Equilibrium MCQs with Answers – Part 7 (NEET Chemistry)

Explanation: Statement I is correct. The ionic product is \[ Q_{sp}=[\mathrm{M^+}][\mathrm{X^-}] \] Substituting, \[ Q_{sp}=(2.00\times10^{-4})(1.00\times10^{-4}) \] \[ Q_{sp}=2.00\times10^{-8} \] The supplied solubility product is \[ K_{sp}=1.00\times10^{-8} \] Therefore, \[ Q_{sp}\gt K_{sp} \] The solution is supersaturated with respect to MX. Statement II is correct. The appearance of a solid precipitate may involve a nucleation process. Seed crystals provide an existing solid surface on which additional solid can form. Their addition can therefore facilitate the approach to precipitation equilibrium. However, the intrinsic Ksp depends on the defined equilibrium and temperature, not on whether seed crystals were initially present. Statement III is incorrect. A clear solution need not be thermodynamically unsaturated. Supersaturated solutions can remain visibly clear for some time if the formation of a new solid phase is kinetically hindered. Visual appearance alone therefore cannot replace comparison of Qsp with Ksp. Statement IV is correct. If pure MX solid and the dissolved ions coexist at equilibrium, the saturation condition is \[ [\mathrm{M^+}]_{\mathrm{eq}}[\mathrm{X^-}]_{\mathrm{eq}}=K_{sp} \] This equality describes the equilibrium composition. It is not a statement about the speed at which that composition is reached. The experiment distinguishes a thermodynamic tendency for precipitation from the kinetic process of forming visible solid. It also shows why a seed crystal can influence the observed onset of precipitation without changing the equilibrium constant. Statements I, II and IV are correct.

Explanation: \(\textbf{Construct the metal-hydroxide equilibrium expression:}\) The dissolution reaction is \[ \mathrm{M(OH)_2(s)\rightleftharpoons M^{2+}+2OH^-} \] Therefore, \[ K_{sp}=[\mathrm{M^{2+}}][\mathrm{OH^-}]^2 \] The exponent of two on hydroxide concentration is determined by the balanced dissolution equation. \(\textbf{Express hydroxide concentration through pH:}\) At \(298\,\mathrm{K}\), \[ pH+pOH=14.00 \] Therefore, \[ pOH=14.00-pH \] and \[ \log_{10}[\mathrm{OH^-}]=pH-14.00 \] where concentrations are expressed numerically relative to the standard molarity. \(\textbf{Rearrange the solubility product:}\) \[ [\mathrm{M^{2+}}]=\frac{K_{sp}}{[\mathrm{OH^-}]^2} \] Taking common logarithms, \[ \log_{10}[\mathrm{M^{2+}}] =\log_{10}K_{sp}-2\log_{10}[\mathrm{OH^-}] \] Substituting, \[ \log_{10}[\mathrm{M^{2+}}] =\log_{10}K_{sp}-2(pH-14.00) \] Thus, \[ \log_{10}[\mathrm{M^{2+}}] =(\log_{10}K_{sp}+28.00)-2pH \] \(\textbf{Identify the graph slope:}\) The coefficient of pH is \[ m=-2 \] This agrees with the two plotted values: \[ m=\frac{-8.00-(-6.00)}{11.00-10.00} \] \[ m=-2 \] \(\textbf{Determine Ksp from one measured point:}\) At pH 10.00, \[ pOH=14.00-10.00 \] \[ pOH=4.00 \] Therefore, \[ [\mathrm{OH^-}]=1.00\times10^{-4}\,\mathrm{M} \] The plotted metal-ion concentration is \[ [\mathrm{M^{2+}}]=1.00\times10^{-6}\,\mathrm{M} \] Substituting, \[ K_{sp}=(10^{-6})(10^{-4})^2 \] \[ K_{sp}=1.00\times10^{-14} \] \(\textbf{Verify using the second plotted point:}\) At pH 11.00, \[ [\mathrm{OH^-}]=10^{-3}\,\mathrm{M} \] and \[ [\mathrm{M^{2+}}]=10^{-8}\,\mathrm{M} \] Therefore, \[ K_{sp}=(10^{-8})(10^{-3})^2 \] \[ K_{sp}=1.00\times10^{-14} \] Both points give the same value. \(\textbf{Interpret the logarithmic graph:}\) Increasing pH by one unit raises the hydroxide-ion concentration tenfold at the specified temperature. Because hydroxide enters Ksp to the second power, the saturated free metal-ion concentration decreases by a factor of one hundred. Its common logarithm therefore decreases by two units. The graph slope directly reflects the hydroxide stoichiometric exponent, while the vertical position of the line determines the numerical Ksp. The analysis connects aqueous pH, a solid dissolution equilibrium and graphical inference. \(\textbf{Final answer:}\) \[ \boxed{m=-2,\qquad K_{sp}=1.00\times10^{-14}} \]

Explanation: \(\textbf{Use electrical neutrality to reconstruct the missing ion:}\) A bulk aqueous solution must contain equal total positive and negative charge. The positive-charge equivalent concentration is \[ C_+=2[\mathrm{M^{2+}}]+[\mathrm{Na^+}] \] Substituting, \[ C_+=2(0.0200)+0.0400 \] \[ C_+=0.0800\,\mathrm{mol\,L^{-1}} \] The negative-charge equivalent concentration is \[ C_-=[\mathrm{Cl^-}]+2[\mathrm{X^{2-}}] \] Therefore, \[ 0.0800=0.0400+2[\mathrm{X^{2-}}] \] \(\textbf{Solve for the missing ionic concentration:}\) \[ 2[\mathrm{X^{2-}}]=0.0400 \] \[ [\mathrm{X^{2-}}]=0.0200\,\mathrm{M} \] \(\textbf{Construct the precipitation reaction:}\) The possible solid is MX: \[ \mathrm{M^{2+}+X^{2-}\rightleftharpoons MX(s)} \] For the dissolution equilibrium, \[ K_{sp}=[\mathrm{M^{2+}}][\mathrm{X^{2-}}] \] Both ions carry charges of magnitude two, but the dissolution coefficients are one each. Their charges do not become powers in the Ksp expression. \(\textbf{Calculate the immediate ionic product:}\) \[ Q_{sp}=(0.0200)(0.0200) \] \[ Q_{sp}=4.00\times10^{-4} \] \(\textbf{Compare with the solubility product:}\) \[ K_{sp}=3.00\times10^{-4} \] Therefore, \[ Q_{sp}\gt K_{sp} \] The mixture is supersaturated and precipitation of MX is thermodynamically favoured. \(\textbf{Distinguish charge balance from precipitation stoichiometry:}\) Electrical neutrality determines the missing X²⁻ concentration because each X²⁻ contributes two units of negative charge. The solubility product, however, uses concentration powers determined by the balanced dissolution equation. These are different uses of stoichiometric information. The \(2+\) and \(2-\) ionic charges do not imply a squared dependence on both ion concentrations. \(\textbf{Interpret the experimental timing:}\) The concentrations are specified immediately after mixing, before substantial solid formation. They can therefore represent a supersaturated transient composition. As precipitation proceeds, the free concentrations of M²⁺ and X²⁻ decrease together. If solid remains at the final equilibrium, their ionic product reaches Ksp. The initial ionic product correctly identifies the thermodynamic direction but does not by itself give the final amount precipitated. The problem combines electrical neutrality, ionic-product construction and interpretation of a changing precipitation system. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{X^{2-}}]=0.0200\,\mathrm{M},\qquad Q_{sp}=4.00\times10^{-4}\gt K_{sp}} \]

