NCERT Solutions · Class 11 Chemistry Equilibrium

Sodium nitrite is the salt of a weak acid \(\displaystyle (\text{HNO}_2)\) and a strong base \(\displaystyle (\text{NaOH})\), so the nitrite ion hydrolyses -- it pulls a proton from water and leaves \(\displaystyle \text{OH}^-\) behind, making the solution basic even though no base was added directly.The hydrolysis reaction is\[\text{NO}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{HNO}_2 + \text{OH}^- \]Step $\displaystyle 1$ -- get the hydrolysis constant \(\displaystyle K_b\) from the given \(\displaystyle K_a\).\(\displaystyle \text{NO}_2^-\) is the conjugate base of \(\displaystyle \text{HNO}_2\), so its equilibrium constant is related to \(\displaystyle K_a\) of the acid through\[K_b = \frac{K_w}{K_a} \]where \(\displaystyle K_w = 1.0\times10^{-14}\) is the ion product of water and \(\displaystyle K_a = 4.5\times10^{-4}\) is the given ionization constant of nitrous acid. This is the step people skip -- you are handed \(\displaystyle K_a\) of the acid, but the species actually reacting with water is the conjugate base, so you must convert to \(\displaystyle K_b\) before setting up the equilibrium table.\[K_b = \frac{1.0\times10^{-14}}{4.5\times10^{-4}} = 2.22\times10^{-11} \]Step $\displaystyle 2$ -- set up the equilibrium in terms of the degree of hydrolysis \(\displaystyle h\).Let \(\displaystyle C = 0.04\ \text{M}\) be the initial concentration of \(\displaystyle \text{NO}_2^-\), and let \(\displaystyle h\) be the fraction of it that hydrolyses. At equilibrium:\[[\text{NO}_2^-] = C(1-h), \qquad [\text{HNO}_2] = Ch, \qquad [\text{OH}^-] = Ch \]so\[K_b = \frac{[\text{HNO}_2][\text{OH}^-]}{[\text{NO}_2^-]} = \frac{Ch\cdot Ch}{C(1-h)} = \frac{Ch^2}{1-h} \]Because \(\displaystyle K_b\) is so small compared with \(\displaystyle C\), \(\displaystyle h \ll 1\), and \(\displaystyle 1-h \approx 1\). The expression simplifies to\[K_b \approx Ch^2 \quad\Rightarrow\quad h = \sqrt{\dfrac{K_b}{C}} \]Step $\displaystyle 3$ -- substitute and solve for \(\displaystyle h\).\[h = \sqrt{\frac{2.22\times10^{-11}}{0.04}} = \sqrt{5.56\times10^{-10}} = 2.36\times10^{-5} \]This is the degree of hydrolysis: only about \(\displaystyle 2.36\times10^{-5}\) (i.e. \(\displaystyle 2.36\times10^{-3}\,\%\)) of the nitrite ions actually hydrolyse -- most of the \(\displaystyle \text{NO}_2^-\) stays untouched, which is exactly why the assumption \(\displaystyle 1-h\approx1\) was safe to make.Step $\displaystyle 4$ -- get \(\displaystyle [\text{OH}^-]\), then \(\displaystyle \text{pOH}\), then \(\displaystyle \text{pH}\).\[[\text{OH}^-] = Ch = 0.04\ \text{M} \times 2.36\times10^{-5} = 9.43\times10^{-7}\ \text{M} \]\[\text{pOH} = -\log[\text{OH}^-] = -\log(9.43\times10^{-7}) = 6.03 \]The last step to pH uses \(\displaystyle \text{pH} + \text{pOH} = 14\) at \(\displaystyle 25^\circ\text{C}\) -- 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.\[\text{pH} = 14 - \text{pOH} = 14 - 6.03 = 7.97 \]The value above $\displaystyle 7$ confirms the chemistry: a salt of a weak acid and a strong base gives a basic solution, and \(\displaystyle \text{pH} = 7.97\) is exactly that -- mildly basic, as expected.Answer: pH ≈ $\displaystyle 7.97$; degree of hydrolysis \(\displaystyle h \approx 2.36\times10^{-5}\) (about \(\displaystyle 2.36\times10^{-3}\,\%\)).

Pyridinium hydrochloride is a salt of a strong acid (HCl) and a weak base (pyridine), so in water it hands you the pyridinium ion, \(\displaystyle \mathrm{C_5H_5NH^+} \) — 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, \(\displaystyle K_b \), takes one more step.Step $\displaystyle 1$ — What's dissolved, and how much of it.\(\displaystyle \mathrm{C_5H_5NHCl} \) is an ionic salt: it dissociates completely in water. \[\mathrm{C_5H_5NHCl(aq) \rightarrow C_5H_5NH^+(aq) + Cl^-(aq)} \] So a $\displaystyle 0.02$ M solution of the salt starts with \(\displaystyle [\mathrm{C_5H_5NH^+}]_0 = 0.02\ \text{mol L}^{-1} \). Chloride is just a spectator ion here.Step $\displaystyle 2$ — Turn the pH into \(\displaystyle [\mathrm{H^+}]\).By definition \(\displaystyle \mathrm{pH} = -\log_{10}[\mathrm{H^+}] \), where \(\displaystyle [\mathrm{H^+}]\) is the equilibrium hydrogen-ion concentration in mol L\(\displaystyle ^{-1}\). So \[[\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-3.44} = 3.63\times10^{-4}\ \text{mol L}^{-1} \]This is the easy step to fumble: the data hands you pH, not a concentration — convert before touching the equilibrium expression.Step $\displaystyle 3$ — Set up the equilibrium that is actually ionizing.The pyridinium ion is a weak acid; it loses a proton to give back pyridine: \[\mathrm{C_5H_5NH^+(aq) \rightleftharpoons C_5H_5N(aq) + H^+(aq)} \]Essentially every \(\displaystyle \mathrm{H^+} \) in solution comes from this ionization (water's own contribution is negligible next to \(\displaystyle 10^{-4}\ \text{mol L}^{-1}\)), so at equilibrium \[[\mathrm{H^+}] = [\mathrm{C_5H_5N}] = x = 3.63\times10^{-4}\ \text{mol L}^{-1}, \qquad [\mathrm{C_5H_5NH^+}] = 0.02 - x \]Step $\displaystyle 4$ — Write \(\displaystyle K_a\) for the pyridinium ion and substitute.Naming the formula: \(\displaystyle K_a = \dfrac{[\mathrm{C_5H_5N}][\mathrm{H^+}]}{[\mathrm{C_5H_5NH^+}]} \) — the ratio of the ionized products to the un-ionized acid, all at equilibrium.\[K_a = \frac{(3.63\times10^{-4})(3.63\times10^{-4})}{0.02 - 3.63\times10^{-4}} = \frac{1.318\times10^{-7}}{0.01964}\ \text{mol L}^{-1} = 6.71\times10^{-6}\ \text{mol L}^{-1} \]Notice the denominator keeps the "\(\displaystyle -x\)" instead of rounding it away: \(\displaystyle x\) is about $\displaystyle 1.8$% of $\displaystyle 0.02$, small but not free to drop, since the extra arithmetic costs nothing here.Step $\displaystyle 5$ — Get from the acid's \(\displaystyle K_a\) to the base's \(\displaystyle K_b\).\(\displaystyle \mathrm{C_5H_5NH^+} \) is the conjugate acid of pyridine, and any conjugate acid–base pair is tied together at a given temperature by \[K_a \times K_b = K_w = 1.0\times10^{-14}\ \text{mol}^2\text{L}^{-2} \quad (\text{at } 298\ \text{K}) \]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 \(\displaystyle K_a\), not report it as if it were already \(\displaystyle K_b\).\[K_b = \frac{K_w}{K_a} = \frac{1.0\times10^{-14}\ \text{mol}^2\text{L}^{-2}}{6.71\times10^{-6}\ \text{mol L}^{-1}} = 1.49\times10^{-9}\ \text{mol L}^{-1} \]Rounding once, to the two significant figures that the pH (given to two decimal places) and the $\displaystyle 0.02$ M concentration support: \[K_b \approx 1.5\times10^{-9}\ \text{mol L}^{-1} \]Answer: \(\displaystyle K_a(\mathrm{C_5H_5NH^+}) \approx 6.7\times10^{-6}\ \text{mol L}^{-1} \), so the ionization constant of pyridine is \(\displaystyle K_b \approx 1.5\times10^{-9}\ \text{mol L}^{-1} \).

