Equilibrium MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: \(\textbf{Determine Ka independently from P:}\) The weak-acid equilibrium is \[ \mathrm{HA\rightleftharpoons H^++A^-} \] Its ionization constant is \[ K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \] For P, \[ K_{a,P}=\frac{(0.0100)(0.100)}{0.100} \] \[ K_{a,P}=0.0100\,\mathrm{M} \] \(\textbf{Verify the independent equilibrium Q:}\) \[ K_{a,Q}=\frac{(0.0200)(0.0500)}{0.100} \] \[ K_{a,Q}=0.0100\,\mathrm{M} \] Both solutions satisfy the same ionization constant. \(\textbf{Determine the immediate mixed concentrations:}\) Equal volumes are combined. Before chemical re-equilibration, the concentration of each species is the arithmetic mean of its concentrations in the two original solutions. Therefore, \[ [\mathrm{H^+}]_{\mathrm{mix}} =\frac{0.0100+0.0200}{2} \] \[ [\mathrm{H^+}]_{\mathrm{mix}}=0.0150\,\mathrm{M} \] Similarly, \[ [\mathrm{HA}]_{\mathrm{mix}} =\frac{0.100+0.100}{2} \] \[ [\mathrm{HA}]_{\mathrm{mix}}=0.100\,\mathrm{M} \] And, \[ [\mathrm{A^-}]_{\mathrm{mix}} =\frac{0.100+0.0500}{2} \] \[ [\mathrm{A^-}]_{\mathrm{mix}}=0.0750\,\mathrm{M} \] \(\textbf{Calculate the instantaneous mixed reaction quotient:}\) \[ Q_{\mathrm{mix}} = \frac{[\mathrm{H^+}]_{\mathrm{mix}} [\mathrm{A^-}]_{\mathrm{mix}}} {[\mathrm{HA}]_{\mathrm{mix}}} \] Substituting, \[ Q_{\mathrm{mix}} = \frac{(0.0150)(0.0750)}{0.100} \] \[ Q_{\mathrm{mix}}=0.01125\,\mathrm{M} \] \(\textbf{Compare with the equilibrium constant:}\) \[ Q_{\mathrm{mix}}=0.01125\gt K_a=0.0100 \] Therefore, the reverse ionization reaction is favoured. Hydrogen ions and A⁻ tend to associate and form additional HA. \(\textbf{Interpret the effect of mixing:}\) Each original solution was independently at equilibrium. However, their hydrogen-ion concentrations differed. Mixing creates a new shared hydrogen-ion concentration and a new conjugate-base-to-acid ratio. These averaged concentrations do not necessarily satisfy the original equilibrium expression. The immediate mixture consequently undergoes chemical re-equilibration. Formation of additional HA consumes hydrogen ions and A⁻, causing the reaction quotient to decrease towards Ka. The example demonstrates why combining separately equilibrated solutions is not equivalent to assuming that the mixture must already be at equilibrium. The correct direction is determined by the mixed reaction quotient, not by the equilibrium status of the separate starting solutions.

Explanation: Statement I is correct. Initially, the solution has \[ pH_i=2.00 \] Therefore, \[ [\mathrm{H^+}]_i=10^{-2}\,\mathrm{M} \] Using Kw, \[ [\mathrm{OH^-}]_i=\frac{10^{-14}}{10^{-2}} \] \[ [\mathrm{OH^-}]_i=10^{-12}\,\mathrm{M} \] At exact equivalence, the solution contains water and nonhydrolysing NaCl. The solution is neutral, so \[ [\mathrm{H^+}]_f=[\mathrm{OH^-}]_f \] At \(298\,\mathrm{K}\), \[ [\mathrm{H^+}]_f=[\mathrm{OH^-}]_f=10^{-7}\,\mathrm{M} \] Consequently, \[ \frac{[\mathrm{OH^-}]_f}{[\mathrm{OH^-}]_i} =\frac{10^{-7}}{10^{-12}} \] \[ =10^5 \] Statement II is also correct. The hydrogen-ion concentration changes from \(10^{-2}\) to \(10^{-7}\,\mathrm{M}\). Its decrease factor is \[ \frac{10^{-2}}{10^{-7}}=10^5 \] Statement III is incorrect. The ionic product of water is fixed by temperature under the stated idealized conditions. It remains \[ K_w=10^{-14} \] both before and after neutralization. The individual hydrogen-ion and hydroxide-ion concentrations change in opposite directions while their product remains constant. Statement IV is correct. At stoichiometric equivalence, the externally introduced strong acid and strong base have neutralized each other. However, water continues to autoionize. The resulting hydronium and hydroxide ions are present at equal, nonzero concentrations. Neutrality does not mean that all hydrogen and hydroxide ions have disappeared. The comparison illustrates how neutralization changes water's equilibrium ionic composition without changing Kw. It also explains why an acidic solution contains a much smaller hydroxide-ion concentration than neutral pure water at the same temperature. Thus, Statements I, II and IV are correct.

Explanation: \(\textbf{Convert the target pH into hydrogen-ion concentration:}\) The required final pH is 2.00. Therefore, \[ [\mathrm{H^+}]=10^{-2.00} \] \[ [\mathrm{H^+}]=0.0100\,\mathrm{M} \] This equals the supplied acid-ionization constant numerically. \(\textbf{Establish the required conjugate-species ratio:}\) The acid-ionization equilibrium is \[ \mathrm{HA\rightleftharpoons H^++A^-} \] Its equilibrium constant is \[ K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \] Therefore, \[ \frac{[\mathrm{A^-}]}{[\mathrm{HA}]} =\frac{K_a}{[\mathrm{H^+}]} \] Substituting, \[ \frac{[\mathrm{A^-}]}{[\mathrm{HA}]} =\frac{0.0100}{0.0100} \] \[ \frac{[\mathrm{A^-}]}{[\mathrm{HA}]}=1 \] Thus, the final free conjugate-base concentration must equal the unionized acid concentration. \(\textbf{Apply electrical neutrality:}\) NaA supplies \[ [\mathrm{Na^+}]=s \] With negligible hydroxide concentration, \[ [\mathrm{A^-}]=[\mathrm{Na^+}]+[\mathrm{H^+}] \] Therefore, \[ [\mathrm{A^-}]=s+0.0100 \] \(\textbf{Apply conservation of acid-derived material:}\) Both HA and NaA contribute A-containing formula units. Their total analytical concentration is \[ C_{\mathrm{total}}=0.100+s \] The material balance requires \[ [\mathrm{HA}]+[\mathrm{A^-}]=0.100+s \] Substituting, \[ [\mathrm{HA}]+s+0.0100=0.100+s \] Therefore, \[ [\mathrm{HA}]=0.0900\,\mathrm{M} \] \(\textbf{Enforce the required equilibrium species equality:}\) Because \[ [\mathrm{A^-}]=[\mathrm{HA}] \] we require \[ s+0.0100=0.0900 \] Therefore, \[ s=0.0800\,\mathrm{M} \] \(\textbf{Verify the final ionization quotient:}\) With \[ [\mathrm{HA}]=[\mathrm{A^-}]=0.0900\,\mathrm{M} \] and \[ [\mathrm{H^+}]=0.0100\,\mathrm{M} \] the quotient is \[ Q_a=\frac{(0.0100)(0.0900)}{0.0900} \] \[ Q_a=0.0100\,\mathrm{M} \] This equals Ka. \(\textbf{Interpret the preparation requirement:}\) Equal equilibrium concentrations of HA and A⁻ do not require equal analytical amounts of initially added HA and NaA. Some of the HA ionizes and supplies A⁻ and hydrogen ions. Therefore, the conjugate-base concentration exceeds the amount supplied directly by NaA. The salt amount must be chosen after accounting for the acid-derived contribution. Simply adding \(0.100\,\mathrm{M}\) NaA because \(0.100\,\mathrm{M}\) HA was initially used would not satisfy the exact requested equilibrium. The correct preparation follows from the target pH, acid-ionization equilibrium, electrical neutrality and material conservation. \(\textbf{Final answer:}\) \[ \boxed{C_{\mathrm{NaA}}=0.0800\,\mathrm{mol\,L^{-1}}} \]