Explanation: \(\textbf{Determine the original dissolved inventory:}\) The salt dissolves as \[ \mathrm{MX(s)\rightleftharpoons M^++X^-} \] In pure water, \[ K_{sp,1}=S_1^2 \] Therefore, \[ S_1=\sqrt{1.00\times10^{-4}} \] \[ S_1=0.0100\,\mathrm{M} \] The initial solution volume is \(1.00\,\mathrm{L}\). Thus, the dissolved amounts are \[ n_{\mathrm{M^+},0}=0.0100\,\mathrm{mol} \] \[ n_{\mathrm{X^-},0}=0.0100\,\mathrm{mol} \] \(\textbf{Account for the added common ion:}\) NaX supplies \[ n_{\mathrm{X^-,added}}=0.0100\,\mathrm{mol} \] The analytical X-derived ionic amount before precipitation is therefore \[ n_{\mathrm{X^-,total}}=0.0100+0.0100 \] \[ n_{\mathrm{X^-,total}}=0.0200\,\mathrm{mol} \] The total M-derived amount remains \(0.0100\,\mathrm{mol}\). \(\textbf{Check the immediate lower-temperature saturation condition:}\) Before precipitation, \[ [\mathrm{M^+}]=0.0100\,\mathrm{M} \] \[ [\mathrm{X^-}]=0.0200\,\mathrm{M} \] Therefore, \[ Q_{sp}=(0.0100)(0.0200) \] \[ Q_{sp}=2.00\times10^{-4} \] At \(T_2\), \[ K_{sp,2}=5.60\times10^{-5} \] Since \[ Q_{sp}\gt K_{sp,2} \] precipitation is favoured. \(\textbf{Represent the final dissolved-ion amounts:}\) Let x moles of MX precipitate. Because the volume is \(1.00\,\mathrm{L}\), \[ [\mathrm{M^+}]_f=0.0100-x \] \[ [\mathrm{X^-}]_f=0.0200-x \] The two concentrations decrease by equal amounts because MX contains one of each ion. \(\textbf{Apply the lower-temperature solubility product:}\) \[ (0.0100-x)(0.0200-x)=5.60\times10^{-5} \] Expanding, \[ 2.00\times10^{-4}-0.0300x+x^2 =5.60\times10^{-5} \] \[ x^2-0.0300x+1.44\times10^{-4}=0 \] Factorising, \[ (x-0.00600)(x-0.0240)=0 \] \(\textbf{Select the chemically feasible root:}\) The solution initially contains only \(0.0100\,\mathrm{mol}\) M-derived material. Therefore, \(0.0240\,\mathrm{mol}\) MX cannot precipitate. The permissible root is \[ x=0.00600\,\mathrm{mol} \] \(\textbf{Recover the final free-ion concentrations:}\) \[ [\mathrm{M^+}]_f=0.0100-0.00600 \] \[ [\mathrm{M^+}]_f=0.00400\,\mathrm{M} \] \[ [\mathrm{X^-}]_f=0.0200-0.00600 \] \[ [\mathrm{X^-}]_f=0.0140\,\mathrm{M} \] \(\textbf{Verify the final equilibrium:}\) \[ Q_{sp,f}=(0.00400)(0.0140) \] \[ Q_{sp,f}=5.60\times10^{-5} \] This equals the lower-temperature Ksp. \(\textbf{Interpret the combined disturbances:}\) Cooling lowers the intrinsic solubility product in the specified system. Introducing NaX simultaneously raises the initial X⁻ concentration. Both changes favour precipitation relative to the original saturated solution. However, the final amount precipitated is not the sum of two separately calculated precipitation amounts. The two disturbances act on the same dissolved-ion material balances and must be treated simultaneously. The amount of MX precipitated is limited by the available M-derived inventory, while excess X⁻ from NaX remains in solution. The correct result therefore requires a changed equilibrium constant, common-ion accounting, precipitation stoichiometry and a physically feasible algebraic root. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{MX,precipitated}}=0.00600\,\mathrm{mol},\qquad [\mathrm{M^+}]_f=0.00400\,\mathrm{M}} \]

Explanation: \(\textbf{Identify the source of hydroxide ions:}\) NaA dissociates completely into Na⁺ and A⁻. Its anion is the conjugate base of a weak acid and undergoes hydrolysis: \[ \mathrm{A^-+H_2O\rightleftharpoons HA+OH^-} \] The hydrolysis constant is \[ K_h=\frac{K_w}{K_a} \] \[ K_h=\frac{10^{-14}}{10^{-9}} \] \[ K_h=1.00\times10^{-5} \] \(\textbf{Estimate the hydroxide concentration before magnesium addition:}\) For small hydrolysis, \[ [\mathrm{OH^-}]\approx\sqrt{K_hC} \] \[ [\mathrm{OH^-}]\approx\sqrt{(1.00\times10^{-5})(0.0400)} \] \[ [\mathrm{OH^-}]\approx\sqrt{4.00\times10^{-7}} \] \[ [\mathrm{OH^-}]\approx6.32\times10^{-4}\,\mathrm{M} \] The estimated hydrolysis fraction is \[ h\approx\frac{6.32\times10^{-4}}{0.0400} \] \[ h\approx0.0158 \] The small-hydrolysis approximation is therefore reasonable. \(\textbf{Calculate the NaA solution pH:}\) \[ pOH\approx-\log_{10}(6.32\times10^{-4}) \] \[ pOH\approx3.199 \] \[ pH\approx14.000-3.199 \] \[ pH\approx10.80 \] \(\textbf{Test magnesium hydroxide saturation in NaA:}\) The ionic product is \[ Q_{sp}=[\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2 \] Immediately after magnesium addition, \[ Q_{sp,\mathrm{NaA}}\approx(10^{-5})(6.32\times10^{-4})^2 \] \[ Q_{sp,\mathrm{NaA}}\approx4.00\times10^{-12} \] Therefore, \[ Q_{sp,\mathrm{NaA}}\gt K_{sp} \] Mg(OH)₂ precipitation is thermodynamically favoured. \(\textbf{Test the NaCl solution separately:}\) NaCl is a salt of a strong acid and strong base. Under the usual \(298\,\mathrm{K}\) treatment, it undergoes negligible hydrolysis and is approximately neutral. Hence, \[ [\mathrm{OH^-}]\approx1.00\times10^{-7}\,\mathrm{M} \] The corresponding ionic product is \[ Q_{sp,\mathrm{NaCl}}\approx(10^{-5})(10^{-7})^2 \] \[ Q_{sp,\mathrm{NaCl}}\approx1.00\times10^{-19} \] Thus, \[ Q_{sp,\mathrm{NaCl}}\lt K_{sp} \] Precipitation is not favoured in the NaCl solution. \(\textbf{Interpret the coupled equilibria:}\) NaA raises the hydroxide concentration through anion hydrolysis. Magnesium hydroxide saturation depends on the square of that concentration. Even though both solutions receive the same magnesium-ion amount, their precipitation tendencies differ markedly. The added magnesium can subsequently consume hydroxide ions through precipitation, so the calculated NaA pH represents the initial hydrolysis condition, not necessarily the final pH after extensive solid formation. The problem links salt hydrolysis, pH and the ionic-product criterion for precipitation. Only the NaA solution is initially supersaturated with respect to Mg(OH)₂.

Explanation: \(\textbf{Identify the hydrolysis equilibrium:}\) The hydrolysing species is A⁻: \[ \mathrm{A^-+H_2O\rightleftharpoons HA+OH^-} \] Its hydrolysis constant is \[ K_h=\frac{K_w}{K_a} \] Although Ka is stipulated to remain unchanged, Kw differs between the two temperatures. Consequently, the hydrolysis constant also changes. \(\textbf{Evaluate the equilibrium at T₁:}\) \[ K_{h,1}=\frac{10^{-14}}{10^{-5}} \] \[ K_{h,1}=1.00\times10^{-9} \] Under the small-hydrolysis approximation, \[ [\mathrm{OH^-}]_1\approx\sqrt{K_{h,1}C} \] \[ [\mathrm{OH^-}]_1\approx\sqrt{(10^{-9})(10^{-2})} \] \[ [\mathrm{OH^-}]_1\approx3.16\times10^{-6}\,\mathrm{M} \] Therefore, \[ pOH_1\approx5.50 \] Since \[ pK_{w,1}=14.00 \] we obtain \[ pH_1\approx14.00-5.50 \] \[ pH_1\approx8.50 \] \(\textbf{Evaluate the changed hydrolysis constant at T₂:}\) \[ K_{h,2}=\frac{10^{-12}}{10^{-5}} \] \[ K_{h,2}=1.00\times10^{-7} \] Thus, \[ [\mathrm{OH^-}]_2\approx\sqrt{(10^{-7})(10^{-2})} \] \[ [\mathrm{OH^-}]_2\approx3.16\times10^{-5}\,\mathrm{M} \] Therefore, \[ pOH_2\approx4.50 \] \(\textbf{Use the correct pKw at T₂:}\) \[ pK_{w,2}=12.00 \] Hence, \[ pH_2\approx12.00-4.50 \] \[ pH_2\approx7.50 \] \(\textbf{Check the hydrolysis approximation:}\) At \(T_2\), the approximate hydrolysis fraction is \[ h_2\approx\frac{3.16\times10^{-5}}{0.0100} \] \[ h_2\approx0.00316 \] Thus, only about \(0.316\%\) of the salt anions undergo hydrolysis, supporting the approximation. \(\textbf{Interpret the opposite numerical changes:}\) The hydroxide-ion concentration increases tenfold on heating under the stated constant-Ka assumption. Nevertheless, the calculated pH decreases by one unit. This occurs because the ionic product of water increases by a factor of one hundred. At \(T_2\), the neutral pH is 6.00 rather than 7.00. The salt solution is therefore still basic at pH 7.50. Its basic character must be assessed relative to the neutral pH at that temperature, not by applying the \(298\,\mathrm{K}\) benchmark indiscriminately. The result connects temperature-dependent water ionization, salt hydrolysis and the definition of pH. \(\textbf{Final answer:}\) \[ \boxed{pH_{T_1}\approx8.50,\qquad pH_{T_2}\approx7.50} \]