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 $\displaystyle 7$; 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: This is the same equilibrium chapter's relation \(\displaystyle K_w = K_a \times K_b \) 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 (\(\displaystyle K_b = K_w/K_a \) for a weak acid's conjugate base, or \(\displaystyle K_a = K_w/K_b \) 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 \(\displaystyle \text{NaNO}_2 \) with \(\displaystyle \text{NaNO}_3 \)-type salts because the formulas look alike. \(\displaystyle \text{HNO}_3 \) (nitric acid) is strong, but \(\displaystyle \text{HNO}_2 \) (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:\(\displaystyle \text{NaCl} \): parent base \(\displaystyle \text{NaOH} \) (strong), parent acid \(\displaystyle \text{HCl} \) (strong). Neither \(\displaystyle \text{Na}^+ \) nor \(\displaystyle \text{Cl}^- \) hydrolyses. → neutral\(\displaystyle \text{KBr} \): parent base \(\displaystyle \text{KOH} \) (strong), parent acid \(\displaystyle \text{HBr} \) (strong). Neither \(\displaystyle \text{K}^+ \) nor \(\displaystyle \text{Br}^- \) hydrolyses. → neutral\(\displaystyle \text{NaCN} \): parent base \(\displaystyle \text{NaOH} \) (strong), parent acid \(\displaystyle \text{HCN} \) (weak). \(\displaystyle \text{Na}^+ \) is inert, but \(\displaystyle \text{CN}^- \) hydrolyses: \[\text{CN}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HCN}(aq) + \text{OH}^-(aq) \] This releases \(\displaystyle \text{OH}^- \). → basic\(\displaystyle \text{NH}_4\text{NO}_3 \): parent base \(\displaystyle \text{NH}_4\text{OH} \) (weak), parent acid \(\displaystyle \text{HNO}_3 \) (strong). \(\displaystyle \text{NO}_3^- \) is inert, but \(\displaystyle \text{NH}_4^+ \) hydrolyses: \[\text{NH}_4^+(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4\text{OH}(aq) + \text{H}^+(aq) \] This releases \(\displaystyle \text{H}^+ \). → acidic\(\displaystyle \text{NaNO}_2 \): parent base \(\displaystyle \text{NaOH} \) (strong), parent acid \(\displaystyle \text{HNO}_2 \) (weak — this is the one that looks like it should pair with strong-acid behaviour but does not). \(\displaystyle \text{Na}^+ \) is inert, but \(\displaystyle \text{NO}_2^- \) hydrolyses: \[\text{NO}_2^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HNO}_2(aq) + \text{OH}^-(aq) \] → basic\(\displaystyle \text{KF} \): parent base \(\displaystyle \text{KOH} \) (strong), parent acid \(\displaystyle \text{HF} \) (weak). \(\displaystyle \text{K}^+ \) is inert, but \(\displaystyle \text{F}^- \) hydrolyses: \[\text{F}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HF}(aq) + \text{OH}^-(aq) \] → basicAnswer: 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).

A weak acid's own ionization can be too large to ignore -- when that happens, the shortcut \(\displaystyle [\mathrm{H^+}] = \sqrt{K_a C} \) is wrong, and you must solve the exact quadratic instead.Part $\displaystyle 1$ -- pH of $\displaystyle 0.1$ M chloroacetic acidChloroacetic acid (\(\displaystyle \mathrm{ClCH_2COOH}\), write it as \(\displaystyle \mathrm{HA}\)) ionizes as\[\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-} \]Let \(\displaystyle x = [\mathrm{H^+}] = [\mathrm{A^-}] \) at equilibrium, starting from \(\displaystyle C = 0.1\ \mathrm{M} \) of \(\displaystyle \mathrm{HA}\), so \(\displaystyle [\mathrm{HA}] = C - x \).The ionization constant \(\displaystyle K_a \) (the equilibrium constant for this proton-loss step) is\[K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} = \frac{x^2}{C-x} = 1.35\times10^{-3} \]Check before approximating: the usual shortcut drops the \(\displaystyle x\) in \(\displaystyle C-x\), valid only when \(\displaystyle x \ll C\). Here \(\displaystyle C/K_a = 0.1/(1.35\times10^{-3}) \approx 74\), well below the \(\displaystyle \gtrsim 400\) benchmark that makes the shortcut safe -- so \(\displaystyle x\) must be kept, and the quadratic solved in full. Skipping this check is exactly where this problem goes wrong.Rearranging:\[x^2 + K_a x - K_a C = 0 \]\[x^2 + (1.35\times10^{-3})x - (1.35\times10^{-4}) = 0 \]By the quadratic formula, \(\displaystyle x = \dfrac{-K_a + \sqrt{K_a^2 + 4K_aC}}{2} \):\[x = \frac{-1.35\times10^{-3} + \sqrt{(1.35\times10^{-3})^2 + 4(1.35\times10^{-3})(0.1)}}{2} \]\[x = \frac{-1.35\times10^{-3} + \sqrt{1.8225\times10^{-6} + 5.4\times10^{-4}}}{2} = \frac{-1.35\times10^{-3} + \sqrt{5.4182\times10^{-4}}}{2} \]\[x = \frac{-1.35\times10^{-3} + 2.3277\times10^{-2}}{2} = 1.096\times10^{-2}\ \mathrm{M} \]So \(\displaystyle [\mathrm{H^+}] = 1.096\times10^{-2}\ \mathrm{M} \) (a check: \(\displaystyle x^2/(C-x) = 1.35\times10^{-3} \), which reproduces \(\displaystyle K_a\) — the fraction ionized here is nearly $\displaystyle 11$%, confirming the shortcut would have understated it).\[\mathrm{pH} = -\log[\mathrm{H^+}] = -\log(1.096\times10^{-2}) = 1.96 \]Part $\displaystyle 2$ -- pH of $\displaystyle 0.1$ M sodium chloroacetate solutionSodium chloroacetate is the salt of a weak acid (\(\displaystyle \mathrm{HA}\)) and a strong base (\(\displaystyle \mathrm{NaOH}\)). In solution it is fully dissociated into \(\displaystyle \mathrm{Na^+}\) and \(\displaystyle \mathrm{A^-}\); \(\displaystyle \mathrm{Na^+}\) is a spectator, but \(\displaystyle \mathrm{A^-}\) 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 $\displaystyle 7$:\[\mathrm{A^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{HA} + \mathrm{OH^-} \]The equilibrium constant for this hydrolysis is the base constant \(\displaystyle K_b \) of \(\displaystyle \mathrm{A^-}\), linked to \(\displaystyle K_a\) of the parent acid through the water constant \(\displaystyle K_w = [\mathrm{H^+}][\mathrm{OH^-}] = 1\times10^{-14}\) (at $\displaystyle 298$ K):\[K_b = \frac{K_w}{K_a} = \frac{1\times10^{-14}}{1.35\times10^{-3}} = 7.407\times10^{-12} \]Let \(\displaystyle y = [\mathrm{OH^-}] = [\mathrm{HA}] \) formed, with initial \(\displaystyle [\mathrm{A^-}] = C = 0.1\ \mathrm{M}\):\[K_b = \frac{[\mathrm{HA}][\mathrm{OH^-}]}{[\mathrm{A^-}]} = \frac{y^2}{C-y} \]Because \(\displaystyle K_b\) is tiny here (unlike \(\displaystyle K_a\) in Part $\displaystyle 1$), \(\displaystyle y\) will be minute next to \(\displaystyle C\), so \(\displaystyle C - y \approx C\) is safe this time -- the same shortcut that failed above is fine here because the size of the equilibrium constant relative to \(\displaystyle C\) is what decides it, not a blanket rule:\[y^2 \approx K_b C = (7.407\times10^{-12})(0.1) = 7.407\times10^{-13} \]\[y = \sqrt{7.407\times10^{-13}} = 8.607\times10^{-7}\ \mathrm{M} \](Indeed \(\displaystyle y/C \approx 8.6\times10^{-6}\), utterly negligible -- the approximation is justified after the fact.)\[\mathrm{pOH} = -\log[\mathrm{OH^-}] = -\log(8.607\times10^{-7}) = 6.07 \]Using \(\displaystyle \mathrm{pH} + \mathrm{pOH} = 14 \) (from \(\displaystyle K_w\) at $\displaystyle 298$ K):\[\mathrm{pH} = 14 - 6.07 = 7.93 \]The salt solution is mildly basic (pH above $\displaystyle 7$), exactly because the anion of a weak acid is a weak base -- not neutral, even though it is "just a salt."Answer: pH of $\displaystyle 0.1$ M chloroacetic acid ≈ $\displaystyle 1.96$ (acidic); pH of $\displaystyle 0.1$ M sodium chloroacetate solution ≈ $\displaystyle 7.93$ (mildly basic).