Explanation: \(\textbf{Determine the hydrogen-ion concentrations:}\) By definition, \[ pH=-\log_{10}[\mathrm{H^+}] \] For solution P, \[ [\mathrm{H^+}]_P=10^{-5}\,\mathrm{M} \] For solution Q, \[ [\mathrm{H^+}]_Q=10^{-6}\,\mathrm{M} \] Thus, P has ten times the hydrogen-ion concentration of Q. \(\textbf{Recover the hydroxide concentration in P:}\) \[ K_w=[\mathrm{H^+}][\mathrm{OH^-}] \] Therefore, \[ [\mathrm{OH^-}]_P=\frac{10^{-14}}{10^{-5}} \] \[ [\mathrm{OH^-}]_P=10^{-9}\,\mathrm{M} \] \(\textbf{Recover the hydroxide concentration in Q:}\) \[ [\mathrm{OH^-}]_Q=\frac{10^{-14}}{10^{-6}} \] \[ [\mathrm{OH^-}]_Q=10^{-8}\,\mathrm{M} \] \(\textbf{Calculate the hydroxide concentration ratio:}\) \[ \frac{[\mathrm{OH^-}]_Q}{[\mathrm{OH^-}]_P} =\frac{10^{-8}}{10^{-9}} \] \[ \frac{[\mathrm{OH^-}]_Q}{[\mathrm{OH^-}]_P}=10 \] \(\textbf{Determine the hydrogen-to-hydroxide ratio in P:}\) \[ R_P=\frac{10^{-5}}{10^{-9}} \] \[ R_P=10^4 \] \(\textbf{Determine the corresponding ratio in Q:}\) \[ R_Q=\frac{10^{-6}}{10^{-8}} \] \[ R_Q=10^2 \] \(\textbf{Calculate the ratio of the two ionic ratios:}\) \[ \frac{R_P}{R_Q}=\frac{10^4}{10^2} \] \[ \frac{R_P}{R_Q}=100 \] \(\textbf{Interpret the logarithmic amplification:}\) A one-unit decrease in pH increases hydrogen-ion concentration by a factor of ten. At fixed Kw, the hydroxide-ion concentration simultaneously decreases by a factor of ten. The ratio of hydrogen ions to hydroxide ions consequently changes by a factor of one hundred. Thus, the hydrogen-to-hydroxide ratio is more sensitive to a pH difference than either ionic concentration considered individually. The calculation distinguishes a concentration ratio from a ratio of two concentration ratios. The ionic product remains constant throughout. \(\textbf{Final answer:}\) \[ \boxed{\frac{R_P}{R_Q}=100,\qquad \frac{[\mathrm{OH^-}]_Q}{[\mathrm{OH^-}]_P}=10} \]

Explanation: Statement I is correct because NaCl is derived from a strong acid and a strong base. Sodium ions do not undergo appreciable acid-base hydrolysis in the usual aqueous treatment. Chloride ions are the conjugate bases of a strong acid and have negligible tendency to accept protons from water. Consequently, an ordinary aqueous NaCl solution is approximately neutral at \(298\,\mathrm{K}\). Statement II is correct because acetate is the conjugate base of weak ethanoic acid. Its hydrolysis is represented by \[ \mathrm{CH_3COO^-+H_2O\rightleftharpoons CH_3COOH+OH^-} \] Formation of hydroxide ions makes the solution basic. Statement III is correct because ammonium ion is the conjugate acid of the weak base ammonia. Its hydrolysis is represented by \[ \mathrm{NH_4^++H_2O\rightleftharpoons NH_3+H_3O^+} \] Formation of hydronium ions makes the solution acidic. Statement IV is incorrect because the salt S contains ions derived from both a weak acid and a weak base. Both ions can participate in hydrolysis, but the relative acid and base strengths determine the solution's acid-base character. For ammonium ion, \[ K_a(\mathrm{NH_4^+})=\frac{K_w}{K_b(\mathrm{NH_3})} \] \[ K_a(\mathrm{NH_4^+})=10^{-10} \] For acetate, \[ K_b(\mathrm{CH_3COO^-}) =\frac{K_w}{K_a(\mathrm{CH_3COOH})} \] \[ K_b(\mathrm{CH_3COO^-})=10^{-9} \] The acetate ion has the greater hydrolytic basicity under the supplied conditions. Therefore, S is basic rather than acidic. The classification of a salt must account for the parent acid and base, the actual hydrolysing ions and the relative equilibrium constants. It cannot be inferred merely from the presence of ammonium or from the fact that both ions undergo hydrolysis. Thus, Statements I, II and III are correct.

Explanation: \(\textbf{Identify the hydrolysing ion:}\) NH₄Cl is derived from a strong acid and a weak base. The chloride ion is effectively a spectator ion. The ammonium ion undergoes hydrolysis: \[ \mathrm{NH_4^++H_2O\rightleftharpoons NH_3+H_3O^+} \] \(\textbf{Determine the hydrolysis constant:}\) The ammonium ion is the conjugate acid of ammonia. Therefore, \[ K_h=K_a(\mathrm{NH_4^+}) \] \[ K_h=\frac{K_w}{K_b(\mathrm{NH_3})} \] Substituting, \[ K_h=\frac{1.00\times10^{-14}}{1.00\times10^{-5}} \] \[ K_h=1.00\times10^{-9} \] \(\textbf{Represent the hydrolysis extent:}\) Let x molarity of ammonium ions undergo hydrolysis. Under the stated assumptions, \[ [\mathrm{NH_4^+}]=0.100-x \] \[ [\mathrm{NH_3}]=x \] \[ [\mathrm{H^+}]\approx x \] \(\textbf{Apply the hydrolysis equilibrium:}\) \[ K_h=\frac{x^2}{0.100-x} \] Since hydrolysis is assumed small, \[ 0.100-x\approx0.100 \] Therefore, \[ x\approx\sqrt{K_hC} \] \(\textbf{Calculate the hydrogen-ion concentration:}\) \[ x\approx\sqrt{(1.00\times10^{-9})(0.100)} \] \[ x\approx\sqrt{1.00\times10^{-10}} \] \[ x\approx1.00\times10^{-5}\,\mathrm{M} \] \(\textbf{Determine the pH:}\) \[ pH=-\log_{10}(10^{-5}) \] \[ pH=5.00 \] \(\textbf{Determine the hydrolysed fraction:}\) \[ h=\frac{x}{C} \] \[ h=\frac{1.00\times10^{-5}}{0.100} \] \[ h=1.00\times10^{-4} \] Converting to percentage, \[ 100h=0.0100\% \] \(\textbf{Check the approximation:}\) Only a very small fraction of ammonium ions undergoes hydrolysis. Thus, \[ x\ll0.100\,\mathrm{M} \] The small-hydrolysis approximation is consistent with the calculated result. \(\textbf{Interpret the acidic character:}\) A salt can be completely dissociated into ions without those ions remaining chemically unreactive. NH₄Cl dissociates completely, but a small fraction of ammonium ions subsequently reacts with water. This generates sufficient hydronium ions to produce an acidic solution. The pH is determined by the hydrolysis equilibrium rather than by treating the salt as a strong acid. \(\textbf{Final answer:}\) \[ \boxed{pH\approx5.00,\qquad h\approx0.0100\%} \]

