Equilibrium MCQs with Answers – Part 4 (NEET Chemistry)

Explanation: Statement I is correct because the central atoms of BF₃ and AlCl₃ can accept an electron pair. BF₃ has an electron-deficient boron centre. AlCl₃ similarly contains an electron-deficient aluminium centre. Each acts as the Lewis acid in the specified interaction. Statement II is correct because fluoride and chloride ions possess available electron pairs. They donate an electron pair to the respective electron-deficient centre. Therefore, F⁻ and Cl⁻ act as Lewis bases in P and Q. Statement III is correct because reaction R involves ammonia accepting a proton. Under the Brønsted–Lowry concept, ammonia is a base because it accepts H⁺. Under the Lewis concept, ammonia is an electron-pair donor, while H⁺ is an electron-pair acceptor. Thus, reaction R can be described using both concepts. Reactions P and Q do not require proton transfer and are naturally described using the Lewis concept. Statement IV is incorrect because proton donation or acceptance is not required for every Lewis acid-base reaction. BF₃ and AlCl₃ illustrate electron-pair acceptors that act as Lewis acids without donating protons. The distinction between Brønsted and Lewis behaviour is therefore chemically meaningful. Lewis acid-base chemistry includes electron-pair interactions that lie outside a strict proton-transfer description. The electron-pair donor and acceptor must be identified from the bonding interaction rather than from the presence of hydrogen in the chemical formula. Thus, Statements I, II and III are correct.

Explanation: \(\textbf{Identify the two simultaneous roles of dihydrogen phosphate:}\) One \(\mathrm{H_2PO_4^-}\) ion donates a proton and becomes \(\mathrm{HPO_4^{2-}}\). The other accepts a proton and becomes \(\mathrm{H_3PO_4}\). Therefore, the displayed reaction illustrates amphiprotic behaviour. \(\textbf{Write the first ionization of phosphoric acid:}\) \[ \mathrm{H_3PO_4\rightleftharpoons H^++H_2PO_4^-} \] Its ionization constant is \[ K_{a1} = \frac{[\mathrm{H^+}][\mathrm{H_2PO_4^-}]}{[\mathrm{H_3PO_4}]} \] \(\textbf{Write the second ionization:}\) \[ \mathrm{H_2PO_4^-\rightleftharpoons H^++HPO_4^{2-}} \] Its constant is \[ K_{a2} = \frac{[\mathrm{H^+}][\mathrm{HPO_4^{2-}}]}{[\mathrm{H_2PO_4^-}]} \] \(\textbf{Construct the self-proton-transfer reaction:}\) Reverse the first ionization: \[ \mathrm{H^++H_2PO_4^-\rightleftharpoons H_3PO_4} \] Its constant is \[ \frac1{K_{a1}} \] Add the reversed first ionization to the forward second ionization. The hydrogen ions cancel, giving \[ \mathrm{2H_2PO_4^-\rightleftharpoons H_3PO_4+HPO_4^{2-}} \] \(\textbf{Calculate the equilibrium constant:}\) \[ K=\frac{K_{a2}}{K_{a1}} \] Substituting, \[ K=\frac{1.00\times10^{-4}}{1.00\times10^{-2}} \] \[ K=0.0100 \] \(\textbf{Verify through the concentration expression:}\) For the displayed reaction, \[ K=\frac{[\mathrm{H_3PO_4}][\mathrm{HPO_4^{2-}}]}{[\mathrm{H_2PO_4^-}]^2} \] Combining the two acid-ionization expressions produces precisely the same relationship. \(\textbf{Interpret the equilibrium preference:}\) Because K is smaller than one, the self-proton-transfer equilibrium is reactant-favoured under the specified standard-state convention. This is consistent with \(\mathrm{H_2PO_4^-}\) being a significant intermediate species between the first and second ionization stages. The reaction cannot be analysed by using only one ionization constant. It requires recognising the ion's acid and base roles and combining the two successive equilibria. \(\textbf{Final answer:}\) \[ \boxed{K=\frac{K_{a2}}{K_{a1}}=0.0100} \]