Explanation: \(\textbf{Identify the separate hydrolysis reactions:}\) Fluoride is the conjugate base of weak acid HF: \[ \mathrm{F^-+H_2O\rightleftharpoons HF+OH^-} \] Ammonium is the conjugate acid of weak base NH₃: \[ \mathrm{NH_4^++H_2O\rightleftharpoons NH_3+H_3O^+} \] The two ions therefore hydrolyse in different directions. \(\textbf{Calculate the pH of NaF solution P:}\) The fluoride hydrolysis constant is \[ K_{h,F}=\frac{K_w}{K_a(\mathrm{HF})} \] \[ K_{h,F}=\frac{10^{-14}}{10^{-4}} \] \[ K_{h,F}=1.00\times10^{-10} \] Therefore, \[ [\mathrm{OH^-}]_P\approx\sqrt{(10^{-10})(0.100)} \] \[ [\mathrm{OH^-}]_P\approx3.16\times10^{-6}\,\mathrm{M} \] \[ pOH_P\approx5.50 \] \[ pH_P\approx8.50 \] \(\textbf{Calculate the pH of NH₄Cl solution Q:}\) For NH₄⁺, \[ K_{h,NH_4}=\frac{K_w}{K_b(\mathrm{NH_3})} \] \[ K_{h,NH_4}=1.00\times10^{-9} \] Therefore, \[ [\mathrm{H^+}]_Q\approx\sqrt{(10^{-9})(0.100)} \] \[ [\mathrm{H^+}]_Q\approx1.00\times10^{-5}\,\mathrm{M} \] \[ pH_Q\approx5.00 \] \(\textbf{Recognize the chemical change on mixing:}\) Mixing introduces NH₄⁺ and F⁻ into the same solution. Both ions can participate in proton-transfer equilibrium: \[ \mathrm{NH_4^++F^-\rightleftharpoons NH_3+HF} \] Their analytical concentrations become approximately equal because equal volumes of equally concentrated salts were mixed. The combined solution therefore behaves, to the usual weak-acid/weak-base salt approximation, like an aqueous ammonium fluoride system. \(\textbf{Determine the mixture pH:}\) For such a system at \(298\,\mathrm{K}\), \[ pH\approx7+\frac12\log_{10} \left(\frac{K_b(\mathrm{NH_3})}{K_a(\mathrm{HF})}\right) \] Substituting, \[ pH_{\mathrm{mix}}\approx7+\frac12\log_{10} \left(\frac{10^{-5}}{10^{-4}}\right) \] \[ pH_{\mathrm{mix}}\approx7+\frac12\log_{10}(10^{-1}) \] \[ pH_{\mathrm{mix}}\approx7-0.50 \] \[ pH_{\mathrm{mix}}\approx6.50 \] \(\textbf{Compare with the arithmetic pH average:}\) The arithmetic mean of the two original pH values is \[ \frac{8.50+5.00}{2}=6.75 \] This differs from the equilibrium result. Hydrogen-ion concentrations are logarithmically related to pH, and the conjugate species react after mixing. Consequently, averaging the original pH values is not a valid substitute for equilibrium calculation. The final mixture is mildly acidic because the relevant parent acid and base strengths are unequal. The calculation connects individual salt hydrolysis, solution mixing and simultaneous weak-acid/weak-base equilibrium. \(\textbf{Final answer:}\) \[ \boxed{pH_P\approx8.50,\qquad pH_Q\approx5.00,\qquad pH_{\mathrm{mix}}\approx6.50} \]

Explanation: \(\textbf{Recover the hydroxide concentration from pH:}\) At \(298\,\mathrm{K}\), \[ pOH=14.00-pH \] \[ pOH=14.00-10.00 \] \[ pOH=4.00 \] Therefore, \[ [\mathrm{OH^-}]=1.00\times10^{-4}\,\mathrm{M} \] Also, \[ [\mathrm{H^+}]=1.00\times10^{-10}\,\mathrm{M} \] \(\textbf{Identify the coupled dissolution and hydrolysis:}\) The salt dissolves as \[ \mathrm{MA(s)\rightleftharpoons M^++A^-} \] The anion hydrolyses: \[ \mathrm{A^-+H_2O\rightleftharpoons HA+OH^-} \] Because HA is electrically neutral, hydrolysis changes the free A⁻ concentration without removing the corresponding dissolved M⁺. \(\textbf{Combine material and charge balances:}\) Let S be the total molar solubility of MA. Every dissolved formula unit provides one M⁺ ion and one acid-derived A unit. Therefore, \[ [\mathrm{M^+}]=S \] \[ S=[\mathrm{A^-}]+[\mathrm{HA}] \] Electrical neutrality requires \[ [\mathrm{M^+}]+[\mathrm{H^+}] =[\mathrm{A^-}]+[\mathrm{OH^-}] \] Subtracting the acid-derived material balance gives \[ [\mathrm{HA}]=[\mathrm{OH^-}]-[\mathrm{H^+}] \] Thus, \[ [\mathrm{HA}]\approx1.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Use the acid-ionization constant to recover free A⁻:}\) \[ K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \] Rearranging, \[ [\mathrm{A^-}]=\frac{K_a[\mathrm{HA}]}{[\mathrm{H^+}]} \] Substituting, \[ [\mathrm{A^-}]\approx \frac{(1.00\times10^{-9})(1.00\times10^{-4})}{1.00\times10^{-10}} \] \[ [\mathrm{A^-}]\approx1.00\times10^{-3}\,\mathrm{M} \] \(\textbf{Determine total salt solubility:}\) \[ S=[\mathrm{A^-}]+[\mathrm{HA}] \] \[ S\approx0.00100+0.000100 \] \[ S\approx0.00110\,\mathrm{M} \] Consequently, \[ [\mathrm{M^+}]\approx0.00110\,\mathrm{M} \] \(\textbf{Calculate Ksp from the free-ion concentrations:}\) \[ K_{sp}=[\mathrm{M^+}][\mathrm{A^-}] \] \[ K_{sp}\approx(0.00110)(0.00100) \] \[ K_{sp}\approx1.10\times10^{-6} \] \(\textbf{Interpret why S² is not the correct numerical substitution:}\) For a nonhydrolysing 1:1 salt dissolved in pure water, \[ K_{sp}=S^2 \] may apply. Here, however, some dissolved A⁻ has been converted into neutral HA. The total dissolved amount of acid-derived material exceeds the free A⁻ concentration. Therefore, \[ [\mathrm{M^+}]\ne[\mathrm{A^-}] \] at saturation. The intrinsic Ksp expression is unchanged, but its evaluation must use the actual free-ion concentrations. The measured pH reveals the extent of hydrolysis, permitting reconstruction of the dissolved species and the solubility product. \(\textbf{Final answer:}\) \[ \boxed{K_{sp}\approx1.10\times10^{-6}} \]