Explanation: \(\textbf{Identify the two hydrolysing ions:}\) The salt contains BH⁺, the conjugate acid of weak base B. It also contains A⁻, the conjugate base of weak acid HA. Both ions can participate in acid-base reactions with water. Their combined proton-transfer process is \[ \mathrm{BH^++A^-\rightleftharpoons B+HA} \] \(\textbf{Construct the overall hydrolysis constant:}\) The acid-ionization constant of BH⁺ is \[ K_a(\mathrm{BH^+})=\frac{K_w}{K_b(\mathrm B)} \] The equilibrium constant for proton transfer from BH⁺ to A⁻ is \[ K_h=\frac{K_a(\mathrm{BH^+})}{K_a(\mathrm{HA})} \] Therefore, \[ K_h=\frac{K_w}{K_a(\mathrm{HA})K_b(\mathrm B)} \] \(\textbf{Calculate the hydrolysis constant:}\) \[ K_h= \frac{1.00\times10^{-14}} {(1.00\times10^{-5})(1.00\times10^{-6})} \] \[ K_h=1.00\times10^{-3} \] \(\textbf{Determine the relative acid and base tendencies:}\) The acid-ionization constant of BH⁺ is \[ K_a(\mathrm{BH^+}) =\frac{10^{-14}}{10^{-6}} \] \[ K_a(\mathrm{BH^+})=10^{-8} \] The base-ionization constant of A⁻ is \[ K_b(\mathrm{A^-}) =\frac{10^{-14}}{10^{-5}} \] \[ K_b(\mathrm{A^-})=10^{-9} \] Therefore, the ammonium-like cation BH⁺ is a stronger acid than A⁻ is a base under the specified comparison. The solution is expected to be acidic. \(\textbf{Apply the weak-acid–weak-base salt relationship:}\) Under the stated approximation, \[ [\mathrm{H^+}] \approx \sqrt{\frac{K_wK_a(\mathrm{HA})}{K_b(\mathrm B)}} \] Substituting, \[ [\mathrm{H^+}] \approx \sqrt{\frac{(10^{-14})(10^{-5})}{10^{-6}}} \] \[ [\mathrm{H^+}] \approx\sqrt{10^{-13}} \] \[ [\mathrm{H^+}] \approx3.162\times10^{-7}\,\mathrm{M} \] \(\textbf{Calculate the pH:}\) \[ pH=-\log_{10}(3.162\times10^{-7}) \] \[ pH\approx6.50 \] \(\textbf{Interpret the two equilibrium quantities:}\) The overall hydrolysis constant measures the tendency of BH⁺ and A⁻ to transfer a proton and form B and HA. The solution pH reflects the relative acid and base tendencies of the two ions. These are related but distinct quantities. A weak-acid–weak-base salt is not automatically neutral merely because both parent substances are weak. The relevant Ka and Kb values determine whether the solution is acidic or basic. \(\textbf{Final answer:}\) \[ \boxed{pH\approx6.50,\qquad K_h=1.00\times10^{-3}} \]

Explanation: \(\textbf{Determine the character of NH₄Cl:}\) NH₄Cl is derived from a strong acid and a weak base. Ammonium ion undergoes acidic hydrolysis. Its acid constant is \[ K_a(\mathrm{NH_4^+}) =\frac{K_w}{K_b(\mathrm{NH_3})} \] \[ K_a(\mathrm{NH_4^+})=\frac{10^{-14}}{10^{-4}} \] \[ K_a(\mathrm{NH_4^+})=10^{-10} \] The approximate hydrogen-ion concentration is \[ [\mathrm{H^+}]_P\approx\sqrt{K_aC} \] \[ [\mathrm{H^+}]_P\approx\sqrt{(10^{-10})(0.100)} \] \[ [\mathrm{H^+}]_P\approx3.162\times10^{-6}\,\mathrm{M} \] Therefore, \[ pH_P\approx5.50 \] \(\textbf{Determine the pH of NaCl:}\) Both sodium and chloride ions are spectators in the usual hydrolysis treatment. Therefore, Q is approximately neutral: \[ pH_Q\approx7.00 \] \(\textbf{Evaluate ammonium acetate:}\) Both ions can undergo hydrolysis. The usual weak-acid–weak-base salt approximation gives \[ pH_R\approx\frac12(pK_w+pK_a-pK_b) \] For ethanoic acid, \[ pK_a=5.00 \] For ammonia, \[ pK_b=4.00 \] Therefore, \[ pH_R\approx\frac12(14.00+5.00-4.00) \] \[ pH_R\approx7.50 \] \(\textbf{Determine the pH of sodium acetate:}\) Acetate undergoes basic hydrolysis: \[ \mathrm{CH_3COO^-+H_2O\rightleftharpoons CH_3COOH+OH^-} \] Its base constant is \[ K_b(\mathrm{CH_3COO^-})=\frac{10^{-14}}{10^{-5}} \] \[ K_b(\mathrm{CH_3COO^-})=10^{-9} \] Therefore, \[ [\mathrm{OH^-}]_S\approx\sqrt{(10^{-9})(0.100)} \] \[ [\mathrm{OH^-}]_S\approx10^{-5}\,\mathrm{M} \] Thus, \[ pOH_S\approx5.00 \] \[ pH_S\approx9.00 \] \(\textbf{Construct the complete order:}\) The estimated pH values are \[ pH_P\approx5.50 \] \[ pH_Q\approx7.00 \] \[ pH_R\approx7.50 \] \[ pH_S\approx9.00 \] Therefore, \[ pH_P\lt pH_Q\lt pH_R\lt pH_S \] \(\textbf{Interpret the comparison:}\) All four salts are assumed completely dissociated, yet their solutions have different acid-base characters. Their pH values depend on whether the dissolved ions undergo acidic hydrolysis, basic hydrolysis, both types or essentially neither. The mixed weak-acid–weak-base salt lies between neutral NaCl and strongly basic sodium acetate under the supplied equilibrium constants. The sequence cannot be predicted from salt concentration alone. The identity and relative strength of the parent acid and base are essential. \(\textbf{Final answer:}\) \[ pH_P\lt pH_Q\lt pH_R\lt pH_S \]