Explanation: \(\textbf{Determine the first species ratio:}\) The first ionization is \[ \mathrm{H_3A\rightleftharpoons H^++H_2A^-} \] Therefore, \[ \frac{[\mathrm{H_2A^-}]}{[\mathrm{H_3A}]} = \frac{K_{a1}}{[\mathrm{H^+}]} \] Substituting, \[ \frac{[\mathrm{H_2A^-}]}{[\mathrm{H_3A}]} = \frac{10^{-2}}{10^{-7}} \] \[ \frac{[\mathrm{H_2A^-}]}{[\mathrm{H_3A}]}=10^5 \] \(\textbf{Determine the second species ratio:}\) For \[ \mathrm{H_2A^-\rightleftharpoons H^++HA^{2-}} \] \[ \frac{[\mathrm{HA^{2-}}]}{[\mathrm{H_2A^-}]} = \frac{K_{a2}}{[\mathrm{H^+}]} \] \[ \frac{[\mathrm{HA^{2-}}]}{[\mathrm{H_2A^-}]} = \frac{10^{-5}}{10^{-7}} \] \[ \frac{[\mathrm{HA^{2-}}]}{[\mathrm{H_2A^-}]}=100 \] \(\textbf{Determine the third species ratio:}\) For \[ \mathrm{HA^{2-}\rightleftharpoons H^++A^{3-}} \] \[ \frac{[\mathrm{A^{3-}}]}{[\mathrm{HA^{2-}}]} = \frac{K_{a3}}{[\mathrm{H^+}]} \] \[ \frac{[\mathrm{A^{3-}}]}{[\mathrm{HA^{2-}}]} = \frac{10^{-9}}{10^{-7}} \] \[ \frac{[\mathrm{A^{3-}}]}{[\mathrm{HA^{2-}}]}=0.0100 \] \(\textbf{Compare the neighbouring species:}\) The second-stage relationship gives \[ [\mathrm{HA^{2-}}]=100[\mathrm{H_2A^-}] \] The third-stage relationship gives \[ [\mathrm{HA^{2-}}]=100[\mathrm{A^{3-}}] \] Thus, \(\mathrm{HA^{2-}}\) is much more abundant than either immediately adjacent protonation state. \(\textbf{Interpret the multistage distribution:}\) The maintained hydrogen-ion concentration is smaller than the first and second ionization constants. Therefore, the first two proton-loss stages are strongly favoured relative to their corresponding protonated forms. However, the hydrogen-ion concentration is larger than the third ionization constant. The third deprotonation is consequently less favoured than retention of the final proton. The predominant species is therefore the doubly deprotonated intermediate, not the fully deprotonated ion. Comparing the hydrogen-ion concentration separately with each successive Ka is essential. It would be incorrect to infer complete deprotonation merely because the first ionization is strongly favoured. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{HA^{2-}}]=100[\mathrm{H_2A^-}]=100[\mathrm{A^{3-}}]} \]

Explanation: \(\textbf{Establish the weak-acid equilibrium:}\) For solution P, \[ \mathrm{HA\rightleftharpoons H^++A^-} \] Let x be the equilibrium hydrogen-ion concentration. Then, \[ K_a=\frac{x^2}{C-x} \] \(\textbf{Substitute the acid data:}\) \[ 0.0100=\frac{x^2}{0.200-x} \] Rearranging, \[ x^2+0.0100x-0.00200=0 \] Factorising, \[ (x-0.0400)(x+0.0500)=0 \] The physically acceptable root is \[ x=0.0400\,\mathrm{M} \] Therefore, \[ [\mathrm{H^+}]_P=0.0400\,\mathrm{M} \] \(\textbf{Calculate the acid degree of ionization:}\) \[ \alpha_P=\frac{x}{C} \] \[ \alpha_P=\frac{0.0400}{0.200} \] \[ \alpha_P=0.200 \] \(\textbf{Calculate the acid pH:}\) \[ pH_P=-\log_{10}(0.0400) \] \[ pH_P=-\log_{10}(4.00\times10^{-2}) \] \[ pH_P=2-\log_{10}4 \] \[ pH_P=2-0.6021 \] \[ pH_P\approx1.40 \] \(\textbf{Apply the corresponding weak-base equilibrium:}\) For solution Q, \[ \mathrm{B+H_2O\rightleftharpoons BH^++OH^-} \] The equilibrium relationship is \[ K_b=\frac{y^2}{C-y} \] where y is the hydroxide-ion concentration. Because the numerical Kb and analytical concentration are identical to the acid's Ka and analytical concentration, the equation for y is identical. Thus, \[ y=0.0400\,\mathrm{M} \] Therefore, \[ [\mathrm{OH^-}]_Q=0.0400\,\mathrm{M} \] And, \[ \alpha_Q=\frac{0.0400}{0.200} \] \[ \alpha_Q=0.200 \] \(\textbf{Calculate the base pOH and pH:}\) \[ pOH_Q=-\log_{10}(0.0400) \] \[ pOH_Q\approx1.40 \] At \(298\,\mathrm{K}\), \[ pH_Q+pOH_Q=14.00 \] Therefore, \[ pH_Q=14.00-1.3979 \] \[ pH_Q\approx12.60 \] \(\textbf{Interpret the acid-base symmetry:}\) The two electrolytes have equal analytical concentrations and numerically equal ionization constants for their respective ionization reactions. Under the stated assumptions, they consequently have equal degrees of ionization. The hydrogen-ion concentration of the weak acid equals the hydroxide-ion concentration of the weak base. However, their pH values are not equal. Instead, their pH values add to fourteen at the specified temperature. The result connects weak-acid and weak-base equilibrium calculations with the ionic product of water. \(\textbf{Final answer:}\) \[ \boxed{pH_P\approx1.40,\quad pH_Q\approx12.60,\quad\alpha_P=\alpha_Q=0.200} \]