Explanation: \(\textbf{Recognize the shared dissolved anion:}\) Both salts release A⁻. The free A⁻ concentration therefore receives contributions from dissolution of both solids. It cannot be assigned separately for each salt as though the two solids occupied independent solutions. \(\textbf{Write the two simultaneous saturation conditions:}\) For MA, \[ K_{sp,MA}=[\mathrm{M^+}][\mathrm{A^-}] \] \[ [\mathrm{M^+}]=\frac{1.00\times10^{-8}}{[\mathrm{A^-}]} \] For NA, \[ K_{sp,NA}=[\mathrm{N^+}][\mathrm{A^-}] \] \[ [\mathrm{N^+}]=\frac{9.00\times10^{-8}}{[\mathrm{A^-}]} \] \(\textbf{Use electrical neutrality:}\) The solution contains M⁺, N⁺ and A⁻ as its significant dissolved ions. Therefore, \[ [\mathrm{A^-}]=[\mathrm{M^+}]+[\mathrm{N^+}] \] Substituting the two saturation relations, \[ [\mathrm{A^-}] =\frac{1.00\times10^{-8}}{[\mathrm{A^-}]} +\frac{9.00\times10^{-8}}{[\mathrm{A^-}]} \] Multiplying by \([\mathrm{A^-}]\), \[ [\mathrm{A^-}]^2=1.00\times10^{-7} \] Therefore, \[ [\mathrm{A^-}]=\sqrt{1.00\times10^{-7}} \] \[ [\mathrm{A^-}]\approx3.162\times10^{-4}\,\mathrm{M} \] \(\textbf{Recover the individual cation concentrations:}\) \[ [\mathrm{M^+}] =\frac{1.00\times10^{-8}}{3.162\times10^{-4}} \] \[ [\mathrm{M^+}]\approx3.162\times10^{-5}\,\mathrm{M} \] Similarly, \[ [\mathrm{N^+}] =\frac{9.00\times10^{-8}}{3.162\times10^{-4}} \] \[ [\mathrm{N^+}]\approx2.846\times10^{-4}\,\mathrm{M} \] \(\textbf{Verify the common-anion balance:}\) \[ [\mathrm{M^+}]+[\mathrm{N^+}] \approx3.162\times10^{-5}+2.846\times10^{-4} \] \[ [\mathrm{M^+}]+[\mathrm{N^+}] \approx3.162\times10^{-4}\,\mathrm{M} \] This equals the calculated A⁻ concentration. \(\textbf{Check both equilibrium products:}\) \[ (3.162\times10^{-5})(3.162\times10^{-4}) \approx1.00\times10^{-8} \] \[ (2.846\times10^{-4})(3.162\times10^{-4}) \approx9.00\times10^{-8} \] Both solid phases satisfy their respective Ksp values. \(\textbf{Interpret simultaneous dissolution:}\) If MA were dissolved alone in pure water, its molar solubility would be \(1.00\times10^{-4}\,\mathrm{M}\). If NA were dissolved alone, its molar solubility would be \(3.00\times10^{-4}\,\mathrm{M}\). Those separate-solution values cannot simply be combined because each salt contributes the same anion. The shared A⁻ concentration suppresses dissolution of both solids relative to their independent pure-water behaviour. This is a common-ion effect generated internally by two sparingly soluble salts rather than by an externally added highly soluble electrolyte. The simultaneous solubility products and electrical neutrality determine the complete composition. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{M^+}]\approx3.16\times10^{-5}\,\mathrm{M},\qquad [\mathrm{N^+}]\approx2.85\times10^{-4}\,\mathrm{M}} \]

Explanation: \(\textbf{Identify the coupled equilibria and finite acid inventory:}\) The dissolution equilibrium is \[ \mathrm{MA(s)\rightleftharpoons M^++A^-} \] The acid-base equilibrium is \[ \mathrm{HA\rightleftharpoons H^++A^-} \] A⁻ can accept hydrogen ions and form HA. However, the initially added HCl provides a finite hydrogen-ion inventory. The hydrogen-ion concentration therefore need not remain equal to its original \(0.00100\,\mathrm{M}\) value. \(\textbf{Express total dissolved salt through its free and protonated anions:}\) Let S denote the total molar amount of MA dissolved per litre. Because every dissolved formula unit supplies one M⁺, \[ [\mathrm{M^+}]=S \] Conservation of acid-derived material gives \[ S=[\mathrm{A^-}]+[\mathrm{HA}] \] \(\textbf{Use the chloride concentration and charge balance:}\) Chloride remains in solution at \[ [\mathrm{Cl^-}]=0.00100\,\mathrm{M} \] Let \[ h=[\mathrm{H^+}] \] Electrical neutrality, neglecting hydroxide ions, requires \[ [\mathrm{M^+}]+h=[\mathrm{A^-}]+0.00100 \] Subtracting the acid-derived material balance yields \[ [\mathrm{HA}]+h=0.00100 \] Therefore, \[ [\mathrm{HA}]=0.00100-h \] This equation records consumption of the initially introduced acid through protonation of A⁻. \(\textbf{Express free A⁻ using Ka:}\) \[ K_a=\frac{h[\mathrm{A^-}]}{[\mathrm{HA}]} \] Hence, \[ [\mathrm{A^-}]=\frac{K_a(0.00100-h)}{h} \] The corresponding total solubility is \[ S=(0.00100-h)\left(1+\frac{K_a}{h}\right) \] \(\textbf{Apply the solubility product:}\) \[ K_{sp}=S[\mathrm{A^-}] \] Substituting the preceding expressions, \[ 8.91\times10^{-8} =(0.00100-h)^2 \left(1+\frac{10^{-5}}{h}\right) \left(\frac{10^{-5}}{h}\right) \] \(\textbf{Check the physically consistent solution:}\) Take \[ h=1.00\times10^{-4}\,\mathrm{M} \] Then, \[ [\mathrm{HA}]=0.00100-0.000100 \] \[ [\mathrm{HA}]=9.00\times10^{-4}\,\mathrm{M} \] The ionization relation gives \[ [\mathrm{A^-}] =\frac{(10^{-5})(9.00\times10^{-4})}{10^{-4}} \] \[ [\mathrm{A^-}]=9.00\times10^{-5}\,\mathrm{M} \] Therefore, \[ S=[\mathrm{HA}]+[\mathrm{A^-}] \] \[ S=9.00\times10^{-4}+9.00\times10^{-5} \] \[ S=9.90\times10^{-4}\,\mathrm{M} \] \(\textbf{Verify both equilibria:}\) The dissolution ionic product is \[ Q_{sp}=(9.90\times10^{-4})(9.00\times10^{-5}) \] \[ Q_{sp}=8.91\times10^{-8}=K_{sp} \] The acid-ionization quotient is \[ Q_a=\frac{(10^{-4})(9.00\times10^{-5})}{9.00\times10^{-4}} \] \[ Q_a=1.00\times10^{-5}=K_a \] Thus, both equilibrium relations are satisfied. \(\textbf{Determine the pH:}\) \[ pH=-\log_{10}(10^{-4}) \] \[ pH=4.00 \] \(\textbf{Interpret the finite-reservoir effect:}\) The initially added strong acid promotes dissolution by protonating A⁻. As dissolution proceeds, some of the original acid is consumed and the free hydrogen-ion concentration decreases. The final pH must therefore be calculated from the coupled material balances rather than held fixed at its initial value. The total solubility includes both free A⁻ and neutral HA derived from dissolved MA. Only free A⁻ appears in Ksp. The result combines acid consumption, protonation equilibrium, electrical neutrality and salt dissolution. \(\textbf{Final answer:}\) \[ \boxed{pH=4.00,\qquad S=9.90\times10^{-4}\,\mathrm{M}} \]