Explanation: \(\textbf{Convert the measured pH into hydroxide concentration:}\) At \(298\,\mathrm{K}\), \[ pH+pOH=14.00 \] Therefore, \[ pOH=14.00-10.00 \] \[ pOH=4.00 \] Thus, \[ [\mathrm{OH^-}]=10^{-4}\,\mathrm{M} \] \(\textbf{Determine the hydrolysis extent:}\) Each A⁻ ion hydrolysed produces one HA molecule and one OH⁻ ion. Under the stated assumptions, \[ x=[\mathrm{HA}]\approx[\mathrm{OH^-}] \] Therefore, \[ x=1.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Recover the remaining anion concentration:}\) The initial salt concentration is \(0.100\,\mathrm{M}\). Consequently, \[ [\mathrm{A^-}]=0.100-x \] \[ [\mathrm{A^-}]=0.100-0.000100 \] \[ [\mathrm{A^-}]=0.0999\,\mathrm{M} \] \(\textbf{Calculate the hydrolysis constant:}\) For the displayed reaction, \[ K_h=\frac{[\mathrm{HA}][\mathrm{OH^-}]}{[\mathrm{A^-}]} \] Substituting, \[ K_h=\frac{(1.00\times10^{-4})^2}{0.0999} \] \[ K_h=\frac{1.00\times10^{-8}}{0.0999} \] \[ K_h\approx1.001\times10^{-7} \] \(\textbf{Recover the parent acid-ionization constant:}\) For conjugate acid HA and base A⁻, \[ K_aK_h=K_w \] Therefore, \[ K_a=\frac{K_w}{K_h} \] Substituting, \[ K_a=\frac{1.00\times10^{-14}}{1.001\times10^{-7}} \] \[ K_a\approx9.99\times10^{-8}\,\mathrm{M} \] \(\textbf{Determine the percentage hydrolysis:}\) \[ h=\frac{x}{C} \] \[ h=\frac{1.00\times10^{-4}}{0.100} \] \[ h=1.00\times10^{-3} \] Therefore, \[ 100h=0.100\% \] \(\textbf{Check the approximation:}\) The hydrolysed concentration is only one-thousandth of the analytical salt concentration. Thus, the assumption of small hydrolysis is justified. The exact remaining A⁻ concentration differs only slightly from \(0.100\,\mathrm{M}\). \(\textbf{Interpret the reverse inference:}\) A measured basic pH reveals the hydroxide concentration generated through hydrolysis. The hydrolysis extent then determines the concentrations of HA and remaining A⁻. These concentrations give the anion's base-ionization constant. The parent acid-ionization constant is obtained from its conjugate relationship with Kw. This sequence is necessary because the measured pH does not directly equal the pKa of the parent weak acid. \(\textbf{Final answer:}\) \[ \boxed{K_a\approx9.99\times10^{-8}\,\mathrm{M},\qquad h=0.100\%} \]

Explanation: \(\textbf{Construct the overall hydrolysis constant:}\) The salt contains a conjugate acid BH⁺ and conjugate base A⁻. Their overall hydrolysis is \[ \mathrm{BH^++A^-\rightleftharpoons B+HA} \] The equilibrium constant is \[ K_h=\frac{K_w}{K_a(\mathrm{HA})K_b(\mathrm B)} \] Substituting, \[ K_h=\frac{10^{-14}}{(10^{-5})(10^{-5})} \] \[ K_h=1.00\times10^{-4} \] \(\textbf{Express the equilibrium through degree of hydrolysis:}\) Let the analytical salt concentration be C and the degree of hydrolysis be h. Then, \[ [\mathrm{BH^+}]=C(1-h) \] \[ [\mathrm{A^-}]=C(1-h) \] \[ [\mathrm B]=Ch \] \[ [\mathrm{HA}]=Ch \] \(\textbf{Apply the overall equilibrium expression:}\) \[ K_h= \frac{[\mathrm B][\mathrm{HA}]} {[\mathrm{BH^+}][\mathrm{A^-}]} \] Substituting, \[ K_h=\frac{(Ch)^2}{[C(1-h)]^2} \] The analytical concentration cancels: \[ K_h=\frac{h^2}{(1-h)^2} \] \(\textbf{Solve for the hydrolysis fraction:}\) Taking the positive square root, \[ \sqrt{K_h}=\frac{h}{1-h} \] \[ 0.0100=\frac{h}{1-h} \] Therefore, \[ h=\frac{0.0100}{1.0100} \] \[ h\approx0.009901 \] \[ 100h\approx0.9901\% \] \(\textbf{Identify the dilution invariant:}\) The expression for h does not contain C. Consequently, the same hydrolysis fraction is obtained at \(0.100\,\mathrm{M}\) and \(0.0100\,\mathrm{M}\) under the stipulated treatment. \(\textbf{Determine the pH:}\) For a weak-acid–weak-base salt, \[ pH\approx\frac12(pK_w+pK_a-pK_b) \] Here, \[ pK_w=14.00 \] \[ pK_a=pK_b=5.00 \] Therefore, \[ pH\approx\frac12(14.00+5.00-5.00) \] \[ pH\approx7.00 \] \(\textbf{Interpret the different dilution behaviour:}\) For a salt containing both a hydrolysing cation and anion, the overall proton-transfer reaction produces one molecule of each neutral parent species. Under the stated assumptions, its degree-of-hydrolysis expression is independent of analytical concentration. This differs from the familiar small-hydrolysis behaviour of salts containing only one significantly hydrolysing ion. The result must not be extrapolated to arbitrarily extreme dilution, where water autoionization can become important relative to the salt concentration. Within the specified regime, both the hydrolysis fraction and pH remain approximately unchanged. \(\textbf{Final answer:}\) \[ \boxed{h_1\approx h_2\approx0.9901\%,\qquad pH_1\approx pH_2\approx7.00} \]