Explanation: \(\textbf{Determine the conjugate-acid ionization constant:}\) B accepts a proton and produces BH⁺. The conjugate pair obeys \[ K_a(\mathrm{BH^+})K_b(\mathrm{B})=K_w \] Therefore, \[ K_a(\mathrm{BH^+})=\frac{K_w}{K_b(\mathrm{B})} \] Substituting, \[ K_a(\mathrm{BH^+})=\frac{1.00\times10^{-14}}{3.00\times10^{-9}} \] \[ K_a(\mathrm{BH^+})=3.33\times10^{-6} \] \(\textbf{Construct the proton-transfer equilibrium constant:}\) The desired reaction is obtained by combining the ionization of HA with the reverse ionization of BH⁺. Therefore, \[ K=\frac{K_a(\mathrm{HA})}{K_a(\mathrm{BH^+})} \] Equivalently, \[ K=\frac{K_a(\mathrm{HA})K_b(\mathrm{B})}{K_w} \] \(\textbf{Calculate the equilibrium constant:}\) \[ K=\frac{(3.00\times10^{-5})(3.00\times10^{-9})}{1.00\times10^{-14}} \] \[ K=9.00 \] \(\textbf{Represent the proton-transfer extent:}\) Let x molarity of HA transfer a proton to B. The equilibrium concentrations become \[ [\mathrm{HA}]=0.100-x \] \[ [\mathrm{B}]=0.100-x \] \[ [\mathrm{A^-}]=x \] \[ [\mathrm{BH^+}]=x \] \(\textbf{Apply the equilibrium expression:}\) \[ K=\frac{[\mathrm{A^-}][\mathrm{BH^+}]}{[\mathrm{HA}][\mathrm{B}]} \] Substituting, \[ 9.00=\frac{x^2}{(0.100-x)^2} \] Taking the positive square root, \[ 3.00=\frac{x}{0.100-x} \] \[ 0.300-3x=x \] \[ 4x=0.300 \] \[ x=0.0750\,\mathrm{M} \] \(\textbf{Determine the converted fraction:}\) \[ f=\frac{x}{C_{\mathrm{HA,initial}}} \] \[ f=\frac{0.0750}{0.100} \] \[ f=0.750 \] \[ f=75.0\% \] \(\textbf{Verify the equilibrium composition:}\) The remaining HA and B concentrations are each \(0.0250\,\mathrm{M}\). The product concentrations are each \(0.0750\,\mathrm{M}\). Therefore, \[ Q=\frac{(0.0750)^2}{(0.0250)^2} \] \[ Q=9.00 \] This agrees with the equilibrium constant. \(\textbf{Interpret the acid-base relationship:}\) The proton transfer is product-favoured because HA is stronger as an acid than BH⁺ under the specified conditions. Nevertheless, the reaction does not require complete conversion of the reactants. The equilibrium constant determines the balance between protonated and deprotonated species. The calculation combines conjugate acid-base relationships with stoichiometric equilibrium composition rather than treating proton transfer as automatically complete. \(\textbf{Final answer:}\) \[ \boxed{f=75.0\%} \]