In pure water, water dissociates into equal amounts of \(\displaystyle \text{H}^+\) and \(\displaystyle \text{OH}^-\) — "neutral" means \(\displaystyle [\text{H}^+] = [\text{OH}^-]\), not that \(\displaystyle [\text{H}^+] = 10^{-7}\ \text{M}\). That value of \(\displaystyle 10^{-7}\) only holds at $\displaystyle 25$°C. At any other temperature, \(\displaystyle K_w\) changes, so the neutral \(\displaystyle [\text{H}^+]\) changes too — but the water is still neutral as long as \(\displaystyle [\text{H}^+]\) equals \(\displaystyle [\text{OH}^-]\).The ionic product of water is \[K_w = [\text{H}^+][\text{OH}^-] \] where \(\displaystyle [\text{H}^+]\) and \(\displaystyle [\text{OH}^-]\) are the equilibrium concentrations (in mol L\(\displaystyle ^{-1}\)) of the hydrogen and hydroxide ions produced by the self-ionisation of water.For neutral water, \(\displaystyle [\text{H}^+] = [\text{OH}^-]\). Call this common value \(\displaystyle x\). Then \[K_w = x \times x = x^2 \]At $\displaystyle 310$ K, \(\displaystyle K_w = 2.7 \times 10^{-14}\), so \[x^2 = 2.7 \times 10^{-14} \] \[x = \sqrt{2.7 \times 10^{-14}} = \sqrt{2.7}\times 10^{-7}\ \text{mol L}^{-1} \]Since \(\displaystyle \sqrt{2.7} = 1.643\), \[x = [\text{H}^+] = 1.643 \times 10^{-7}\ \text{mol L}^{-1} \]Now find the pH using \[\text{pH} = -\log_{10}[\text{H}^+] \] where \(\displaystyle [\text{H}^+]\) is in mol L\(\displaystyle ^{-1}\).Substituting: \[\text{pH} = -\log_{10}(1.643 \times 10^{-7}) \] \[= -\left[\log_{10}(1.643) + \log_{10}(10^{-7})\right] \] \[= -\left[0.216 - 7\right] \] \[= 7 - 0.216 \] \[= 6.784 \]The data (\(\displaystyle 2.7 \times 10^{-14}\)) has two significant figures, so round the final pH to two decimal places: \(\displaystyle \text{pH} = 6.78\).This pH is below $\displaystyle 7$, but the water is still perfectly neutral — pH $\displaystyle 7$ is only the neutral point at $\displaystyle 25$°C. At $\displaystyle 310$ K the neutral point itself has shifted, because \(\displaystyle K_w\) increases with temperature (water's self-ionisation is endothermic), which pushes \(\displaystyle [\text{H}^+]\) up and pH down even though \(\displaystyle [\text{H}^+]\) still equals \(\displaystyle [\text{OH}^-]\).Answer: pH = $\displaystyle 6.78$ (with \(\displaystyle [\text{H}^+] = [\text{OH}^-] = 1.64 \times 10^{-7}\ \text{mol L}^{-1}\)); the water is still neutral even though this pH is less than $\displaystyle 7$, since neutrality requires \(\displaystyle [\text{H}^+]=[\text{OH}^-]\), not pH = 7.

Every OH⁻ or H⁺ ion counts once it is free in solution — Ca(OH)₂ releases $\displaystyle 2$ mol OH⁻ per mole, and H₂SO₄ releases $\displaystyle 2$ mol H⁺ per mole. Missing that factor of $\displaystyle 2$ is the single most common error in this question.The tool you need throughout is\[n = C \times V \]where \(\displaystyle n\) is moles, \(\displaystyle C\) is molarity (mol/L), and \(\displaystyle V\) is volume in litres. After mixing, whichever ion is left over gets diluted into the total volume of the mixture (not the volume it came in), and then\[\text{pH} = -\log_{10}[\text{H}^+], \qquad \text{pOH} = -\log_{10}[\text{OH}^-], \qquad \text{pH} + \text{pOH} = 14 \ (\text{at }25^\circ\text{C}) \](a) $\displaystyle 10$ mL of $\displaystyle 0.2$ M Ca(OH)₂ + $\displaystyle 25$ mL of $\displaystyle 0.1$ M HClMoles of Ca(OH)₂ taken: \[n(\text{Ca(OH)}_2) = 0.010\ \text{L} \times 0.2\ \text{mol L}^{-1} = 2.0\times10^{-3}\ \text{mol} \] Each formula unit gives $\displaystyle 2$ OH⁻, so \[n(\text{OH}^-) = 2 \times 2.0\times10^{-3} = 4.0\times10^{-3}\ \text{mol} \]Moles of HCl ($\displaystyle 1$ H⁺ per formula unit): \[n(\text{H}^+) = 0.025\ \text{L} \times 0.1\ \text{mol L}^{-1} = 2.5\times10^{-3}\ \text{mol} \]H⁺ neutralises an equal amount of OH⁻ (\(\displaystyle \text{H}^+ + \text{OH}^- \to \text{H}_2\text{O}\)); the base is in excess: \[n(\text{OH}^-)_{\text{left}} = 4.0\times10^{-3} - 2.5\times10^{-3} = 1.5\times10^{-3}\ \text{mol} \]This leftover OH⁻ is now spread through both volumes combined, \(\displaystyle 10 + 25 = 35\ \text{mL} = 0.035\ \text{L}\) — using only $\displaystyle 25$ mL or only $\displaystyle 10$ mL here is the second common slip: \[[\text{OH}^-] = \frac{1.5\times10^{-3}\ \text{mol}}{0.035\ \text{L}} = 4.286\times10^{-2}\ \text{mol L}^{-1} \]\[\text{pOH} = -\log_{10}(4.286\times10^{-2}) = 1.368 \] \[\text{pH} = 14 - 1.368 = 12.632 \](b) $\displaystyle 10$ mL of $\displaystyle 0.01$ M H₂SO₄ + $\displaystyle 10$ mL of $\displaystyle 0.01$ M Ca(OH)₂Moles of H₂SO₄: \[n(\text{H}_2\text{SO}_4) = 0.010\ \text{L}\times0.01\ \text{mol L}^{-1} = 1.0\times10^{-4}\ \text{mol} \] $\displaystyle 2$ H⁺ per formula unit: \[n(\text{H}^+) = 2 \times 1.0\times10^{-4} = 2.0\times10^{-4}\ \text{mol} \]Moles of Ca(OH)₂: \[n(\text{Ca(OH)}_2) = 0.010\ \text{L}\times0.01\ \text{mol L}^{-1} = 1.0\times10^{-4}\ \text{mol} \] $\displaystyle 2$ OH⁻ per formula unit: \[n(\text{OH}^-) = 2 \times 1.0\times10^{-4} = 2.0\times10^{-4}\ \text{mol} \]\(\displaystyle n(\text{H}^+) = n(\text{OH}^-) = 2.0\times10^{-4}\ \text{mol}\) exactly — every H⁺ is used up by an OH⁻ and nothing is left over. The resulting solution contains only water and dissolved Ca²⁺/SO₄²⁻ (neither of which reacts with water here), so \[[\text{H}^+] = [\text{OH}^-] = 1.0\times10^{-7}\ \text{mol L}^{-1} \quad\Rightarrow\quad \text{pH} = 7.00 \](c) $\displaystyle 10$ mL of $\displaystyle 0.1$ M H₂SO₄ + $\displaystyle 10$ mL of $\displaystyle 0.1$ M KOHMoles of H₂SO₄: \[n(\text{H}_2\text{SO}_4) = 0.010\ \text{L}\times0.1\ \text{mol L}^{-1} = 1.0\times10^{-3}\ \text{mol} \] \[n(\text{H}^+) = 2 \times 1.0\times10^{-3} = 2.0\times10^{-3}\ \text{mol} \]Moles of KOH ($\displaystyle 1$ OH⁻ per formula unit — this is where part (c) differs from part (a)'s Ca(OH)₂): \[n(\text{KOH}) = n(\text{OH}^-) = 0.010\ \text{L}\times0.1\ \text{mol L}^{-1} = 1.0\times10^{-3}\ \text{mol} \]Acid is in excess: \[n(\text{H}^+)_{\text{left}} = 2.0\times10^{-3} - 1.0\times10^{-3} = 1.0\times10^{-3}\ \text{mol} \]Total volume \(\displaystyle = 10 + 10 = 20\ \text{mL} = 0.020\ \text{L}\): \[[\text{H}^+] = \frac{1.0\times10^{-3}\ \text{mol}}{0.020\ \text{L}} = 5.0\times10^{-2}\ \text{mol L}^{-1} \]\[\text{pH} = -\log_{10}(5.0\times10^{-2}) = 1.301 \]Each result is reported to the precision the input data ($\displaystyle 2$ significant figures on the concentrations) support.Answer: (a) pH ≈ $\displaystyle 12.63$ (basic, excess OH⁻); (b) pH = $\displaystyle 7.00$ (exact neutralisation); (c) pH ≈ $\displaystyle 1.30$ (acidic, excess H⁺)