Explanation: \(\textbf{Determine the original ionic inventories:}\) MCl supplies M⁺ in a \(1:1\) ratio. Therefore, \[ n_{\mathrm{M^+},0}=(0.100)(0.1000) \] \[ n_{\mathrm{M^+},0}=0.0100\,\mathrm{mol} \] NaX similarly supplies X⁻: \[ n_{\mathrm{X^-},0}=(0.0600)(0.1000) \] \[ n_{\mathrm{X^-},0}=0.00600\,\mathrm{mol} \] \(\textbf{Account for the final solution volume:}\) The mixed volume is \[ V_f=0.1000+0.1000 \] \[ V_f=0.2000\,\mathrm{L} \] The ionic concentrations immediately after mixing, before precipitation, are \[ [\mathrm{M^+}]_0=\frac{0.0100}{0.2000} \] \[ [\mathrm{M^+}]_0=0.0500\,\mathrm{M} \] \[ [\mathrm{X^-}]_0=\frac{0.00600}{0.2000} \] \[ [\mathrm{X^-}]_0=0.0300\,\mathrm{M} \] \(\textbf{Translate the collected solid into ion consumption:}\) The precipitation reaction consumes one M⁺ and one X⁻ per formula unit of MX. The collected amount is \[ n_{\mathrm{MX}}=0.00200\,\mathrm{mol} \] Therefore, \[ n_{\mathrm{M^+,consumed}}=n_{\mathrm{X^-,consumed}}=0.00200\,\mathrm{mol} \] \(\textbf{Determine the final dissolved amounts:}\) \[ n_{\mathrm{M^+},f}=0.0100-0.00200 \] \[ n_{\mathrm{M^+},f}=0.00800\,\mathrm{mol} \] \[ n_{\mathrm{X^-},f}=0.00600-0.00200 \] \[ n_{\mathrm{X^-},f}=0.00400\,\mathrm{mol} \] \(\textbf{Convert to the equilibrium concentrations:}\) \[ [\mathrm{M^+}]_f=\frac{0.00800}{0.2000} \] \[ [\mathrm{M^+}]_f=0.0400\,\mathrm{M} \] \[ [\mathrm{X^-}]_f=\frac{0.00400}{0.2000} \] \[ [\mathrm{X^-}]_f=0.0200\,\mathrm{M} \] \(\textbf{Recover the solubility product:}\) Because solid MX remains at equilibrium, \[ K_{sp}=[\mathrm{M^+}]_f[\mathrm{X^-}]_f \] Therefore, \[ K_{sp}=(0.0400)(0.0200) \] \[ K_{sp}=8.00\times10^{-4} \] \(\textbf{Check the initial precipitation tendency:}\) The initial ionic product was \[ Q_{sp,0}=(0.0500)(0.0300) \] \[ Q_{sp,0}=1.50\times10^{-3} \] Thus, \[ Q_{sp,0}\gt K_{sp} \] Precipitation from the initial mixture was thermodynamically consistent. \(\textbf{Interpret the reverse equilibrium determination:}\) The measured solid amount supplies information about how many moles of each dissolved ion have been removed. Those material changes must be applied before calculating Ksp. The original mixed concentrations give the initial reaction quotient, not the equilibrium constant. The final free-ion concentrations are unequal because the original ionic inventories were unequal. The correct Ksp follows from the saturated solution remaining after precipitation. The problem reverses the usual calculation: experimental precipitate quantity is used to infer the equilibrium constant rather than using Ksp to predict precipitate quantity. \(\textbf{Final answer:}\) \[ \boxed{K_{sp}=8.00\times10^{-4}} \]

Explanation: \(\textbf{Determine the precipitation-onset hydroxide concentration:}\) The metal hydroxide dissolves according to \[ \mathrm{M(OH)_2(s)\rightleftharpoons M^{2+}+2OH^-} \] At the onset of precipitation, \[ [\mathrm{M^{2+}}][\mathrm{OH^-}]^2=K_{sp} \] The metal concentration is approximately unchanged immediately before precipitation: \[ [\mathrm{M^{2+}}]\approx0.0100\,\mathrm{M} \] Therefore, \[ [\mathrm{OH^-}]_{\mathrm{onset}} =\sqrt{\frac{1.00\times10^{-12}}{0.0100}} \] \[ [\mathrm{OH^-}]_{\mathrm{onset}} =1.00\times10^{-5}\,\mathrm{M} \] \(\textbf{Convert the threshold to pH:}\) \[ pOH_{\mathrm{onset}}=5.00 \] At \(298\,\mathrm{K}\), \[ pH_{\mathrm{onset}}=14.00-5.00 \] \[ pH_{\mathrm{onset}}=9.00 \] \(\textbf{Identify the ammonia-buffer ratio at this pH:}\) For ammonia and ammonium, \[ pK_b=5.00 \] \[ pK_a(\mathrm{NH_4^+})=pK_w-pK_b \] \[ pK_a(\mathrm{NH_4^+})=14.00-5.00 \] \[ pK_a(\mathrm{NH_4^+})=9.00 \] The buffer relation is \[ pH=pK_a+\log_{10}\frac{n_{\mathrm{NH_3}}}{n_{\mathrm{NH_4^+}}} \] At pH 9.00, \[ \log_{10}\frac{n_{\mathrm{NH_3}}}{n_{\mathrm{NH_4^+}}}=0 \] Therefore, \[ n_{\mathrm{NH_3}}=n_{\mathrm{NH_4^+}} \] \(\textbf{Determine the required neutralization:}\) Added hydroxide reacts with ammonium: \[ \mathrm{NH_4^++OH^-\rightarrow NH_3+H_2O} \] Let x moles of NaOH be added. Before precipitation begins, \[ n_{\mathrm{NH_3}}=0.0300+x \] \[ n_{\mathrm{NH_4^+}}=0.0700-x \] At the threshold, these amounts must be equal: \[ 0.0300+x=0.0700-x \] \[ 2x=0.0400 \] \[ x=0.0200\,\mathrm{mol} \] \(\textbf{Verify the buffer composition at onset:}\) After adding \(0.0200\,\mathrm{mol}\) NaOH, \[ n_{\mathrm{NH_3}}=0.0500\,\mathrm{mol} \] \[ n_{\mathrm{NH_4^+}}=0.0500\,\mathrm{mol} \] Their ratio is one. Therefore, the buffer pH equals the conjugate-acid pKa of 9.00, as required for the precipitation threshold. \(\textbf{Interpret the coupled buffer–solubility problem:}\) The amount of added NaOH cannot be inferred merely from the hydroxide concentration required for precipitation. Most of the added strong base is consumed by ammonium ions while the buffer composition changes. The precipitation threshold first fixes a pH through Ksp. That pH then fixes the required NH₃/NH₄⁺ ratio. Finally, neutralization stoichiometry determines the NaOH amount needed to reach that ratio. Once M(OH)₂ begins precipitating, further addition of base can be affected by hydroxide consumption into the solid, so the pre-onset buffer calculation should not be extrapolated indefinitely. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{NaOH,onset}}\approx0.0200\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the contributions to dissolution enthalpy:}\) The lattice dissociation enthalpy represents the energy absorbed when one mole of an ionic solid separates into its constituent gaseous ions. The total ionic hydration enthalpy represents the energy change when those gaseous ions become hydrated in water. Lattice dissociation is endothermic, whereas hydration is generally exothermic. Under the stated assumptions, the molar enthalpy change of dissolution is obtained by adding these contributions: \[ \Delta_{\mathrm{sol}}H =\Delta_{\mathrm{lattice}}H+\Delta_{\mathrm{hyd}}H \] \(\textbf{Evaluate Statement I:}\) For solid P (MX), \[ \Delta_{\mathrm{sol}}H_P=800+(-820) \] \[ \Delta_{\mathrm{sol}}H_P=-20\,\mathrm{kJ\,mol^{-1}} \] The negative enthalpy change indicates that dissolution of P is exothermic. Therefore, Statement I is correct. \(\textbf{Evaluate Statement II:}\) For solid Q (NY), \[ \Delta_{\mathrm{sol}}H_Q=900+(-850) \] \[ \Delta_{\mathrm{sol}}H_Q=+50\,\mathrm{kJ\,mol^{-1}} \] The positive enthalpy change indicates that dissolution of Q is endothermic. Therefore, Statement II is correct. \(\textbf{Evaluate Statement III:}\) The lattice dissociation enthalpy alone does not determine the equilibrium solubility of an ionic solid. Although P has a smaller lattice dissociation enthalpy than Q, hydration and entropy effects must also be considered. The equilibrium constant of dissolution is governed by the standard Gibbs energy change: \[ \Delta_{\mathrm{sol}}G^\circ =\Delta_{\mathrm{sol}}H^\circ-T\Delta_{\mathrm{sol}}S^\circ \] The relationship between standard Gibbs energy and the equilibrium constant is \[ \Delta_{\mathrm{sol}}G^\circ=-RT\ln K \] Thus, the relative solubilities cannot be determined solely from the supplied lattice dissociation enthalpies. The available information does not establish that P is more soluble than Q at every temperature. Therefore, Statement III is incorrect. \(\textbf{Evaluate Statement IV:}\) The temperature dependence of an equilibrium constant is related to the standard enthalpy change through the van't Hoff equation: \[ \frac{d\ln K}{dT} =\frac{\Delta_{\mathrm{sol}}H^\circ}{RT^2} \] For dissolution of P, \[ \Delta_{\mathrm{sol}}H_P=-20\,\mathrm{kJ\,mol^{-1}} \] Because the dissolution is exothermic, increasing temperature tends to decrease its dissolution equilibrium constant. For dissolution of Q, \[ \Delta_{\mathrm{sol}}H_Q=+50\,\mathrm{kJ\,mol^{-1}} \] Because the dissolution is endothermic, increasing temperature tends to increase its dissolution equilibrium constant. These conclusions apply over the small temperature interval for which the stated enthalpy assumptions remain valid. Therefore, Statement IV is correct. \(\textbf{Final conclusion:}\) Statements I and II correctly identify the signs of the dissolution enthalpies. Statement III is incorrect because lattice dissociation enthalpy alone cannot establish a universal solubility ranking. Statement IV correctly predicts the opposite temperature dependences of the two dissolution equilibrium constants. Hence, Statements I, II and IV are correct.