Explanation: \(\textbf{Identify the governing constants for P:}\) A⁻ is the conjugate base of HA. Therefore, \[ K_{h,P}=\frac{K_w}{K_a(\mathrm{HA})} \] At \(T_1\), \[ K_{h,P,1}=\frac{10^{-14}}{10^{-5}} \] \[ K_{h,P,1}=1.00\times10^{-9} \] At \(T_2\), \[ K_{h,P,2}=\frac{4.00\times10^{-14}}{1.00\times10^{-4}} \] \[ K_{h,P,2}=4.00\times10^{-10} \] \(\textbf{Calculate the change factor for P:}\) \[ \frac{K_{h,P,2}}{K_{h,P,1}} = \frac{4.00\times10^{-10}}{1.00\times10^{-9}} \] \[ \frac{K_{h,P,2}}{K_{h,P,1}}=0.400 \] The basic hydrolysis constant decreases. \(\textbf{Identify the governing constants for Q:}\) BH⁺ is the conjugate acid of B. Therefore, \[ K_{h,Q}=\frac{K_w}{K_b(\mathrm B)} \] At \(T_1\), \[ K_{h,Q,1}=\frac{10^{-14}}{10^{-4}} \] \[ K_{h,Q,1}=1.00\times10^{-10} \] At \(T_2\), \[ K_{h,Q,2}=\frac{4.00\times10^{-14}}{2.00\times10^{-4}} \] \[ K_{h,Q,2}=2.00\times10^{-10} \] \(\textbf{Calculate the change factor for Q:}\) \[ \frac{K_{h,Q,2}}{K_{h,Q,1}} = \frac{2.00\times10^{-10}}{1.00\times10^{-10}} \] \[ \frac{K_{h,Q,2}}{K_{h,Q,1}}=2.00 \] The acidic hydrolysis constant doubles. \(\textbf{Interpret the contrasting temperature responses:}\) Both hydrolysis constants involve Kw, which increases on moving to \(T_2\). However, each also depends on the ionization constant of a different parent species. The Ka of HA increases tenfold, more rapidly than Kw. Consequently, the hydrolysis constant of A⁻ decreases. The Kb of B increases only twofold, less rapidly than Kw. Consequently, the hydrolysis constant of BH⁺ increases. A rise in Kw therefore does not require every hydrolysis constant to increase. The temperature dependence of the appropriate conjugate equilibrium must also be included. \(\textbf{Final answer:}\) \[ \boxed{\frac{K_{h,P,2}}{K_{h,P,1}}=0.400,\qquad \frac{K_{h,Q,2}}{K_{h,Q,1}}=2.00} \]

Explanation: \(\textbf{Identify the relevant conjugate acid-base pair:}\) The hydrolysing ion is A²⁻. It accepts a proton from water and becomes HA⁻. Therefore, the relevant conjugate acid is HA⁻, not H₂A. Its acid-ionization constant is the second ionization constant of the diprotic acid. \(\textbf{Construct the hydrolysis constant:}\) \[ K_h=K_b(\mathrm{A^{2-}}) \] \[ K_h=\frac{K_w}{K_{a2}} \] Substituting, \[ K_h=\frac{1.00\times10^{-14}}{1.00\times10^{-10}} \] \[ K_h=1.00\times10^{-4} \] \(\textbf{Represent the hydrolysis extent:}\) Let x molarity of A²⁻ react with water. Then, \[ [\mathrm{A^{2-}}]=0.0930-x \] \[ [\mathrm{HA^-}]=x \] \[ [\mathrm{OH^-}]\approx x \] \(\textbf{Apply the hydrolysis equilibrium:}\) \[ K_h=\frac{[\mathrm{HA^-}][\mathrm{OH^-}]}{[\mathrm{A^{2-}}]} \] Therefore, \[ 1.00\times10^{-4}=\frac{x^2}{0.0930-x} \] \(\textbf{Solve the exact equation:}\) \[ 1.00\times10^{-4}(0.0930-x)=x^2 \] \[ 9.30\times10^{-6}-1.00\times10^{-4}x=x^2 \] \[ x^2+1.00\times10^{-4}x-9.30\times10^{-6}=0 \] The physically acceptable positive root is approximately \[ x=3.000\times10^{-3}\,\mathrm{M} \] The small difference from the rounded value does not affect the displayed answer precision. \(\textbf{Determine the hydroxide concentration:}\) \[ [\mathrm{OH^-}]\approx0.00300\,\mathrm{M} \] \(\textbf{Calculate the pOH:}\) \[ pOH=-\log_{10}(3.00\times10^{-3}) \] \[ pOH=3-\log_{10}3 \] \[ pOH=3-0.4771 \] \[ pOH=2.5229 \] \(\textbf{Calculate the pH:}\) \[ pH=14.00-2.5229 \] \[ pH=11.4771 \] \[ pH\approx11.477 \] \(\textbf{Check the dominant-hydrolysis assumption:}\) The fraction converted into HA⁻ is approximately \[ h=\frac{0.00300}{0.0930} \] \[ h\approx0.0323 \] Thus, only about \(3.23\%\) of the original A²⁻ has undergone the dominant hydrolysis. At the resulting basic pH, further protonation of HA⁻ to H₂A is strongly disfavoured relative to the first hydrolysis stage. \(\textbf{Interpret the successive-ionization connection:}\) A²⁻ is the conjugate base formed in the second ionization of H₂A. Its tendency to accept a proton must therefore be compared with Ka₂. Using Ka₁ would incorrectly associate the hydrolysis with the H₂A/HA⁻ pair. The calculation connects multistage acid ionization, conjugate-base hydrolysis, exact equilibrium composition and pH. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{OH^-}]\approx0.00300\,\mathrm{M},\qquad pH\approx11.477} \]

Explanation: \(\textbf{Establish the hydrolysis behaviour of P:}\) A weak-acid–strong-base salt contains a basic anion A⁻. Its hydrolysis produces hydroxide ions: \[ \mathrm{A^-+H_2O\rightleftharpoons HA+OH^-} \] For small hydrolysis, \[ [\mathrm{OH^-}]\approx\sqrt{K_bC} \] where \[ K_b=\frac{K_w}{K_a} \] \(\textbf{Transform the relationship into pOH:}\) Taking negative common logarithms, \[ pOH\approx\frac12(pK_b+pC) \] Because \[ pH=pK_w-pOH \] we obtain \[ pH_P\approx pK_w-\frac12pK_b-\frac12pC \] Therefore, the slope for P is \[ m_P=-\frac12 \] \(\textbf{Establish the hydrolysis behaviour of Q:}\) A strong-acid–weak-base salt contains an acidic cation BH⁺. Its hydrolysis produces hydronium ions: \[ \mathrm{BH^++H_2O\rightleftharpoons B+H_3O^+} \] For small hydrolysis, \[ [\mathrm{H^+}]\approx\sqrt{K_a(\mathrm{BH^+})C} \] Taking negative logarithms, \[ pH_Q\approx\frac12\left[pK_a(\mathrm{BH^+})+pC\right] \] Therefore, \[ m_Q=+\frac12 \] \(\textbf{Establish the behaviour of R:}\) For the weak-acid–weak-base salt, \[ \mathrm{BH^++A^-\rightleftharpoons B+HA} \] The usual approximate pH expression is \[ pH_R\approx\frac12(pK_w+pK_a-pK_b) \] Under the stated assumptions, this expression does not contain the analytical salt concentration. Therefore, \[ m_R\approx0 \] \(\textbf{Interpret the three graphical responses:}\) As pC increases, the analytical salt concentration decreases. For P, dilution reduces the hydroxide-ion concentration, causing pH to decrease towards neutrality. For Q, dilution reduces the hydrogen-ion concentration, causing pH to increase towards neutrality. For R, the usual hydrolysis approximation predicts little change in pH across the specified range. The three salt classes therefore have different pH responses to dilution even though all are assumed completely dissociated. The slopes follow from their different hydrolysis equilibria rather than from dissociation of the salts themselves. These straight-line relations are approximations and should not be extrapolated into extremely dilute conditions where water autoionization becomes dominant. \(\textbf{Final answer:}\) \[ \boxed{m_P=-\frac12,\qquad m_Q=+\frac12,\qquad m_R\approx0} \]