The solubility product tells you how the ions split up when a salt dissolves — and the exponents on the ion concentrations follow directly from the stoichiometry, so a salt like \(\displaystyle \mathrm{Ag_2CrO_4} \) does NOT dissolve with equal amounts of both ions.Call the molar solubility of each salt \(\displaystyle s \) (mol dissolved per litre of saturated solution). For a salt \(\displaystyle \mathrm{A}_x\mathrm{B}_y \) that dissociates as \[\mathrm{A}_x\mathrm{B}_y \;\rightleftharpoons\; x\,\mathrm{A}^{y+} + y\,\mathrm{B}^{x-} \] the ion concentrations at equilibrium are \(\displaystyle [\mathrm{A}^{y+}] = xs\) and \(\displaystyle [\mathrm{B}^{x-}] = ys\), so the solubility product is \[K_{sp} = [\mathrm{A}^{y+}]^{x}[\mathrm{B}^{x-}]^{y} = (xs)^x(ys)^y . \] This is where people slip: the exponent on each ion is its own stoichiometric coefficient, not "$\displaystyle 1$" by default — miss that and \(\displaystyle \mathrm{Ag_2CrO_4} \) gets solved as if it were a $\displaystyle 1$:$\displaystyle 1$ salt.Using the \(\displaystyle K_{sp} \) values from Table $\displaystyle 6.9$ at $\displaystyle 298$ K:1. Silver chromate, \(\displaystyle \mathrm{Ag_2CrO_4} \) \(\displaystyle (K_{sp} = 1.1\times10^{-12})\)\[\mathrm{Ag_2CrO_4} \rightleftharpoons 2\mathrm{Ag}^+ + \mathrm{CrO_4^{2-}} \] With solubility \(\displaystyle s \): \(\displaystyle [\mathrm{Ag}^+] = 2s\), \(\displaystyle [\mathrm{CrO_4^{2-}}] = s\). \[K_{sp} = (2s)^2(s) = 4s^3 \] \[4s^3 = 1.1\times10^{-12} \quad\Rightarrow\quad s^3 = 2.75\times10^{-13} \] \[s = \left(2.75\times10^{-13}\right)^{1/3} = 6.5\times10^{-5}\ \text{mol L}^{-1} \] So the individual ion molarities are \[[\mathrm{Ag}^+] = 2s = 1.3\times10^{-4}\ \text{mol L}^{-1}, \qquad [\mathrm{CrO_4^{2-}}] = s = 6.5\times10^{-5}\ \text{mol L}^{-1}. \]2. Barium chromate, \(\displaystyle \mathrm{BaCrO_4} \) \(\displaystyle (K_{sp} = 1.2\times10^{-10})\)\[\mathrm{BaCrO_4} \rightleftharpoons \mathrm{Ba}^{2+} + \mathrm{CrO_4^{2-}} \] This is a $\displaystyle 1$:$\displaystyle 1$ salt, so \(\displaystyle [\mathrm{Ba}^{2+}] = [\mathrm{CrO_4^{2-}}] = s\) and \[K_{sp} = s^2 \quad\Rightarrow\quad s = \sqrt{1.2\times10^{-10}} = 1.1\times10^{-5}\ \text{mol L}^{-1}. \] So \(\displaystyle [\mathrm{Ba}^{2+}] = [\mathrm{CrO_4^{2-}}] = 1.1\times10^{-5}\ \text{mol L}^{-1}\).3. Ferric hydroxide, \(\displaystyle \mathrm{Fe(OH)_3} \) \(\displaystyle (K_{sp} = 1.0\times10^{-38})\)\[\mathrm{Fe(OH)_3} \rightleftharpoons \mathrm{Fe}^{3+} + 3\mathrm{OH}^- \] With \(\displaystyle [\mathrm{Fe}^{3+}] = s\) and \(\displaystyle [\mathrm{OH}^-] = 3s\): \[K_{sp} = (s)(3s)^3 = 27s^4 \] \[27s^4 = 1.0\times10^{-38} \quad\Rightarrow\quad s^4 = 3.7\times10^{-40} \] \[s = \left(3.7\times10^{-40}\right)^{1/4} = 1.4\times10^{-10}\ \text{mol L}^{-1} \] So \[[\mathrm{Fe}^{3+}] = s = 1.4\times10^{-10}\ \text{mol L}^{-1}, \qquad [\mathrm{OH}^-] = 3s = 4.2\times10^{-10}\ \text{mol L}^{-1}. \]4. Lead chloride, \(\displaystyle \mathrm{PbCl_2} \) \(\displaystyle (K_{sp} = 1.6\times10^{-5})\)\[\mathrm{PbCl_2} \rightleftharpoons \mathrm{Pb}^{2+} + 2\mathrm{Cl}^- \] With \(\displaystyle [\mathrm{Pb}^{2+}] = s\) and \(\displaystyle [\mathrm{Cl}^-] = 2s\): \[K_{sp} = (s)(2s)^2 = 4s^3 \] \[4s^3 = 1.6\times10^{-5} \quad\Rightarrow\quad s^3 = 4.0\times10^{-6} \] \[s = \left(4.0\times10^{-6}\right)^{1/3} = 1.6\times10^{-2}\ \text{mol L}^{-1} \] So \[[\mathrm{Pb}^{2+}] = s = 1.6\times10^{-2}\ \text{mol L}^{-1}, \qquad [\mathrm{Cl}^-] = 2s = 3.2\times10^{-2}\ \text{mol L}^{-1}. \]5. Mercurous iodide, \(\displaystyle \mathrm{Hg_2I_2} \) \(\displaystyle (K_{sp} = 4.5\times10^{-29})\)The mercurous ion exists as the dimer \(\displaystyle \mathrm{Hg_2^{2+}}\), not as separate \(\displaystyle \mathrm{Hg}^+\) ions — this is the other place this problem catches people out. \[\mathrm{Hg_2I_2} \rightleftharpoons \mathrm{Hg_2^{2+}} + 2\mathrm{I}^- \] With \(\displaystyle [\mathrm{Hg_2^{2+}}] = s\) and \(\displaystyle [\mathrm{I}^-] = 2s\): \[K_{sp} = (s)(2s)^2 = 4s^3 \] \[4s^3 = 4.5\times10^{-29} \quad\Rightarrow\quad s^3 = 1.125\times10^{-29} \] \[s = \left(1.125\times10^{-29}\right)^{1/3} = 2.2\times10^{-10}\ \text{mol L}^{-1} \] So \[[\mathrm{Hg_2^{2+}}] = s = 2.2\times10^{-10}\ \text{mol L}^{-1}, \qquad [\mathrm{I}^-] = 2s = 4.5\times10^{-10}\ \text{mol L}^{-1}. \]Each \(\displaystyle K_{sp} \) was given to two significant figures, so every solubility above is rounded to two significant figures as well — no rounding was done in the middle of any calculation.Answer: Molar solubilities (at $\displaystyle 298$ K) — \(\displaystyle \mathrm{Ag_2CrO_4}: 6.5\times10^{-5}\ \text{mol L}^{-1} \); \(\displaystyle \mathrm{BaCrO_4}: 1.1\times10^{-5}\ \text{mol L}^{-1} \); \(\displaystyle \mathrm{Fe(OH)_3}: 1.4\times10^{-10}\ \text{mol L}^{-1} \); \(\displaystyle \mathrm{PbCl_2}: 1.6\times10^{-2}\ \text{mol L}^{-1} \); \(\displaystyle \mathrm{Hg_2I_2}: 2.2\times10^{-10}\ \text{mol L}^{-1} \). Ion molarities — \(\displaystyle [\mathrm{Ag}^+]=1.3\times10^{-4}\ \text{M}\), \(\displaystyle [\mathrm{CrO_4^{2-}}]=6.5\times10^{-5}\ \text{M}\); \(\displaystyle [\mathrm{Ba}^{2+}]=[\mathrm{CrO_4^{2-}}]=1.1\times10^{-5}\ \text{M}\); \(\displaystyle [\mathrm{Fe}^{3+}]=1.4\times10^{-10}\ \text{M}\), \(\displaystyle [\mathrm{OH}^-]=4.2\times10^{-10}\ \text{M}\); \(\displaystyle [\mathrm{Pb}^{2+}]=1.6\times10^{-2}\ \text{M}\), \(\displaystyle [\mathrm{Cl}^-]=3.2\times10^{-2}\ \text{M}\); \(\displaystyle [\mathrm{Hg_2^{2+}}]=2.2\times10^{-10}\ \text{M}\), \(\displaystyle [\mathrm{I}^-]=4.5\times10^{-10}\ \text{M}\).