Explanation: \(\textbf{Establish the initial saturation condition:}\) AgCl dissolves according to \[ \mathrm{AgCl(s)\rightleftharpoons Ag^++Cl^-} \] Its solubility product is \[ K_{sp}=[\mathrm{Ag^+}][\mathrm{Cl^-}] \] The added NaCl supplies approximately \[ [\mathrm{Cl^-}]_1=0.0100\,\mathrm{M} \] The contribution from dissolved AgCl is negligible relative to this concentration. \(\textbf{Determine the original saturated silver-ion concentration:}\) \[ [\mathrm{Ag^+}]_1=\frac{K_{sp}}{[\mathrm{Cl^-}]_1} \] \[ [\mathrm{Ag^+}]_1=\frac{1.00\times10^{-10}}{0.0100} \] \[ [\mathrm{Ag^+}]_1=1.00\times10^{-8}\,\mathrm{M} \] \(\textbf{Calculate the original dissolved silver amount:}\) \[ n_{\mathrm{Ag^+},1}=[\mathrm{Ag^+}]_1V_1 \] \[ n_{\mathrm{Ag^+},1}=(1.00\times10^{-8})(0.1000) \] \[ n_{\mathrm{Ag^+},1}=1.00\times10^{-9}\,\mathrm{mol} \] \(\textbf{Determine the chloride concentration after dilution:}\) The sodium chloride amount remains approximately constant. The solution volume increases tenfold. Therefore, \[ [\mathrm{Cl^-}]_2\approx0.0100\frac{0.1000}{1.00} \] \[ [\mathrm{Cl^-}]_2\approx0.00100\,\mathrm{M} \] \(\textbf{Apply saturation at the new volume:}\) The undissolved AgCl remains present, so the final solution must satisfy Ksp. Therefore, \[ [\mathrm{Ag^+}]_2\approx\frac{1.00\times10^{-10}}{0.00100} \] \[ [\mathrm{Ag^+}]_2\approx1.00\times10^{-7}\,\mathrm{M} \] \(\textbf{Determine the final dissolved silver amount:}\) \[ n_{\mathrm{Ag^+},2}=[\mathrm{Ag^+}]_2V_2 \] \[ n_{\mathrm{Ag^+},2}\approx(1.00\times10^{-7})(1.00) \] \[ n_{\mathrm{Ag^+},2}\approx1.00\times10^{-7}\,\mathrm{mol} \] \(\textbf{Compare concentrations and amounts:}\) \[ \frac{[\mathrm{Ag^+}]_2}{[\mathrm{Ag^+}]_1} =\frac{10^{-7}}{10^{-8}} \] \[ \frac{[\mathrm{Ag^+}]_2}{[\mathrm{Ag^+}]_1}=10 \] For total dissolved silver, \[ \frac{n_{\mathrm{Ag^+},2}}{n_{\mathrm{Ag^+},1}} =\frac{10^{-7}}{10^{-9}} \] \[ \frac{n_{\mathrm{Ag^+},2}}{n_{\mathrm{Ag^+},1}}=100 \] \(\textbf{Interpret the counterintuitive concentration increase:}\) Adding water normally decreases solute concentrations immediately. However, the diluted solution remains in contact with undissolved AgCl. Dilution reduces the concentration of the externally supplied common chloride ion. Additional AgCl consequently dissolves until saturation is restored. The equilibrium Ag⁺ concentration rises despite the increase in solution volume. The total dissolved amount increases even more strongly because both concentration and volume increase. This distinction would not hold if the undissolved solid were absent or insufficient to maintain saturation. The problem links common-ion suppression, dissolution following dilution and the difference between concentration and total amount. \(\textbf{Final answer:}\) \[ \boxed{\frac{[\mathrm{Ag^+}]_2}{[\mathrm{Ag^+}]_1}\approx10,\qquad \frac{n_{\mathrm{Ag^+},2}}{n_{\mathrm{Ag^+},1}}\approx100} \]

Explanation: \(\textbf{Calculate the original saturation concentration:}\) For the 1:1 salt, \[ \mathrm{MX(s)\rightleftharpoons M^++X^-} \] In pure water with no side reactions, \[ K_{sp}=S^2 \] At \(T_1\), \[ S_1=\sqrt{1.00\times10^{-4}} \] \[ S_1=0.0100\,\mathrm{M} \] \(\textbf{Determine the original dissolved amount:}\) The solution volume is \(1.00\,\mathrm{L}\). Therefore, \[ n_{\mathrm{dissolved},1}=S_1V \] \[ n_{\mathrm{dissolved},1}=(0.0100)(1.00) \] \[ n_{\mathrm{dissolved},1}=0.0100\,\mathrm{mol} \] \(\textbf{Determine the total available salt inventory:}\) The vessel also contains \[ n_{\mathrm{solid},1}=0.00200\,\mathrm{mol} \] The total amount of MX formula units is therefore \[ n_{\mathrm{total}}=0.0100+0.00200 \] \[ n_{\mathrm{total}}=0.0120\,\mathrm{mol} \] \(\textbf{Calculate the new saturation demand:}\) At \(T_2\), \[ S_{\mathrm{sat},2}=\sqrt{2.25\times10^{-4}} \] \[ S_{\mathrm{sat},2}=0.0150\,\mathrm{M} \] A saturated \(1.00\,\mathrm{L}\) solution at \(T_2\) would therefore contain \[ n_{\mathrm{sat},2}=(0.0150)(1.00) \] \[ n_{\mathrm{sat},2}=0.0150\,\mathrm{mol} \] \(\textbf{Compare the required and available amounts:}\) The extra amount needed to reach the new saturation condition is \[ 0.0150-0.0100 \] \[ =0.00500\,\mathrm{mol} \] However, only \(0.00200\,\mathrm{mol}\) undissolved MX is available. Therefore, complete dissolution of the solid cannot restore saturation. \(\textbf{Calculate the actual final concentration:}\) All \(0.0120\,\mathrm{mol}\) of MX is eventually dissolved. Thus, \[ S_{\mathrm{actual},2}=\frac{0.0120}{1.00} \] \[ S_{\mathrm{actual},2}=0.0120\,\mathrm{M} \] \(\textbf{Check the final ionic product:}\) \[ Q_{sp,2}=(0.0120)^2 \] \[ Q_{sp,2}=1.44\times10^{-4} \] Since \[ Q_{sp,2}\lt K_{sp,2} \] the final solution is unsaturated with respect to MX. \(\textbf{Interpret the finite-solid constraint:}\) An increased Ksp permits greater dissolution at the higher temperature. However, the actual dissolved concentration cannot exceed the amount of salt available in the closed vessel divided by the solution volume. The new saturation concentration describes what could dissolve if enough material were present. Here, the solid inventory is exhausted before that concentration is reached. Consequently, the final dissolved concentration is \(0.0120\,\mathrm{M}\), not \(0.0150\,\mathrm{M}\). The distinction is between an equilibrium saturation limit and an actual concentration constrained by the total material inventory. The calculation integrates temperature-dependent Ksp, material conservation and solid-phase exhaustion. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{M^+}]_f=[\mathrm{X^-}]_f=0.0120\,\mathrm{M},\qquad Q_{sp,f}=1.44\times10^{-4}\lt K_{sp,2}} \]