Explanation: \(\textbf{Determine the initial dissolved amount:}\) Both vessels initially contain one litre of saturated solution. Therefore, \[ n_{\mathrm{dissolved,initial}}=SV \] \[ n_{\mathrm{dissolved,initial}}=0.0100(1.00) \] \[ n_{\mathrm{dissolved,initial}}=0.0100\,\mathrm{mol} \] \(\textbf{Determine the amount required for saturation after dilution:}\) The final aqueous volume is \(2.00\,\mathrm{L}\). At unchanged temperature, the molar solubility remains \(0.0100\,\mathrm{M}\). Therefore, a saturated final solution would require \[ n_{\mathrm{dissolved,saturated}}=0.0100(2.00) \] \[ n_{\mathrm{dissolved,saturated}}=0.0200\,\mathrm{mol} \] The additional amount required to restore saturation is \[ \Delta n=0.0200-0.0100 \] \[ \Delta n=0.0100\,\mathrm{mol} \] \(\textbf{Evaluate vessel P:}\) P initially contains \[ n_{\mathrm{solid},P}=0.0150\,\mathrm{mol} \] This exceeds the additional \(0.0100\,\mathrm{mol}\) required. Therefore, enough solid dissolves to restore saturation. The remaining undissolved solid is \[ n_{\mathrm{solid,final},P}=0.0150-0.0100 \] \[ n_{\mathrm{solid,final},P}=0.00500\,\mathrm{mol} \] The dissolved concentration remains at the saturation value: \[ C_P=0.0100\,\mathrm{M} \] \(\textbf{Evaluate vessel Q:}\) Q initially contains only \[ n_{\mathrm{solid},Q}=0.00500\,\mathrm{mol} \] This is insufficient to restore saturation in two litres. All its remaining solid therefore dissolves. The maximum dissolved amount is \[ n_{\mathrm{dissolved,final},Q} =0.0100+0.00500 \] \[ n_{\mathrm{dissolved,final},Q}=0.0150\,\mathrm{mol} \] \(\textbf{Determine Q's final concentration:}\) \[ C_Q=\frac{0.0150}{2.00} \] \[ C_Q=0.00750\,\mathrm{M} \] Because \[ 0.00750\lt 0.0100 \] Q is unsaturated. \(\textbf{Interpret the role of undissolved solid:}\) The equilibrium dissolved concentration of a saturated solution is fixed by temperature and the salt-solvent system under the stated conditions. However, sufficient undissolved solid must be available to restore saturation after dilution. Vessel P has enough solid, while Q does not. Adding more solid to an already saturated solution does not increase its equilibrium dissolved concentration. Nevertheless, the amount of solid present can determine whether saturation survives a later dilution. This distinction requires both the solubility limit and the complete material inventory. \(\textbf{Final answer:}\) \[ \boxed{C_P=0.0100\,\mathrm{M},\qquad C_Q=0.00750\,\mathrm{M}} \]

Explanation: \(\textbf{Interpret the ionic dissolution stoichiometry:}\) Each mole of A₂B₃ that dissolves produces two moles of A³⁺ and three moles of B²⁻. Let its molar solubility be S. Then, \[ [\mathrm{A^{3+}}]=2S \] \[ [\mathrm{B^{2-}}]=3S \] \(\textbf{Recover the molar solubility:}\) The measured cation concentration is \[ 2S=0.0200\,\mathrm{M} \] Therefore, \[ S=\frac{0.0200}{2} \] \[ S=0.0100\,\mathrm{M} \] \(\textbf{Determine the anion concentration:}\) \[ [\mathrm{B^{2-}}]=3S \] \[ [\mathrm{B^{2-}}]=3(0.0100) \] \[ [\mathrm{B^{2-}}]=0.0300\,\mathrm{M} \] \(\textbf{Construct the solubility-product expression:}\) The dissolution reaction is \[ \mathrm{A_2B_3(s)\rightleftharpoons2A^{3+}+3B^{2-}} \] The pure solid does not appear in the equilibrium-constant expression. The ionic stoichiometric coefficients become exponents: \[ K_{sp}=[\mathrm{A^{3+}}]^2[\mathrm{B^{2-}}]^3 \] \(\textbf{Substitute the equilibrium concentrations:}\) \[ K_{sp}=(0.0200)^2(0.0300)^3 \] \[ (0.0200)^2=4.00\times10^{-4} \] \[ (0.0300)^3=2.70\times10^{-5} \] Therefore, \[ K_{sp}=(4.00\times10^{-4})(2.70\times10^{-5}) \] \[ K_{sp}=1.08\times10^{-8}\,\mathrm{M^5} \] \(\textbf{Verify the charge balance:}\) The positive charge-equivalent concentration is \[ 3[\mathrm{A^{3+}}]=3(0.0200) \] \[ =0.0600\,\mathrm{M} \] The negative charge-equivalent concentration is \[ 2[\mathrm{B^{2-}}]=2(0.0300) \] \[ =0.0600\,\mathrm{M} \] Thus, the dissolved ions satisfy electrical neutrality. \(\textbf{Interpret the concentration powers:}\) The molar solubility is not equal to either ionic concentration because the dissolution produces multiple ions per formula unit. Similarly, the solubility product is not obtained by multiplying ionic concentrations without considering the reaction coefficients. The exponents arise from the balanced dissolution equation. The conventional concentration expression has units of \(\mathrm{M^5}\), while the corresponding thermodynamic equilibrium constant is dimensionless when written with activities. \(\textbf{Final answer:}\) \[ \boxed{S=0.0100\,\mathrm{M},\qquad K_{sp}=1.08\times10^{-8}\,\mathrm{M^5}} \]

Explanation: \(\textbf{Determine the solubility relationship for P:}\) The salt AB dissolves according to \[ \mathrm{AB(s)\rightleftharpoons A^++B^-} \] If its molar solubility is \(S_P\), \[ [\mathrm{A^+}]=[\mathrm{B^-}]=S_P \] Therefore, \[ K_{sp,P}=S_P^2 \] \(\textbf{Calculate P's molar solubility:}\) \[ S_P=\sqrt{K_{sp,P}} \] \[ S_P=\sqrt{1.00\times10^{-8}} \] \[ S_P=1.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Determine the solubility relationship for Q:}\) The second dissolution reaction is \[ \mathrm{CD_2(s)\rightleftharpoons C^{2+}+2D^-} \] For molar solubility \(S_Q\), \[ [\mathrm{C^{2+}}]=S_Q \] \[ [\mathrm{D^-}]=2S_Q \] Thus, \[ K_{sp,Q}=[\mathrm{C^{2+}}][\mathrm{D^-}]^2 \] \[ K_{sp,Q}=S_Q(2S_Q)^2 \] \[ K_{sp,Q}=4S_Q^3 \] \(\textbf{Calculate Q's molar solubility:}\) \[ 4.00\times10^{-9}=4S_Q^3 \] \[ S_Q^3=1.00\times10^{-9} \] \[ S_Q=1.00\times10^{-3}\,\mathrm{M} \] \(\textbf{Compare the molar solubilities:}\) \[ \frac{S_Q}{S_P} = \frac{1.00\times10^{-3}}{1.00\times10^{-4}} \] \[ \frac{S_Q}{S_P}=10.0 \] \(\textbf{Interpret the apparently reversed ordering:}\) The numerical solubility product of Q is smaller than that of P. Nevertheless, Q has the greater molar solubility. The difference arises because the salts have different dissolution stoichiometries. For AB, Ksp varies as the square of molar solubility. For CD₂, Ksp varies as the cube of molar solubility multiplied by four. Therefore, numerical Ksp values cannot be used as a direct solubility ranking across different salt formulas. The balanced dissolution reactions must first be translated into the appropriate molar-solubility expressions. \(\textbf{Final answer:}\) \[ \boxed{S_P=1.00\times10^{-4}\,\mathrm{M},\quad S_Q=1.00\times10^{-3}\,\mathrm{M},\quad\frac{S_Q}{S_P}=10.0} \]