The solubility product links back to molarity through the dissociation stoichiometry — and Ag₂CrO₄ does NOT dissociate $\displaystyle 1$:$\displaystyle 1$ like AgBr does, so you cannot just take a square root for both salts.Step $\displaystyle 1$: Write the dissociation and the \(\displaystyle \mathrm{K_{sp}}\) expression for each salt.For silver chromate: \[\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq) \] If the molar solubility of \(\displaystyle \text{Ag}_2\text{CrO}_4\) is \(\displaystyle s_1\) (mol L\(\displaystyle ^{-1}\)), then at equilibrium \(\displaystyle [\text{Ag}^+] = 2s_1\) and \(\displaystyle [\text{CrO}_4^{2-}] = s_1\). Here \(\displaystyle \mathrm{K_{sp}}\) is the solubility product constant — the product of the ion concentrations, each raised to the power of its stoichiometric coefficient, once the solid has stopped dissolving further.\[K_{sp}(\text{Ag}_2\text{CrO}_4) = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = (2s_1)^2(s_1) = 4s_1^3 \]This is the step people skip: because two Ag⁺ ions come out per formula unit, the Ag⁺ concentration is \(\displaystyle 2s_1\), not \(\displaystyle s_1\), and that factor of $\displaystyle 2$ gets squared inside \(\displaystyle \mathrm{K_{sp}}\) — it does not cancel.For silver bromide: \[\text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq) \] If the molar solubility of \(\displaystyle \text{AgBr}\) is \(\displaystyle s_2\), then \(\displaystyle [\text{Ag}^+] = [\text{Br}^-] = s_2\), and\[K_{sp}(\text{AgBr}) = [\text{Ag}^+][\text{Br}^-] = s_2^2 \]Step $\displaystyle 2$: Solve for \(\displaystyle s_1\).\[4s_1^3 = K_{sp}(\text{Ag}_2\text{CrO}_4) = 1.1 \times 10^{-12} \] \[s_1^3 = \frac{1.1 \times 10^{-12}}{4} = 2.75 \times 10^{-13} \] \[s_1 = (2.75 \times 10^{-13})^{1/3} \]Writing \(\displaystyle 2.75 \times 10^{-13} = 275 \times 10^{-15}\) so the exponent is a multiple of $\displaystyle 3$: \[s_1 = (275)^{1/3} \times 10^{-5} = 6.50 \times 10^{-5}\ \text{mol L}^{-1} \]Step $\displaystyle 3$: Solve for \(\displaystyle s_2\).\[s_2^2 = K_{sp}(\text{AgBr}) = 5.0 \times 10^{-13} \] \[s_2 = (5.0 \times 10^{-13})^{1/2} \]Writing \(\displaystyle 5.0 \times 10^{-13} = 50 \times 10^{-14}\) so the exponent is even: \[s_2 = \sqrt{50} \times 10^{-7} = 7.07 \times 10^{-7}\ \text{mol L}^{-1} \]Step $\displaystyle 4$: Take the ratio.\[\frac{s_1}{s_2} = \frac{6.50 \times 10^{-5}\ \text{mol L}^{-1}}{7.07 \times 10^{-7}\ \text{mol L}^{-1}} = 91.9 \approx 92 \]The units of molarity cancel in the ratio, leaving a pure number — this ratio only compares molar solubility (mol of formula units dissolved per litre), not the concentration of either individual ion.The molarity of the saturated \(\displaystyle \text{Ag}_2\text{CrO}_4\) solution is about $\displaystyle 92$ times that of the saturated AgBr solution, even though its \(\displaystyle \mathrm{K_{sp}}\) is only about twice as large — because \(\displaystyle \mathrm{K_{sp}}\) for a $\displaystyle 2$:$\displaystyle 1$ salt scales as the cube of solubility, while for a $\displaystyle 1$:$\displaystyle 1$ salt it scales as the square, a small \(\displaystyle \mathrm{K_{sp}}\) ratio does not translate into a small solubility ratio.Answer: \(\displaystyle s(\text{Ag}_2\text{CrO}_4) : s(\text{AgBr}) \approx 92 : 1\)

Mixing equal volumes of two solutions dilutes each one to exactly half its original concentration — you cannot skip that halving and plug the labelled concentrations straight into \(\displaystyle K_{sp}\).Step $\displaystyle 1$: Find the concentration of each ion right after mixingSodium iodate, \(\displaystyle \text{NaIO}_3\), gives one \(\displaystyle \text{IO}_3^-\) ion per formula unit: \[\text{NaIO}_3 \rightarrow \text{Na}^+ + \text{IO}_3^- \]Cupric chlorate, \(\displaystyle \text{Cu(ClO}_3)_2\), gives one \(\displaystyle \text{Cu}^{2+}\) and two \(\displaystyle \text{ClO}_3^-\) per formula unit: \[\text{Cu(ClO}_3)_2 \rightarrow \text{Cu}^{2+} + 2\text{ClO}_3^- \]Both starting solutions are \(\displaystyle 0.002\ \text{M}\). When you mix equal volumes, say \(\displaystyle V\) of each, the total volume becomes \(\displaystyle 2V\), so each solute's moles are now spread over twice the volume — every concentration is cut in half: \[\text{new concentration} = \frac{0.002\ \text{M} \times V}{2V} = 0.001\ \text{M} \]So immediately after mixing (before any reaction): \[[\text{IO}_3^-] = 0.001\ \text{M}, \qquad [\text{Cu}^{2+}] = 0.001\ \text{M} \]This $\displaystyle 50$% dilution is the step people skip — using the original $\displaystyle 0.002$ M values here would give an answer four times too large for the ionic product below.Step $\displaystyle 2$: Write the dissolution equilibrium for the solid that might formCopper iodate is \(\displaystyle \text{Cu(IO}_3)_2\): \[\text{Cu(IO}_3)_2 (s) \rightleftharpoons \text{Cu}^{2+}(aq) + 2\,\text{IO}_3^-(aq) \]Its solubility product expression (the equilibrium law for a sparingly soluble salt) is \[K_{sp} = [\text{Cu}^{2+}][\text{IO}_3^-]^2 \] where \(\displaystyle [\text{Cu}^{2+}]\) and \(\displaystyle [\text{IO}_3^-]\) are the equilibrium (saturation) concentrations of the ions, and \(\displaystyle \mathrm{K_{sp}}\) is the fixed value that the product of ion concentrations equals only when the solution is exactly saturated.Step $\displaystyle 3$: Compute the ionic product \(\displaystyle Q_{sp}\) for the concentrations you actually haveSince the solution has just been mixed, the ions are not yet necessarily at equilibrium, so you compute the ionic product \(\displaystyle Q_{sp}\) using the same expression as \(\displaystyle \mathrm{K_{sp}}\), but with the concentrations present right now: \[Q_{sp} = [\text{Cu}^{2+}][\text{IO}_3^-]^2 \]Substituting the post-mixing concentrations from Step $\displaystyle 1$: \[Q_{sp} = (0.001)(0.001)^2 = (1\times10^{-3})(1\times10^{-3})^2 \] \[Q_{sp} = (1\times10^{-3})(1\times10^{-6}) = 1\times10^{-9} \]Step $\displaystyle 4$: Compare \(\displaystyle Q_{sp}\) with \(\displaystyle \mathrm{K_{sp}}\) to decide if precipitation happensThe rule connecting the two: Here, \[Q_{sp} = 1\times10^{-9} \qquad \text{and} \qquad K_{sp} = 7.4\times10^{-8} \]Since \(\displaystyle 1\times10^{-9} < 7.4\times10^{-8}\), we have \(\displaystyle Q_{sp} < K_{sp}\).It is easy to compare these wrong because the exponents look close (\(\displaystyle -9\) vs \(\displaystyle -8\)); writing both in the same power of ten shows \(\displaystyle Q_{sp} = 0.1\times10^{-8}\), clearly smaller than \(\displaystyle K_{sp} = 7.4\times10^{-8}\) — about $\displaystyle 74$ times smaller.Because the ionic product is well below the solubility product, the solution remains unsaturated with respect to copper iodate, and no precipitate can form.Answer: No — since \(\displaystyle Q_{sp} = 1\times10^{-9}\) is less than \(\displaystyle K_{sp} = 7.4\times10^{-8}\) for \(\displaystyle \text{Cu(IO}_3)_2\), the solution stays unsaturated and copper iodate does not precipitate.

In a buffer, H⁺ keeps pulling free benzoate ion back into un‑ionized benzoic acid, so less of the dissolved silver salt survives as the ion \(\displaystyle \mathrm{C_6H_5COO^-}\) and to keep \(\displaystyle [Ag^+][C_6H_5COO^-]\) equal to \(\displaystyle \mathrm{K_{sp}}\), more silver benzoate has to dissolve.Silver benzoate dissolves as\[AgOOCC_6H_5(s) \rightleftharpoons Ag^+(aq) + C_6H_5COO^-(aq),\qquad K_{sp}=[Ag^+][C_6H_5COO^-] \]Step $\displaystyle 1$: Solubility in pure water, \(\displaystyle S_0\)With no extra source of H⁺ around, every mole of salt that dissolves gives one mole of \(\displaystyle Ag^+\) and one mole of \(\displaystyle C_6H_5COO^-\), so \(\displaystyle [Ag^+]=[C_6H_5COO^-]=S_0\).\[K_{sp}=S_0^{\,2}\ \implies\ S_0=\sqrt{K_{sp}}=\sqrt{2.5\times10^{-13}}=5.0\times10^{-7}\ mol\,L^{-1} \]Step $\displaystyle 2$: Solubility in the pH $\displaystyle 3.19$ buffer, \(\displaystyle S\)This is the part that trips people up: in the buffer, not all the dissolved benzoate stays as the ion — some of it is protonated to benzoic acid, governed by benzoic acid's own ionization equilibrium\[C_6H_5COOH \rightleftharpoons H^+ + C_6H_5COO^-,\qquad K_a=\frac{[H^+][C_6H_5COO^-]}{[C_6H_5COOH]}=6.46\times10^{-5} \]Let \(\displaystyle S\) be the total amount of silver benzoate dissolved in the buffer. All the silver has nowhere else to go, so \(\displaystyle [Ag^+]=S\). But the benzoate that comes off the salt splits between the ion and the acid form:\[[C_6H_5COO^-]+[C_6H_5COOH]=S \]From the \(\displaystyle K_a\) expression, \(\displaystyle [C_6H_5COOH]=\dfrac{[H^+][C_6H_5COO^-]}{K_a}\). Substitute this into the mass balance and solve for the ion:\[[C_6H_5COO^-]\left(1+\frac{[H^+]}{K_a}\right)=S\ \implies\ [C_6H_5COO^-]=\frac{S\,K_a}{K_a+[H^+]} \]Put this into the solubility product, with \(\displaystyle [Ag^+]=S\):\[K_{sp}=S\times\frac{S\,K_a}{K_a+[H^+]}=\frac{S^{2}K_a}{K_a+[H^+]}\ \implies\ S^{2}=K_{sp}\left(1+\frac{[H^+]}{K_a}\right) \]Now get \(\displaystyle [H^+]\) from the buffer's pH:\[[H^+]=10^{-pH}=10^{-3.19}=6.46\times10^{-4}\ mol\,L^{-1} \]so\[\frac{[H^+]}{K_a}=\frac{6.46\times10^{-4}}{6.46\times10^{-5}}=10 \](pH $\displaystyle 3.19$ is exactly one unit below benzoic acid's \(\displaystyle pK_a\) — that's why \(\displaystyle [H^+]\) comes out to precisely \(\displaystyle 10\,K_a\)).\[S^{2}=K_{sp}(1+10)=11\,K_{sp}=11\times2.5\times10^{-13}=2.75\times10^{-12}\ mol^{2}\,L^{-2} \]\[S=\sqrt{2.75\times10^{-12}}=1.66\times10^{-6}\ mol\,L^{-1} \]Step $\displaystyle 3$: How many times more soluble\[\frac{S}{S_0}=\frac{\sqrt{11\,K_{sp}}}{\sqrt{K_{sp}}}=\sqrt{11}=3.32 \]Checking directly with the numbers: \(\displaystyle \dfrac{1.66\times10^{-6}}{5.0\times10^{-7}}=3.32\) — the same ratio, because it only ever depended on \(\displaystyle 1+[H^+]/K_a\), not on \(\displaystyle \mathrm{K_{sp}}\) itself. Rounding to the precision the data supports (three significant figures, matching \(\displaystyle K_a\)):Answer: Silver benzoate is about $\displaystyle 3.32$ times (\(\displaystyle \sqrt{11}\)) more soluble in the pH $\displaystyle 3.19$ buffer than in pure water — its solubility rises from \(\displaystyle 5.0\times10^{-7}\ mol\,L^{-1}\) to \(\displaystyle 1.66\times10^{-6}\ mol\,L^{-1}\).