Explanation: Statement I is correct because ammonia possesses a lone pair of electrons on nitrogen. In reaction P, this lone pair participates in forming a coordinate bond to the silver(I) centre. Ammonia is therefore functioning as a Lewis base. Statement II is correct because the metal centre accepts electron-pair donation from ammonia. It acts as the Lewis-acid centre in the coordination interaction. The reaction involves replacement of coordinated water molecules by ammonia ligands. The water molecules displaced from the coordination sphere become free solvent molecules. Statement III is correct because reaction Q involves a direct proton transfer. Hydronium ion donates a proton and becomes water. Ammonia accepts that proton and becomes ammonium ion. Thus, hydronium is the Brønsted–Lowry acid and ammonia is the Brønsted–Lowry base. The same reaction can also be interpreted through electron-pair donation from ammonia to the proton. Statement IV is incorrect because the displayed coordination reaction does not require proton transfer. Ammonia substitutes for coordinated water through a ligand-exchange process. The silver(I) centre accepts electron-pair donation without functioning as a proton donor. The important distinction is between coordination to an electron-deficient centre and acceptance of a proton. Both interactions can involve the ammonia nitrogen lone pair, but they are not chemically identical. This example demonstrates that Lewis basicity can explain reactions beyond the proton-transfer framework. The correct statements are I, II and III.

Explanation: \(\textbf{Identify the initial ionic contributions:}\) BHCl supplies BH⁺ and Cl⁻ ions. NaOH supplies Na⁺ and OH⁻ ions. The initial concentrations contributed by these strong electrolytes are \[ [\mathrm{BH^+}]_0=0.0300\,\mathrm{M} \] \[ [\mathrm{OH^-}]_0=0.0100\,\mathrm{M} \] The molecular base concentration is initially \(0.200\,\mathrm{M}\). \(\textbf{Infer the net ionization extent:}\) The measured final hydroxide concentration is \(0.0200\,\mathrm{M}\). Therefore, the net increase in hydroxide concentration is \[ x=0.0200-0.0100 \] \[ x=0.0100\,\mathrm{M} \] The forward ionization produces one BH⁺ ion for each additional OH⁻ ion. \(\textbf{Determine the equilibrium base concentration:}\) \[ [\mathrm B]_{\mathrm{eq}}=0.200-x \] \[ [\mathrm B]_{\mathrm{eq}}=0.200-0.0100 \] \[ [\mathrm B]_{\mathrm{eq}}=0.190\,\mathrm{M} \] \(\textbf{Determine the equilibrium conjugate-acid concentration:}\) \[ [\mathrm{BH^+}]_{\mathrm{eq}}=0.0300+x \] \[ [\mathrm{BH^+}]_{\mathrm{eq}}=0.0300+0.0100 \] \[ [\mathrm{BH^+}]_{\mathrm{eq}}=0.0400\,\mathrm{M} \] \(\textbf{Apply the base-ionization expression:}\) \[ K_b=\frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm B]} \] Substituting, \[ K_b=\frac{(0.0400)(0.0200)}{0.190} \] \[ K_b=\frac{0.000800}{0.190} \] \[ K_b=4.21\times10^{-3}\,\mathrm{M} \] \(\textbf{Verify electrical neutrality:}\) The positive ions include BH⁺ and Na⁺. Their total concentration is \[ 0.0400+0.0100=0.0500\,\mathrm{M} \] The negative ions include Cl⁻ and OH⁻. Their total concentration is \[ 0.0300+0.0200=0.0500\,\mathrm{M} \] The charge balance is satisfied. \(\textbf{Interpret the coupled ion sources:}\) The equilibrium hydroxide concentration is not equal to the amount of base that ionized because NaOH already supplied hydroxide ions. Similarly, the final BH⁺ concentration is not solely produced by B because BHCl supplied an initial amount. The ionization extent must be recovered by subtracting the strong-electrolyte contribution. Only then can the equilibrium concentrations be substituted into the Kb expression. The independent electrical-neutrality check confirms the result. \(\textbf{Final answer:}\) \[ \boxed{K_b=4.21\times10^{-3}\,\mathrm{mol\,L^{-1}}} \]