Explanation: \(\textbf{Determine the immediate post-mixing silver concentration:}\) Equal volumes are mixed. Before appreciable precipitation, the silver-ion concentration is halved: \[ [\mathrm{Ag^+}]_{\mathrm{mix}} =\frac{2.00\times10^{-10}}2 \] \[ [\mathrm{Ag^+}]_{\mathrm{mix}} =1.00\times10^{-10}\,\mathrm{M} \] \(\textbf{Determine the halide concentrations:}\) Both halide ions originate from solution Q and are diluted twofold. Therefore, \[ [\mathrm{Cl^-}]_{\mathrm{mix}}=\frac{0.0400}{2} \] \[ [\mathrm{Cl^-}]_{\mathrm{mix}}=0.0200\,\mathrm{M} \] And, \[ [\mathrm{Br^-}]_{\mathrm{mix}}=\frac{0.00400}{2} \] \[ [\mathrm{Br^-}]_{\mathrm{mix}}=0.00200\,\mathrm{M} \] \(\textbf{Evaluate the ionic product for AgCl:}\) For \[ \mathrm{AgCl(s)\rightleftharpoons Ag^++Cl^-} \] the instantaneous ionic product is \[ Q_{sp,\mathrm{AgCl}} =[\mathrm{Ag^+}][\mathrm{Cl^-}] \] Substituting, \[ Q_{sp,\mathrm{AgCl}} =(1.00\times10^{-10})(0.0200) \] \[ Q_{sp,\mathrm{AgCl}}=2.00\times10^{-12}\,\mathrm{M^2} \] Comparing, \[ Q_{sp,\mathrm{AgCl}}\lt K_{sp,\mathrm{AgCl}} \] Therefore, AgCl precipitation is not favoured. \(\textbf{Evaluate the ionic product for AgBr:}\) For \[ \mathrm{AgBr(s)\rightleftharpoons Ag^++Br^-} \] \[ Q_{sp,\mathrm{AgBr}} =[\mathrm{Ag^+}][\mathrm{Br^-}] \] Substituting, \[ Q_{sp,\mathrm{AgBr}} =(1.00\times10^{-10})(0.00200) \] \[ Q_{sp,\mathrm{AgBr}}=2.00\times10^{-13}\,\mathrm{M^2} \] Here, \[ Q_{sp,\mathrm{AgBr}}\gt K_{sp,\mathrm{AgBr}} \] Therefore, AgBr precipitation is favoured. \(\textbf{Interpret the selective tendency:}\) Although chloride is present at ten times the bromide concentration, AgBr has the much smaller solubility product. Each possible precipitate must be evaluated using its own ionic product and solubility product. Comparing only halide concentrations would be insufficient. Likewise, comparing Ksp values without calculating the post-mixing ionic products would be insufficient. The decision concerns the immediate thermodynamic tendency following mixing. It does not require calculation of the final equilibrium amount of precipitate. \(\textbf{Final answer:}\) AgBr alone is precipitation-favoured because \(Q_{sp,\mathrm{AgBr}}\gt K_{sp,\mathrm{AgBr}}\), while \(Q_{sp,\mathrm{AgCl}}\lt K_{sp,\mathrm{AgCl}}\).

Explanation: \(\textbf{Determine which silver salt precipitates first:}\) Before precipitation, the halide concentrations are equal. The threshold silver-ion concentration for each salt follows from \[ K_{sp}=[\mathrm{Ag^+}][\mathrm{X^-}] \] For AgBr, \[ [\mathrm{Ag^+}]_{\mathrm{onset,Br}} = \frac{10^{-13}}{0.0100} \] \[ [\mathrm{Ag^+}]_{\mathrm{onset,Br}}=10^{-11}\,\mathrm{M} \] For AgCl, \[ [\mathrm{Ag^+}]_{\mathrm{onset,Cl}} = \frac{10^{-10}}{0.0100} \] \[ [\mathrm{Ag^+}]_{\mathrm{onset,Cl}}=10^{-8}\,\mathrm{M} \] Therefore, AgBr begins precipitating first. \(\textbf{Identify the silver concentration when AgCl begins:}\) Before AgCl precipitation starts, its dissolved chloride concentration remains approximately \(0.0100\,\mathrm{M}\). At the onset of AgCl precipitation, \[ [\mathrm{Ag^+}][\mathrm{Cl^-}]=K_{sp}(\mathrm{AgCl}) \] Therefore, \[ [\mathrm{Ag^+}] = \frac{1.00\times10^{-10}}{0.0100} \] \[ [\mathrm{Ag^+}]=1.00\times10^{-8}\,\mathrm{M} \] \(\textbf{Determine the remaining dissolved bromide:}\) AgBr has already begun precipitating. The solution remains in equilibrium with solid AgBr, so \[ [\mathrm{Ag^+}][\mathrm{Br^-}]=K_{sp}(\mathrm{AgBr}) \] Therefore, \[ [\mathrm{Br^-}] = \frac{1.00\times10^{-13}}{1.00\times10^{-8}} \] \[ [\mathrm{Br^-}]=1.00\times10^{-5}\,\mathrm{M} \] \(\textbf{Calculate the fraction of bromide remaining dissolved:}\) \[ f_{\mathrm{remaining}} = \frac{[\mathrm{Br^-}]_{\mathrm{remaining}}} {[\mathrm{Br^-}]_0} \] \[ f_{\mathrm{remaining}} = \frac{1.00\times10^{-5}}{0.0100} \] \[ f_{\mathrm{remaining}}=1.00\times10^{-3} \] \(\textbf{Determine the fraction already precipitated:}\) \[ f_{\mathrm{precipitated}}=1-f_{\mathrm{remaining}} \] \[ f_{\mathrm{precipitated}}=1-0.00100 \] \[ f_{\mathrm{precipitated}}=0.999 \] Converting to percentage, \[ 100f_{\mathrm{precipitated}}=99.9\% \] \(\textbf{Interpret the selective precipitation:}\) The much smaller Ksp of AgBr allows it to precipitate at a much lower free silver-ion concentration. As additional silver nitrate is introduced, the bromide concentration progressively decreases while AgBr remains at saturation. By the time free silver ions reach the threshold for AgCl precipitation, only one-thousandth of the original bromide remains dissolved. The onset of the second precipitate therefore provides an indirect measure of how much of the first ion has been removed. The calculation combines two distinct saturation thresholds with a material-balance fraction. \(\textbf{Final answer:}\) \[ \boxed{99.9\%\ \text{of the initial bromide has precipitated}} \]

