\ \text{M} \]Rounded to three significant figures:\[K_a = 2.01\times10^{-3} \](\(\displaystyle K_a\) is dimensionless in the thermodynamic sense, but on the molarity scale it is often quoted with units of mol L\(\displaystyle ^{-1}\); either convention is accepted.)Step $\displaystyle 4$: Find pKaThe formula for pKa is\[\text{p}K_a = -\log_{10}K_a \]\[\text{p}K_a = -\log_{10}(2.01\times10^{-3}) = -(\log_{10}2.01 - 3) = -(0.3032 - 3) = 2.6968 \]Rounded to three significant figures:\[\text{p}K_a = 2.70 \]Answer: pH = $\displaystyle 1.88$ and pKa = $\displaystyle 2.70$ (with \(\displaystyle K_a = 2.01\times10^{-3}\)).