Explanation: \(\textbf{Identify the conjugate acid-base relationship:}\) For a conjugate acid-base pair in aqueous solution, the acid-ionization constant of the conjugate acid and the base-ionization constant of its conjugate base satisfy \[ K_a(\mathrm{BH^+})K_b(\mathrm{B})=K_w \] This relationship applies at each temperature using the corresponding value of \(K_w\). \(\textbf{Calculate the base-ionization constant at }T_1\textbf{:}\) Rearranging, \[ K_{b,1}=\frac{K_{w,1}}{K_{a,1}} \] Substituting the given values, \[ K_{b,1}=\frac{1.00\times10^{-14}}{1.00\times10^{-5}} \] \[ K_{b,1}=1.00\times10^{-9} \] \(\textbf{Calculate the base-ionization constant at }T_2\textbf{:}\) At the higher temperature, \[ K_{b,2}=\frac{K_{w,2}}{K_{a,2}} \] Substituting, \[ K_{b,2}=\frac{4.00\times10^{-14}}{1.00\times10^{-4}} \] \[ K_{b,2}=4.00\times10^{-10} \] \(\textbf{Compare the two constants:}\) \[ \frac{K_{b,2}}{K_{b,1}} =\frac{4.00\times10^{-10}}{1.00\times10^{-9}} \] \[ \frac{K_{b,2}}{K_{b,1}}=0.400 \] Thus, the base-ionization constant at \(T_2\) is \(40.0\%\) of its value at \(T_1\). Equivalently, it decreases by a factor of \(2.50\). \(\textbf{Interpret the temperature dependence:}\) The ionic product of water increases fourfold between the two temperatures. However, the acid-ionization constant of \(\mathrm{BH^+}\) increases tenfold. Because the base-ionization constant is the ratio \(K_w/K_a\), its value decreases despite the increase in \(K_w\). The temperature dependence of \(K_b\) cannot therefore be determined from the change in \(K_w\) alone. The appropriate acid-ionization constant and ionic product of water must always correspond to the same temperature. \(\textbf{Final answer:}\) \[ \boxed{K_{b,1}=1.00\times10^{-9},\quad K_{b,2}=4.00\times10^{-10}} \]

Explanation: \(\textbf{Account for the two initial hydrogen-ion sources:}\) Complete ionization of HCl produces \[ [\mathrm{H^+}]_{\mathrm{HCl}}=0.0400\,\mathrm{M} \] The first ionization of sulphuric acid produces another \(0.0200\,\mathrm{M}\) of hydrogen ions. Therefore, before its second-stage ionization, \[ [\mathrm{H^+}]_0=0.0400+0.0200 \] \[ [\mathrm{H^+}]_0=0.0600\,\mathrm{M} \] The initial hydrogen sulphate concentration is \[ [\mathrm{HSO_4^-}]_0=0.0200\,\mathrm{M} \] \(\textbf{Represent the second ionization extent:}\) Let x molarity of hydrogen sulphate undergo second-stage ionization. Then, \[ [\mathrm{HSO_4^-}]=0.0200-x \] \[ [\mathrm{SO_4^{2-}}]=x \] \[ [\mathrm{H^+}]=0.0600+x \] \(\textbf{Apply the equilibrium constant:}\) \[ K_{a2}= \frac{[\mathrm{H^+}][\mathrm{SO_4^{2-}}]}{[\mathrm{HSO_4^-}]} \] Substituting, \[ 0.0160=\frac{(0.0600+x)x}{0.0200-x} \] \(\textbf{Solve the exact equation:}\) \[ 0.0160(0.0200-x)=0.0600x+x^2 \] \[ 0.000320-0.0160x=0.0600x+x^2 \] \[ x^2+0.0760x-0.000320=0 \] Factorising, \[ (x-0.00400)(x+0.0800)=0 \] The physically acceptable root is \[ x=0.00400\,\mathrm{M} \] \(\textbf{Determine the final ionic concentrations:}\) \[ [\mathrm{SO_4^{2-}}]=0.00400\,\mathrm{M} \] \[ [\mathrm{H^+}]=0.0600+0.00400 \] \[ [\mathrm{H^+}]=0.0640\,\mathrm{M} \] \(\textbf{Check the ionic charge balance:}\) The chloride concentration is \(0.0400\,\mathrm{M}\). The remaining hydrogen sulphate concentration is \(0.0160\,\mathrm{M}\). Therefore, the total negative charge-equivalent concentration is \[ 0.0400+0.0160+2(0.00400) \] \[ =0.0640\,\mathrm{M} \] This agrees with the hydrogen-ion concentration. \(\textbf{Interpret the coupled acid contributions:}\) The hydrogen ions supplied by HCl and the first ionization of sulphuric acid both affect the second ionization equilibrium. They appear in the initial hydrogen-ion concentration, not as separate equilibrium constants. Their presence suppresses further dissociation of hydrogen sulphate. A calculation treating the second proton of sulphuric acid as completely ionized would therefore overestimate the final hydrogen-ion concentration. The correct method combines the strong-acid contribution with an exact second-stage equilibrium balance. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{H^+}]=0.0640\,\mathrm{M},\quad[\mathrm{SO_4^{2-}}]=0.00400\,\mathrm{M}} \]