Explanation: \(\textbf{Establish the solubility-product expression:}\) For \[ \mathrm{MX_2(s)\rightleftharpoons M^{2+}+2X^-} \] \[ K_{sp}=[\mathrm{M^{2+}}][\mathrm{X^-}]^2 \] \(\textbf{Determine the pure-water solubility:}\) If the molar solubility is \(S_0\), \[ [\mathrm{M^{2+}}]=S_0 \] \[ [\mathrm{X^-}]=2S_0 \] Therefore, \[ K_{sp}=S_0(2S_0)^2 \] \[ K_{sp}=4S_0^3 \] Substituting, \[ 4.00\times10^{-12}=4S_0^3 \] \[ S_0^3=1.00\times10^{-12} \] \[ S_0=1.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Evaluate P with added divalent cation:}\) The added cation concentration is much greater than the amount produced by dissolution. Therefore, \[ [\mathrm{M^{2+}}]\approx0.0100\,\mathrm{M} \] If the new molar solubility is \(S_P\), \[ [\mathrm{X^-}]=2S_P \] Substituting into Ksp, \[ 4.00\times10^{-12} \approx(0.0100)(2S_P)^2 \] \[ 4.00\times10^{-12} \approx0.0400S_P^2 \] \[ S_P^2=1.00\times10^{-10} \] \[ S_P=1.00\times10^{-5}\,\mathrm{M} \] \(\textbf{Evaluate Q with added monovalent anion:}\) Here, \[ [\mathrm{X^-}]\approx0.0100\,\mathrm{M} \] If its molar solubility is \(S_Q\), \[ [\mathrm{M^{2+}}]=S_Q \] Therefore, \[ 4.00\times10^{-12} \approx S_Q(0.0100)^2 \] \[ 4.00\times10^{-12} \approx S_Q(1.00\times10^{-4}) \] \[ S_Q=4.00\times10^{-8}\,\mathrm{M} \] \(\textbf{Check the assumed common-ion dominance:}\) In P, \[ S_P=1.00\times10^{-5}\ll0.0100 \] In Q, the dissolution contribution to X⁻ is \[ 2S_Q=8.00\times10^{-8}\,\mathrm{M} \] This is also much smaller than \(0.0100\,\mathrm{M}\). Both approximations are therefore justified. \(\textbf{Interpret the different suppression factors:}\) The added divalent cation appears to the first power in the Ksp expression. The added monovalent anion appears to the second power. Consequently, equal numerical common-ion concentrations do not produce equal solubility suppression. The effect depends on the dissolution stoichiometry and the exponent of the added ion in the solubility-product expression. \(\textbf{Final answer:}\) \[ \boxed{S_P=1.00\times10^{-5}\,\mathrm{M},\qquad S_Q=4.00\times10^{-8}\,\mathrm{M}} \]

Explanation: \(\textbf{Determine the reference solubility without appreciable hydrolysis:}\) For the dissolution \[ \mathrm{MA(s)\rightleftharpoons M^++A^-} \] let the reference molar solubility be \(S_0\). Then, \[ [\mathrm{M^+}]=[\mathrm{A^-}]=S_0 \] Therefore, \[ K_{sp}=S_0^2 \] Substituting, \[ S_0=\sqrt{1.00\times10^{-8}} \] \[ S_0=1.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Identify the additional reaction in the acidic solution:}\) The dissolved A⁻ ions can accept protons and form HA. The acid equilibrium is \[ \mathrm{HA\rightleftharpoons H^++A^-} \] The external reservoir maintains the stated hydrogen-ion concentration. Consequently, the ratio of A⁻ to HA is fixed by Ka. \(\textbf{Calculate the protonation-state ratio:}\) \[ K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \] Rearranging, \[ \frac{[\mathrm{HA}]}{[\mathrm{A^-}]} =\frac{[\mathrm{H^+}]}{K_a} \] Substituting, \[ \frac{[\mathrm{HA}]}{[\mathrm{A^-}]} = \frac{9.90\times10^{-4}}{1.00\times10^{-5}} \] \[ \frac{[\mathrm{HA}]}{[\mathrm{A^-}]}=99.0 \] Therefore, \[ [\mathrm{HA}]=99[\mathrm{A^-}] \] \(\textbf{Apply the dissolved material balance:}\) Every dissolved MA formula unit supplies one M⁺ ion and one A-containing unit. Let the molar solubility in the acid-maintained solution be S. Then, \[ [\mathrm{M^+}]=S \] The total dissolved A-containing concentration is \[ [\mathrm{A^-}]+[\mathrm{HA}]=S \] Substituting the species ratio, \[ [\mathrm{A^-}]+99[\mathrm{A^-}]=S \] \[ 100[\mathrm{A^-}]=S \] Therefore, \[ [\mathrm{A^-}]=\frac S{100} \] \(\textbf{Apply the solubility-product equilibrium:}\) The dissolution equilibrium is governed by free ion concentrations: \[ K_{sp}=[\mathrm{M^+}][\mathrm{A^-}] \] Substituting, \[ 1.00\times10^{-8}=S\left(\frac S{100}\right) \] \[ 1.00\times10^{-8}=\frac{S^2}{100} \] \[ S^2=1.00\times10^{-6} \] \[ S=1.00\times10^{-3}\,\mathrm{M} \] \(\textbf{Determine the solubility enhancement:}\) \[ \frac S{S_0} = \frac{1.00\times10^{-3}}{1.00\times10^{-4}} \] \[ \frac S{S_0}=10.0 \] \(\textbf{Verify the dissolved species:}\) The free anion concentration is \[ [\mathrm{A^-}]=\frac{10^{-3}}{100} \] \[ [\mathrm{A^-}]=1.00\times10^{-5}\,\mathrm{M} \] The protonated concentration is \[ [\mathrm{HA}]=99(1.00\times10^{-5}) \] \[ [\mathrm{HA}]=9.90\times10^{-4}\,\mathrm{M} \] Their sum is \[ 1.00\times10^{-5}+9.90\times10^{-4} \] \[ =1.00\times10^{-3}\,\mathrm{M} \] The Ksp check gives \[ (1.00\times10^{-3})(1.00\times10^{-5}) =1.00\times10^{-8} \] \(\textbf{Interpret the coupled equilibria:}\) Acid protonates a large fraction of the dissolved A⁻ ions. This lowers the concentration of free A⁻ relative to the total amount of dissolved salt. Additional MA can consequently dissolve while maintaining the same solubility product. The intrinsic Ksp does not increase. Instead, acid-base protonation changes the distribution of dissolved A-containing species. The resulting molar solubility is greater because the free-ion concentration required by Ksp represents only a small fraction of the total dissolved anion-derived material. The result integrates solubility equilibrium, weak-acid ionization and a material balance under an imposed hydrogen-ion condition. \(\textbf{Final answer:}\) \[ \boxed{S=1.00\times10^{-3}\,\mathrm{M},\qquad\frac S{S_0}=10.0} \]