Explanation: \(\textbf{Identify the initial buffer condition:}\) The initial HA and A⁻ amounts are equal. Therefore, under the buffer approximation, \[ pH_i=pK_a \] Since \[ K_a=1.00\times10^{-5} \] we have \[ pK_a=5.000 \] Thus, the initial pH is approximately 5.000. \(\textbf{Express the first neutralization reaction:}\) Added HCl supplies H⁺, which reacts with the conjugate base: \[ \mathrm{H^++A^-\rightarrow HA} \] After adding \(0.0100\,\mathrm{mol}\) HCl, \[ n_{\mathrm{A^-},1}=n-0.0100 \] \[ n_{\mathrm{HA},1}=n+0.0100 \] \(\textbf{Use the observed pH to recover the buffer ratio:}\) The buffer relation gives \[ pH_1=pK_a+\log_{10}\frac{n_{\mathrm{A^-},1}}{n_{\mathrm{HA},1}} \] Substituting the measurement, \[ 4.824\approx5.000+\log_{10} \frac{n-0.0100}{n+0.0100} \] Therefore, \[ \log_{10}\frac{n-0.0100}{n+0.0100}\approx-0.1761 \] Using the supplied logarithm, \[ \frac{n-0.0100}{n+0.0100}=\frac23 \] \(\textbf{Solve for the original component amount:}\) \[ 3(n-0.0100)=2(n+0.0100) \] \[ 3n-0.0300=2n+0.0200 \] \[ n=0.0500\,\mathrm{mol} \] \(\textbf{Determine the buffer inventory after HCl addition:}\) \[ n_{\mathrm{A^-},1}=0.0500-0.0100 \] \[ n_{\mathrm{A^-},1}=0.0400\,\mathrm{mol} \] \[ n_{\mathrm{HA},1}=0.0500+0.0100 \] \[ n_{\mathrm{HA},1}=0.0600\,\mathrm{mol} \] \(\textbf{Apply the subsequent NaOH addition:}\) Hydroxide reacts with the weak acid: \[ \mathrm{HA+OH^-\rightarrow A^-+H_2O} \] The added amount is \(0.0200\,\mathrm{mol}\). Thus, \[ n_{\mathrm{HA},f}=0.0600-0.0200 \] \[ n_{\mathrm{HA},f}=0.0400\,\mathrm{mol} \] \[ n_{\mathrm{A^-},f}=0.0400+0.0200 \] \[ n_{\mathrm{A^-},f}=0.0600\,\mathrm{mol} \] \(\textbf{Calculate the final pH:}\) \[ pH_f=5.000+\log_{10}\frac{0.0600}{0.0400} \] \[ pH_f=5.000+\log_{10}\frac32 \] \[ pH_f=5.000+0.1761 \] \[ pH_f=5.1761 \] \(\textbf{Interpret the reverse inference and sequence:}\) The observed pH after acid addition determines the unknown starting component amounts. The subsequent base addition must be applied to the already changed buffer composition. It cannot be treated as an independent addition to the original buffer. The final ratio is the reciprocal of the ratio after the first addition. Consequently, the final pH is approximately as far above pKa as the first measured pH was below it. This symmetry follows from the selected amounts of acid and base, not from a general rule that equal or successive additions always cancel. The calculation combines reverse buffer composition inference and sequential reaction stoichiometry. \(\textbf{Final answer:}\) \[ \boxed{n=0.0500\,\mathrm{mol}\text{ of each component},\qquad pH_f\approx5.176} \]

Explanation: \(\textbf{Determine P's original conjugate ratio:}\) The acidic-buffer relation is \[ pH=pK_a+\log_{10}\frac{n_{\mathrm{base}}}{n_{\mathrm{acid}}} \] For P, \[ pH=pK_a=5.00 \] Therefore, \[ \frac{n_{\mathrm{P^-}}}{n_{\mathrm{HP}}}=1 \] Because the total amount is \(0.220\,\mathrm{mol}\), \[ n_{\mathrm{P^-},0}=n_{\mathrm{HP},0}=0.110\,\mathrm{mol} \] \(\textbf{Calculate P's composition after HCl addition:}\) Hydrogen ions consume the conjugate base: \[ \mathrm{H^++P^-\rightarrow HP} \] Thus, \[ n_{\mathrm{P^-},f}=0.110-0.0200 \] \[ n_{\mathrm{P^-},f}=0.0900\,\mathrm{mol} \] \[ n_{\mathrm{HP},f}=0.110+0.0200 \] \[ n_{\mathrm{HP},f}=0.130\,\mathrm{mol} \] \(\textbf{Calculate P's final pH:}\) \[ pH_P=5.00+\log_{10}\frac{0.0900}{0.130} \] \[ pH_P=5.00+\log_{10}\frac9{13} \] \[ pH_P=5.00-0.1597 \] \[ pH_P=4.8403 \] \(\textbf{Recover Q's different original component ratio:}\) For Q, \[ pH-pK_a=5.00-4.00 \] \[ pH-pK_a=1.00 \] Therefore, \[ \frac{n_{\mathrm{Q^-}}}{n_{\mathrm{HQ}}}=10 \] The combined amount remains \(0.220\,\mathrm{mol}\). Let \(n_{\mathrm{HQ}}=x\). Then, \[ x+10x=0.220 \] \[ 11x=0.220 \] \[ x=0.0200\,\mathrm{mol} \] Thus, \[ n_{\mathrm{HQ},0}=0.0200\,\mathrm{mol} \] \[ n_{\mathrm{Q^-},0}=0.200\,\mathrm{mol} \] \(\textbf{Calculate Q's composition after HCl addition:}\) \[ n_{\mathrm{Q^-},f}=0.200-0.0200 \] \[ n_{\mathrm{Q^-},f}=0.180\,\mathrm{mol} \] \[ n_{\mathrm{HQ},f}=0.0200+0.0200 \] \[ n_{\mathrm{HQ},f}=0.0400\,\mathrm{mol} \] \(\textbf{Calculate Q's final pH:}\) \[ pH_Q=4.00+\log_{10}\frac{0.180}{0.0400} \] \[ pH_Q=4.00+\log_{10}4.5 \] \[ pH_Q=4.00+0.6532 \] \[ pH_Q=4.6532 \] \(\textbf{Compare resistance to the same acid disturbance:}\) Both buffers began at pH 5.00 and contained the same total number of conjugate-pair formula units. Nevertheless, their starting acid/base ratios were different because their pKa values differed. P initially contained equal amounts of the two components. Q contained ten times as much conjugate base as weak acid. Although Q began with more conjugate base available to consume acid, the same acid addition changed its logarithmic component ratio more strongly. Its resulting pH decrease was therefore larger. The comparison shows why choosing a buffer with pKa close to the desired operating pH is useful for effective resistance to small disturbances. Equal initial pH and equal total analytical concentration do not guarantee equal buffering performance. \(\textbf{Final answer:}\) \[ \boxed{pH_P\approx4.840,\qquad pH_Q\approx4.653} \]