Explanation: \(\textbf{Identify hydrogen carbonate's role in P:}\) In reaction P, hydrogen carbonate loses a proton and becomes carbonate. It therefore acts as the Brønsted–Lowry acid. Ammonia accepts the proton and acts as the base. \(\textbf{Construct the equilibrium constant for P:}\) For the proton-transfer reaction, \[ \mathrm{HCO_3^-+NH_3\rightleftharpoons CO_3^{2-}+NH_4^+} \] the equilibrium constant is \[ K_P=\frac{K_a(\mathrm{HCO_3^-})}{K_a(\mathrm{NH_4^+})} \] Substituting, \[ K_P=\frac{1.00\times10^{-10}}{1.00\times10^{-9}} \] \[ K_P=0.100 \] \(\textbf{Identify hydrogen carbonate's role in Q:}\) In reaction Q, hydrogen carbonate accepts a proton from nitrous acid. It becomes carbonic acid. Therefore, hydrogen carbonate acts as the Brønsted–Lowry base. Nitrous acid acts as the proton donor. \(\textbf{Construct the equilibrium constant for Q:}\) The reaction is \[ \mathrm{HCO_3^-+HNO_2\rightleftharpoons H_2CO_3+NO_2^-} \] The equilibrium constant is \[ K_Q=\frac{K_a(\mathrm{HNO_2})}{K_a(\mathrm{H_2CO_3})} \] Substituting, \[ K_Q=\frac{1.00\times10^{-4}}{1.00\times10^{-6}} \] \[ K_Q=100 \] \(\textbf{Compare the thermodynamic preferences:}\) For P, \[ K_P\lt 1 \] The reaction is reactant-favoured under the stated standard-state comparison. For Q, \[ K_Q\gt 1 \] The reaction is product-favoured. \(\textbf{Interpret the role reversal:}\) Hydrogen carbonate can donate or accept a proton depending on the other reacting species. Its net negative charge does not prevent proton donation. Likewise, the presence of a transferable proton does not prevent proton acceptance. The two reactions demonstrate amphiprotic behaviour through different conjugate pairs and different equilibrium preferences. Correctly predicting the products requires identifying the actual proton transfer before constructing the equilibrium constant. \(\textbf{Final answer:}\) \[ \boxed{K_P=0.100,\quad K_Q=100} \]

Explanation: \(\textbf{Identify the two proton-release stages:}\) The acid undergoes \[ \mathrm{H_2A\rightleftharpoons H^++HA^-} \] followed by \[ \mathrm{HA^-\rightleftharpoons H^++A^{2-}} \] Formation of each HA⁻ ion represents the loss of one proton from an original acid molecule. Formation of each A²⁻ ion represents the loss of two protons. \(\textbf{Use the proton-equivalent balance:}\) With no other significant ions, \[ [\mathrm{H^+}]=[\mathrm{HA^-}]+2[\mathrm{A^{2-}}] \] The measured proton-release concentration is \[ [\mathrm{H^+}]=0.0900\,\mathrm{M} \] \(\textbf{Apply the measured anion ratio:}\) Let \[ [\mathrm{HA^-}]=x \] Then, \[ [\mathrm{A^{2-}}]=0.250x \] Substituting into the charge balance, \[ 0.0900=x+2(0.250x) \] \[ 0.0900=1.50x \] \[ x=0.0600\,\mathrm{M} \] Therefore, \[ [\mathrm{HA^-}]=0.0600\,\mathrm{M} \] \[ [\mathrm{A^{2-}}]=0.0150\,\mathrm{M} \] \(\textbf{Recover the unionized acid concentration:}\) The analytical material balance is \[ C=[\mathrm{H_2A}]+[\mathrm{HA^-}]+[\mathrm{A^{2-}}] \] Thus, \[ 0.100=[\mathrm{H_2A}]+0.0600+0.0150 \] \[ [\mathrm{H_2A}]=0.0250\,\mathrm{M} \] \(\textbf{Calculate the second ionization constant:}\) \[ K_{a2}=\frac{[\mathrm{H^+}][\mathrm{A^{2-}}]}{[\mathrm{HA^-}]} \] Substituting, \[ K_{a2}=\frac{0.0900(0.0150)}{0.0600} \] \[ K_{a2}=0.0225\,\mathrm{M} \] \(\textbf{Check the first ionization relationship:}\) Although not required, the first ionization constant can be evaluated to verify that the stages have distinct constants. \[ K_{a1}=\frac{0.0900(0.0600)}{0.0250} \] \[ K_{a1}=0.216\,\mathrm{M} \] Thus, \[ K_{a1}\gt K_{a2} \] \(\textbf{Interpret the proton balance:}\) The hydrogen-ion concentration is not equal to the concentration of either individual anion. Each completely deprotonated A²⁻ ion accounts for two released protons. The anion ratio, charge balance and material balance must therefore be combined before calculating the second ionization constant. A one-stage weak-acid approximation would be inappropriate because the measured doubly charged species is appreciable. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{H_2A}]=0.0250\,\mathrm{M},\quad K_{a2}=0.0225\,\mathrm{M}} \]