The trap here is the mixing step: the moment you pour the two solutions together, every concentration is halved. The solubility product test is applied to the ions in the mixture, but the question asks for the concentration of the bottles you started with — so you must work back up by a factor of $\displaystyle 2$ at the end.Step $\displaystyle 1$ — what is in each solution. Both salts are strong electrolytes, so each dissociates completely: \[\text{FeSO}_4 \longrightarrow \text{Fe}^{2+} + \text{SO}_4^{2-} \qquad \text{Na}_2\text{S} \longrightarrow 2\,\text{Na}^{+} + \text{S}^{2-} \]Let the (equal) concentration of each stock solution be \(\displaystyle c\) mol L\(\displaystyle ^{-1}\).A quick aside on a common slip: in \(\displaystyle \text{Na}_2\text{S}\) the subscript $\displaystyle 2$ sits on sodium, not on sulphide. One formula unit releases one \(\displaystyle \text{S}^{2-}\), so the sulphide concentration in that bottle is \(\displaystyle c\), not \(\displaystyle 2c\). Likewise the ferrous solution gives \(\displaystyle [\text{Fe}^{2+}] = c\).Step $\displaystyle 2$ — dilution on mixing. Take a volume \(\displaystyle V\) of each and mix. Moles are conserved; the volume doubles: \[[\text{Fe}^{2+}]_{\text{mix}} = \frac{c \times V}{V+V} = \frac{c}{2}, \qquad [\text{S}^{2-}]_{\text{mix}} = \frac{c \times V}{V+V} = \frac{c}{2} \]This is the step people skip. Writing \(\displaystyle (c)(c) = K_{sp}\) treats the ions as if they were never diluted, and that answer is wrong by a factor of two.Step $\displaystyle 3$ — the no-precipitation condition. For the equilibrium \[\text{FeS(s)} \rightleftharpoons \text{Fe}^{2+}(aq) + \text{S}^{2-}(aq), \qquad K_{sp} = [\text{Fe}^{2+}][\text{S}^{2-}] \] the ionic product \(\displaystyle Q = [\text{Fe}^{2+}][\text{S}^{2-}]\) is the same product of concentrations evaluated at whatever concentrations you actually have. Here \(\displaystyle \mathrm{K_{sp}}\) is the solubility product of FeS — the largest value that product can reach while everything stays dissolved. "Maximum concentration with no precipitation" therefore means setting \(\displaystyle Q\) exactly equal to \(\displaystyle \mathrm{K_{sp}}\): \[\left(\frac{c}{2}\right)\left(\frac{c}{2}\right) = K_{sp} = 6.3 \times 10^{-18}\ \text{mol}^2\,\text{L}^{-2} \]Step $\displaystyle 4$ — solve for \(\displaystyle c\). \[\frac{c^{2}}{4} = 6.3 \times 10^{-18}\ \text{mol}^2\,\text{L}^{-2} \] \[c^{2} = 4 \times 6.3 \times 10^{-18} = 2.52 \times 10^{-17}\ \text{mol}^2\,\text{L}^{-2} \] \[c = \sqrt{2.52 \times 10^{-17}} = \sqrt{25.2 \times 10^{-18}} = 5.0200 \times 10^{-9}\ \text{mol L}^{-1} \]Note how the units behave: \(\displaystyle \mathrm{K_{sp}}\) carries mol\(\displaystyle ^2\) L\(\displaystyle ^{-2}\) because it is a product of two concentrations, so its square root comes out in mol L\(\displaystyle ^{-1}\) — a concentration, as it must.Step $\displaystyle 5$ — round, and check. \(\displaystyle \mathrm{K_{sp}}\) is quoted to two significant figures, so the answer is justified to two: \[c \approx 5.0 \times 10^{-9}\ \text{mol L}^{-1} \]Check by putting it back: on mixing, each ion sits at \(\displaystyle c/2 = 2.51 \times 10^{-9}\) mol L\(\displaystyle ^{-1}\), and \[Q = (2.51 \times 10^{-9})(2.51 \times 10^{-9}) = 6.3 \times 10^{-18} = K_{sp} \ \checkmark \] Exactly saturated — nothing precipitates, and any stronger stock solution would tip \(\displaystyle Q\) above \(\displaystyle K_{sp}\).A note on the printed answer. Some copies of the answer key give \(\displaystyle 2.5 \times 10^{-9}\) M. That number is \(\displaystyle \sqrt{K_{sp}}\), which is the concentration of \(\displaystyle \text{Fe}^{2+}\) (and of \(\displaystyle \text{S}^{2-}\)) in the mixture after dilution — it is the correct saturation concentration for the mixed solution, but it is not what the question asks for. The question asks for the concentration of the two equimolar solutions that are going to be mixed, and those must be twice as concentrated, since mixing equal volumes halves them. Doubling back gives \(\displaystyle 5.0 \times 10^{-9}\) M. You can settle it by testing the printed value directly: if each bottle were \(\displaystyle 2.5 \times 10^{-9}\) M, then after mixing each ion would be \(\displaystyle 1.255 \times 10^{-9}\) M and \(\displaystyle Q = 1.6 \times 10^{-18}\), well below \(\displaystyle \mathrm{K_{sp}}\) — so that solution is not the maximum, and a more concentrated pair still gives no precipitate.NCERT's answer key prints \(\displaystyle 2.5\times10^{-9}\ \mathrm{M}\), which is the concentration AFTER mixing, not before. \(\displaystyle \sqrt{K_{sp}} = \sqrt{6.3\times10^{-18}} = 2.51\times10^{-9}\) is what each ion may reach in the mixed solution. But the question asks for the concentration of the solutions — and mixing equal volumes halves each concentration, so the stock must be twice that. The key stops one line short. Answer: \(\displaystyle 5.0 \times 10^{-9}\ \text{mol L}^{-1}\) (i.e. \(\displaystyle 5.02 \times 10^{-9}\) M before rounding to two significant figures, set by the two figures in \(\displaystyle K_{sp}\)) for each of the FeSO\(\displaystyle _4\) and Na\(\displaystyle _2\)S solutions before mixing.

The "minimum volume" question is really a solubility question in disguise: find the saturation concentration first, then ask how much solvent is needed to hold a fixed mass of solute at that concentration.Step $\displaystyle 1$ — Convert the mass of \(\displaystyle \mathrm{CaSO_4}\) to moles.Molar mass of \(\displaystyle \mathrm{CaSO_4}\): \[M = 40 + 32 + 4(16) = 136\ \text{g mol}^{-1} \] (Ca = $\displaystyle 40$, S = $\displaystyle 32$, O\(\displaystyle _4\) = 64.)\[n(\mathrm{CaSO_4}) = \frac{1\ \text{g}}{136\ \text{g mol}^{-1}} = 7.353 \times 10^{-3}\ \text{mol} \]Step $\displaystyle 2$ — Find the molar solubility \(\displaystyle s\) from \(\displaystyle K_{sp}\).\(\displaystyle \mathrm{CaSO_4}\) dissociates as \[\mathrm{CaSO_4(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)} \] If the molar solubility is \(\displaystyle s\) mol L\(\displaystyle ^{-1}\), then at equilibrium \(\displaystyle [\mathrm{Ca^{2+}}] = s\) and \(\displaystyle [\mathrm{SO_4^{2-}}] = s\), because the two ions come out in a $\displaystyle 1$:$\displaystyle 1$ ratio. The solubility product is \[K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{SO_4^{2-}}] = s \times s = s^2 \] Here \(\displaystyle \mathrm{K_{sp}}\) is the equilibrium constant for the dissolution reaction — the product of ion concentrations at saturation, not before it. Solving for \(\displaystyle s\): \[s = \sqrt{K_{sp}} = \sqrt{9.1 \times 10^{-6}} = 3.017 \times 10^{-3}\ \text{mol L}^{-1} \]This \(\displaystyle s\) is the maximum concentration of dissolved \(\displaystyle \mathrm{CaSO_4}\) that a solution can hold at $\displaystyle 298$ K — pack in a mole fraction any denser than this and solid \(\displaystyle \mathrm{CaSO_4}\) starts precipitating back out. That's exactly why it sets the minimum volume: any less water and this same \(\displaystyle 1\) g would exceed \(\displaystyle s\), and some of it would come out of solution rather than dissolve.Step $\displaystyle 3$ — Use \(\displaystyle \text{concentration} = \dfrac{\text{moles}}{\text{volume}} \) to get the volume.The people-get-this-wrong step: read the formula as volume of solution, not volume of water added — but for a solid dissolving into water, the two are treated as the same here since we're finding the minimum water needed, and the dissolved solid contributes negligible volume.\[s = \frac{n(\mathrm{CaSO_4})}{V} \quad\Longrightarrow\quad V = \frac{n(\mathrm{CaSO_4})}{s} \]Substituting: \[V = \frac{7.353 \times 10^{-3}\ \text{mol}}{3.017 \times 10^{-3}\ \text{mol L}^{-1}} = 2.438\ \text{L} \]The data (\(\displaystyle K_{sp}\) given to $\displaystyle 2$ significant figures) justifies rounding to $\displaystyle 3$ significant figures at most.Answer: The minimum volume of water required is \(\displaystyle V \approx 2.44\ \text{L}\) (about $\displaystyle 2440$ mL).