Explanation: \(\textbf{Recover the hydroxide concentration from the precipitation threshold:}\) The magnesium hydroxide dissolution equilibrium is \[ \mathrm{Mg(OH)_2(s)\rightleftharpoons Mg^{2+}+2OH^-} \] At the onset of precipitation, \[ [\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2=K_{sp} \] Therefore, \[ [\mathrm{OH^-}]_{\mathrm{onset}} =\sqrt{\frac{K_{sp}}{[\mathrm{Mg^{2+}}]}} \] Substituting, \[ [\mathrm{OH^-}]_{\mathrm{onset}} =\sqrt{\frac{1.00\times10^{-11}}{1.00\times10^{-3}}} \] \[ [\mathrm{OH^-}]_{\mathrm{onset}} =1.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Identify the origin of hydroxide ions:}\) NaA dissociates into Na⁺ and A⁻. The anion then hydrolyses: \[ \mathrm{A^-+H_2O\rightleftharpoons HA+OH^-} \] Its hydrolysis constant is \[ K_h=\frac{K_w}{K_a} \] Thus, knowledge of the hydrolysis-produced hydroxide concentration allows the parent acid's Ka to be inferred. \(\textbf{Use the salt material balance:}\) Let x be the concentration of A⁻ hydrolysed. With water autoionization negligible, \[ [\mathrm{OH^-}]=x \] \[ [\mathrm{HA}]=x \] \[ [\mathrm{A^-}]=0.100-x \] At precipitation onset, \[ x=1.00\times10^{-4}\,\mathrm{M} \] Therefore, \[ [\mathrm{A^-}]=0.100-0.000100 \] \[ [\mathrm{A^-}]=0.0999\,\mathrm{M} \] \(\textbf{Calculate the hydrolysis constant:}\) \[ K_h=\frac{[\mathrm{HA}][\mathrm{OH^-}]}{[\mathrm{A^-}]} \] \[ K_h=\frac{(1.00\times10^{-4})^2}{0.0999} \] \[ K_h\approx1.001\times10^{-7} \] \(\textbf{Recover the parent acid-ionization constant:}\) \[ K_a=\frac{K_w}{K_h} \] Substituting, \[ K_a=\frac{1.00\times10^{-14}}{1.001\times10^{-7}} \] \[ K_a\approx9.99\times10^{-8} \] Thus, \[ K_a\approx1.00\times10^{-7} \] \(\textbf{Check the approximation:}\) The hydrolysis fraction at onset is \[ h=\frac{1.00\times10^{-4}}{0.100} \] \[ h=0.00100 \] Only \(0.100\%\) of the initial A⁻ inventory is hydrolysed. The small-hydrolysis approximation is therefore well supported. \(\textbf{Interpret the reverse inference:}\) The magnesium hydroxide precipitation threshold determines the hydroxide-ion concentration. That concentration reveals how much conjugate-base hydrolysis has occurred. The hydrolysis equilibrium then gives Kh, from which the parent weak-acid constant follows. This reverses the more usual sequence in which Ka is supplied and precipitation is predicted. The calculation is valid for the stated pre-precipitation condition, before appreciable hydroxide removal changes the hydrolysis equilibrium. It combines solubility-product onset, hydrolysis stoichiometry, acid-base conjugacy and a final approximation check. \(\textbf{Final answer:}\) \[ \boxed{K_a(\mathrm{HA})\approx1.00\times10^{-7}} \]

Explanation: \(\textbf{Identify the required conjugate pair:}\) The desired buffer contains undissociated HA and its conjugate base A⁻. The conjugate base can be produced by partial neutralization: \[ \mathrm{HA+OH^-\rightarrow A^-+H_2O} \] Each mole of added hydroxide converts one mole of HA into one mole of A⁻. Complete neutralization would leave essentially no HA and would not give the intended conjugate-pair buffer. \(\textbf{Determine the acid dissociation index:}\) \[ pK_a=-\log_{10}K_a \] \[ pK_a=-\log_{10}(10^{-5}) \] \[ pK_a=5.000 \] \(\textbf{Recover the required component ratio:}\) The acidic-buffer relation is \[ pH=pK_a+\log_{10}\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}} \] Substituting the target pH, \[ 5.3010=5.000+\log_{10}\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}} \] \[ \log_{10}\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}}=0.3010 \] Therefore, \[ \frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}}=2.00 \] \(\textbf{Connect the ratio with neutralization stoichiometry:}\) Let x moles of NaOH be added. The component amounts after reaction are \[ n_{\mathrm{A^-}}=x \] \[ n_{\mathrm{HA}}=0.150-x \] The required ratio gives \[ \frac{x}{0.150-x}=2.00 \] \[ x=0.300-2x \] \[ 3x=0.300 \] \[ x=0.100\,\mathrm{mol} \] \(\textbf{Verify the prepared composition:}\) The remaining HA amount is \[ n_{\mathrm{HA}}=0.150-0.100 \] \[ n_{\mathrm{HA}}=0.0500\,\mathrm{mol} \] The conjugate-base amount is \[ n_{\mathrm{A^-}}=0.100\,\mathrm{mol} \] Therefore, \[ pH=5.000+\log_{10}\frac{0.100}{0.0500} \] \[ pH=5.3010 \] \(\textbf{Interpret the preparation strategy:}\) The initial solution contains only the weak acid. Partial neutralization creates the conjugate-base component needed for buffering. The target pH determines the ratio of the two components, not their individual concentrations directly. The initial acid inventory then converts that ratio into a required NaOH amount. The final mixture contains appreciable quantities of both conjugate partners. Consequently, it can consume small additions of either strong acid or strong base. The calculation combines buffer selection, acid-base stoichiometry and a target-pH constraint. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{NaOH}}=0.100\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the base and its conjugate acid:}\) Ammonia is the weak-base component. Ammonium ions supplied by NH₄Cl are its conjugate-acid component. The relevant equilibrium is \[ \mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-} \] The added strong acid reacts predominantly with ammonia: \[ \mathrm{NH_3+H^+\rightarrow NH_4^+} \] \(\textbf{Apply the neutralization inventory:}\) Each mole of added HCl consumes one mole of NH₃ and produces one mole of NH₄⁺. Therefore, \[ n_{\mathrm{NH_3},f}=0.120-0.0300 \] \[ n_{\mathrm{NH_3},f}=0.0900\,\mathrm{mol} \] \[ n_{\mathrm{NH_4^+},f}=0.0800+0.0300 \] \[ n_{\mathrm{NH_4^+},f}=0.110\,\mathrm{mol} \] Both buffer components remain present in appreciable quantities. \(\textbf{Construct the basic-buffer expression:}\) For ammonia, \[ pK_b=5.000 \] The appropriate buffer relation is \[ pOH=pK_b+\log_{10} \frac{n_{\mathrm{NH_4^+}}}{n_{\mathrm{NH_3}}} \] The conjugate acid appears in the numerator because this expression is written for pOH. \(\textbf{Calculate the changed pOH:}\) \[ pOH_f=5.000+\log_{10}\frac{0.110}{0.0900} \] \[ pOH_f=5.000+\log_{10}\frac{11}{9} \] \[ pOH_f=5.000+0.0872 \] \[ pOH_f=5.0872 \] \(\textbf{Convert to pH:}\) At \(298\,\mathrm{K}\), \[ pH+pOH=14.000 \] Therefore, \[ pH_f=14.000-5.0872 \] \[ pH_f=8.9128 \] \(\textbf{Check the direction of the disturbance:}\) Before HCl addition, ammonia exceeds ammonium: \[ \frac{n_{\mathrm{NH_3}}}{n_{\mathrm{NH_4^+}}}=\frac{0.120}{0.0800} \] After acid addition, the ratio is \[ \frac{0.0900}{0.110}\lt 1 \] The added acid has reduced the weak-base/conjugate-acid ratio. Therefore, pOH increases and pH decreases. This is consistent with the calculated result. \(\textbf{Interpret the buffer action:}\) HCl is consumed primarily by the NH₃ component rather than remaining as free strong acid. The corresponding increase in NH₄⁺ alters the buffer equilibrium composition. The pH does not remain exactly unchanged, but the change is much smaller than it would be in pure water receiving the same amount of HCl. The pOH equation and the pH equation use opposite component-ratio orientations. Confusing these orientations would incorrectly predict that adding HCl raises the pH. The calculation combines basic-buffer composition, neutralization stoichiometry and logarithmic pH conversion. \(\textbf{Final answer:}\) \[ \boxed{pOH_f\approx5.087,\qquad pH_f\approx8.913} \]