Explanation: \(\textbf{Express the weak-acid ionization relationship:}\) For a weak monobasic acid of analytical concentration C and degree of ionization \(\alpha\), \[ K_a=\frac{C\alpha^2}{1-\alpha} \] The relationship follows from the concentrations of unionized acid and its corresponding ions. \(\textbf{Apply the proposed equality of analytical concentrations:}\) Condition I requires \[ C_{\mathrm{HA}}=C_{\mathrm{HB}}=C \] \(\textbf{Apply the proposed equality of ionization fractions:}\) Condition II requires \[ \alpha_{\mathrm{HA}}=\alpha_{\mathrm{HB}}=\alpha \] with \[ 0\lt \alpha\lt 1 \] \(\textbf{Compare the resulting ionization constants:}\) For HA, \[ K_a(\mathrm{HA})=\frac{C\alpha^2}{1-\alpha} \] For HB, \[ K_a(\mathrm{HB})=\frac{C\alpha^2}{1-\alpha} \] Therefore, the proposed conditions imply \[ K_a(\mathrm{HA})=K_a(\mathrm{HB}) \] This contradicts the supplied constants. \(\textbf{Examine the hydrogen-ion condition:}\) Under the stated assumptions, \[ [\mathrm{H^+}]=C\alpha \] Thus, equal analytical concentrations and equal degrees of ionization would indeed produce equal hydrogen-ion concentrations. However, this mathematical consequence does not make the original proposal chemically possible. The equilibrium constants prevent the two equalities from holding simultaneously. \(\textbf{Reject the fixed-factor misconception:}\) A tenfold difference between ionization constants does not generally produce a tenfold difference in hydrogen-ion concentrations at equal analytical concentration. The hydrogen-ion concentration depends nonlinearly on the ionization constant through the equilibrium expression. The exact concentration ratio also depends on C. \(\textbf{Interpret the consistency test:}\) The conditions must be checked against the governing equilibrium relationships before accepting their combined consequences. Two measured quantities may appear mutually compatible while being inconsistent with independently known equilibrium constants. This problem illustrates the distinction between algebraic consequences and physically realizable equilibrium states. Only option C identifies the contradiction correctly.

Explanation: \(\textbf{Recover the ionization constant from the first solution:}\) For a weak monobasic acid, \[ K_a=\frac{C\alpha^2}{1-\alpha} \] Substituting the first solution's measurements, \[ K_a=\frac{0.200(0.100)^2}{1-0.100} \] \[ K_a=\frac{0.00200}{0.900} \] \[ K_a=2.222\times10^{-3}\,\mathrm{M} \] \(\textbf{Determine the ionized acid in the second solution:}\) The second solution has \[ [\mathrm{H^+}]=0.00500\,\mathrm{M} \] The added NaA contributes A⁻ but no hydrogen ions. Therefore, neglecting water ionization, \(0.00500\,\mathrm{M}\) of HA is ionized. \(\textbf{Determine the remaining unionized acid:}\) \[ [\mathrm{HA}]=0.100-0.00500 \] \[ [\mathrm{HA}]=0.0950\,\mathrm{M} \] \(\textbf{Calculate the required total A⁻ concentration:}\) The equilibrium expression is \[ K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \] Rearranging, \[ [\mathrm{A^-}]=\frac{K_a[\mathrm{HA}]}{[\mathrm{H^+}]} \] Substituting, \[ [\mathrm{A^-}] =\frac{(2.222\times10^{-3})(0.0950)}{0.00500} \] \[ [\mathrm{A^-}]=0.04222\,\mathrm{M} \] \(\textbf{Separate the two A⁻ sources:}\) The acid contributes \(0.00500\,\mathrm{M}\) of A⁻. The remaining A⁻ comes from the completely dissociated salt. Therefore, \[ C_{\mathrm{NaA}}=0.04222-0.00500 \] \[ C_{\mathrm{NaA}}=0.03722\,\mathrm{M} \] \(\textbf{Verify the common-ion contribution:}\) The concentration \(0.04222\,\mathrm{M}\) is the total A⁻ concentration at equilibrium. Only \(0.00500\,\mathrm{M}\) of that comes from the dissociation of HA. The remainder is \(0.03722\,\mathrm{M}\) from the dissociated NaA. Keeping the \(1-\alpha\) factor in the first solution produces a result more accurate than the simple small-ionization approximation. This value appears in option D. \(\textbf{Final answer:}\) \[ \boxed{C_{\mathrm{NaA}}\approx0.03722\,\mathrm{M}} \]