A salt precipitates when the ionic product \(\displaystyle Q_{sp}\) of the ions you actually have in the mixed solution exceeds that salt's solubility product \(\displaystyle \mathrm{K_{sp}}\) — so you must dilute BOTH ions first, then compare. The single mistake that sinks this question is comparing \(\displaystyle 0.04\ \text{M}\) and \(\displaystyle 1.0\times10^{-19}\ \text{M}\) directly. Those are the concentrations before mixing. The moment the two solutions are poured together, each one is diluted by the other.Step $\displaystyle 1$ — the volume after mixing.\[V_{\text{total}} = 10\ \text{mL} + 5\ \text{mL} = 15\ \text{mL}\]Step $\displaystyle 2$ — dilute each ion into that $\displaystyle 15$ mL.Dilution keeps the number of moles fixed and enlarges the volume, so \[C_2 = C_1\times\frac{V_1}{V_2}\] where \(\displaystyle C_1, V_1\) are the concentration and volume before mixing and \(\displaystyle C_2, V_2\) after.The sulphide ion came in the $\displaystyle 10$ mL of H₂S-saturated $\displaystyle 0.1$ M HCl: \[[\mathrm{S^{2-}}] = \left(1.0\times10^{-19}\ \text{M}\right)\times\frac{10\ \text{mL}}{15\ \text{mL}} = 6.67\times10^{-20}\ \text{M}\]Each metal ion came in the $\displaystyle 5$ mL of $\displaystyle 0.04$ M salt solution. Every one of FeSO₄, MnCl₂, ZnCl₂ and CdCl₂ gives one M²⁺ per formula unit, so the metal-ion concentration equals the salt concentration: \[[\mathrm{M^{2+}}] = \left(0.04\ \text{M}\right)\times\frac{5\ \text{mL}}{15\ \text{mL}} = 1.33\times10^{-2}\ \text{M}\]An aside on the step people get wrong: the sulphide is diluted by \(\displaystyle 10/15\) and the metal ion by \(\displaystyle 5/15\) — different factors, because the two solutions contributed different volumes. Using \(\displaystyle 1/2\) for both, or forgetting the dilution altogether, is the usual slip here.Step $\displaystyle 3$ — the ionic product, the same for all four.For a $\displaystyle 1$:$\displaystyle 1$ sulphide, \(\displaystyle \mathrm{MS(s)} \rightleftharpoons \mathrm{M^{2+}(aq)} + \mathrm{S^{2-}(aq)}\), so \[Q_{sp} = [\mathrm{M^{2+}}][\mathrm{S^{2-}}]\] \[Q_{sp} = \left(1.33\times10^{-2}\ \text{M}\right)\times\left(6.67\times10^{-20}\ \text{M}\right) = 8.89\times10^{-22}\ \text{M}^2\]Because all four salts are $\displaystyle 0.04$ M and all four are $\displaystyle 1$:$\displaystyle 1$, this one number is the ionic product in every test tube. Only the \(\displaystyle \mathrm{K_{sp}}\) changes from tube to tube.Step $\displaystyle 4$ — compare with each \(\displaystyle K_{sp}\). Taking the values from Table $\displaystyle 6.9$ of the textbook ($\displaystyle 298$ K):\[\text{FeS:}\quad K_{sp} = 6.3\times10^{-18}\ \text{M}^2 \quad\Rightarrow\quad \frac{Q_{sp}}{K_{sp}} = \frac{8.89\times10^{-22}}{6.3\times10^{-18}} = 1.4\times10^{-4} < 1\]\[\text{MnS:}\quad K_{sp} = 2.5\times10^{-13}\ \text{M}^2 \quad\Rightarrow\quad \frac{Q_{sp}}{K_{sp}} = \frac{8.89\times10^{-22}}{2.5\times10^{-13}} = 3.6\times10^{-9} < 1\]\[\text{ZnS:}\quad K_{sp} = 1.6\times10^{-24}\ \text{M}^2 \quad\Rightarrow\quad \frac{Q_{sp}}{K_{sp}} = \frac{8.89\times10^{-22}}{1.6\times10^{-24}} = 5.6\times10^{2} > 1\]\[\text{CdS:}\quad K_{sp} = 8.0\times10^{-27}\ \text{M}^2 \quad\Rightarrow\quad \frac{Q_{sp}}{K_{sp}} = \frac{8.89\times10^{-22}}{8.0\times10^{-27}} = 1.1\times10^{5} > 1\]\(\displaystyle Q_{sp} < K_{sp}\) means the solution is still unsaturated in that sulphide — no solid appears. \(\displaystyle Q_{sp} > K_{sp}\) means it is supersaturated, and solid drops out until \(\displaystyle Q_{sp}\) falls back to \(\displaystyle K_{sp}\).So FeS and MnS stay dissolved; ZnS and CdS precipitate. Rounding once, at the end: \(\displaystyle Q_{sp} = 8.9\times10^{-22}\ \text{M}^2\), to two significant figures, which is all the two-figure data ($\displaystyle 0.04$ M, \(\displaystyle 1.0\times10^{-19}\) M) supports.A note on the printed answer. The answer key at the back of the book names only the cadmium chloride solution. That is incomplete, and the book's own Table $\displaystyle 6.9$ is what shows it: with \(\displaystyle K_{sp}(\mathrm{ZnS}) = 1.6\times10^{-24}\ \text{M}^2\), the ionic product \(\displaystyle 8.9\times10^{-22}\ \text{M}^2\) is about $\displaystyle 560$ times too large for zinc sulphide to remain in solution. There is no reading of the numbers that rescues the key — leaving out the dilution entirely only raises \(\displaystyle Q_{sp}\) to \(\displaystyle 4\times10^{-21}\ \text{M}^2\) and makes ZnS precipitate even more decisively.Where the key probably comes from is the qualitative-analysis habit: in the lab, H₂S passed through a solution acidified with dilute HCl throws down CdS (a Group II sulphide) but leaves Zn²⁺ in solution, and Zn²⁺ is only caught later as ZnS in Group IV. That is true at the sulphide concentration acidified H₂S actually delivers — but this question hands you \(\displaystyle [\mathrm{S^{2-}}] = 1.0\times10^{-19}\ \text{M}\), and you must answer with the number given, not with the lab habit. Work from \(\displaystyle Q_{sp}\) versus \(\displaystyle \mathrm{K_{sp}}\), as above.NCERT's answer key names only cadmium chloride; zinc sulphide precipitates too. After mixing, \(\displaystyle [\mathrm{S^{2-}}] = 1.0\times10^{-19}\times\tfrac{10}{15} = 6.7\times10^{-20}\) and each metal ion is \(\displaystyle 0.04\times\tfrac{5}{15} = 1.33\times10^{-2}\), so the ionic product is \(\displaystyle Q = 8.9\times10^{-22}\) for every one of the four. Compare against each \(\displaystyle \mathrm{K_{sp}}\): FeS and MnS are far larger than \(\displaystyle Q\), so nothing forms; but \(\displaystyle Q\) exceeds the \(\displaystyle \mathrm{K_{sp}}\) of ZnS as well as that of CdS. The key is incomplete rather than incorrect — what it names does precipitate. Answer: Precipitation takes place in the ZnCl₂ and CdCl₂ solutions — ZnS and CdS come down, since \(\displaystyle Q_{sp} = 8.9\times10^{-22}\ \text{M}^2\) exceeds \(\displaystyle K_{sp}(\mathrm{ZnS}) = 1.6\times10^{-24}\ \text{M}^2\) and \(\displaystyle K_{sp}(\mathrm{CdS}) = 8.0\times10^{-27}\ \text{M}^2\); FeSO₄ and MnCl₂ give no precipitate, because \(\displaystyle 8.9\times10^{-22}\ \text{M}^2\) is far below \(\displaystyle K_{sp}(\mathrm{FeS}) = 6.3\times10^{-18}\ \text{M}^2\) and \(\displaystyle K_{sp}(\mathrm{MnS}) = 2.5\times10^{-13}\ \text{M}^2\).