Explanation: \(\textbf{Identify the proton-transfer direction:}\) Ammonium ion donates a proton to A⁻. Therefore, ammonium is the Brønsted–Lowry acid and A⁻ is the base. The products are ammonia and HA. \(\textbf{Construct the equilibrium constant from the acid constants:}\) For ammonium ion, \[ \mathrm{NH_4^+\rightleftharpoons H^++NH_3} \] For HA, \[ \mathrm{HA\rightleftharpoons H^++A^-} \] Reversing the second ionization and adding it to the first gives the desired proton-transfer reaction. Therefore, \[ K=\frac{K_a(\mathrm{NH_4^+})}{K_a(\mathrm{HA})} \] \(\textbf{Calculate the equilibrium constant:}\) \[ K=\frac{1.00\times10^{-9}}{1.00\times10^{-5}} \] \[ K=1.00\times10^{-4} \] \(\textbf{Calculate the instantaneous reaction quotient:}\) \[ Q=\frac{[\mathrm{NH_3}][\mathrm{HA}]}{[\mathrm{NH_4^+}][\mathrm{A^-}]} \] Substituting, \[ Q=\frac{(0.200)(0.0500)}{(0.100)(0.0200)} \] \[ Q=\frac{0.0100}{0.00200} \] \[ Q=5.00 \] \(\textbf{Compare the quotient with equilibrium:}\) \[ \frac QK=\frac{5.00}{1.00\times10^{-4}} \] \[ \frac QK=5.00\times10^4 \] Thus, \[ Q\gt K \] The reverse reaction is strongly favoured. \(\textbf{Interpret acid strength and composition together:}\) HA is much stronger than ammonium ion according to the supplied acid constants. The forward proton transfer produces the stronger acid HA from the weaker acid ammonium. Consequently, the equilibrium constant for that direction is very small. The supplied composition also contains a relatively large product-to-reactant ratio. Both considerations are consistent with a reverse reaction tendency. However, the decisive test for an instantaneous nonequilibrium mixture remains the comparison of Q with K. The standard-state equilibrium preference and the actual composition must be distinguished. \(\textbf{Final answer:}\) \[ \boxed{K=1.00\times10^{-4},\quad\frac QK=5.00\times10^4} \]

Explanation: \(\textbf{Convert the target pH into hydrogen-ion concentration:}\) At the required final equilibrium, \[ [\mathrm{H^+}]=10^{-4.00} \] \[ [\mathrm{H^+}]=1.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Determine the first ionization ratio:}\) For \[ \mathrm{H_2A\rightleftharpoons H^++HA^-} \] the equilibrium expression is \[ K_{a1}=\frac{[\mathrm{H^+}][\mathrm{HA^-}]}{[\mathrm{H_2A}]} \] Therefore, \[ \frac{[\mathrm{HA^-}]}{[\mathrm{H_2A}]} =\frac{K_{a1}}{[\mathrm{H^+}]} \] Substituting, \[ \frac{[\mathrm{HA^-}]}{[\mathrm{H_2A}]} =\frac{10^{-3}}{10^{-4}} \] \[ \frac{[\mathrm{HA^-}]}{[\mathrm{H_2A}]}=10 \] \(\textbf{Determine the second ionization ratio:}\) For \[ \mathrm{HA^-\rightleftharpoons H^++A^{2-}} \] \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]} =\frac{K_{a2}}{[\mathrm{H^+}]} \] \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]} =\frac{10^{-5}}{10^{-4}} \] \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]}=0.100 \] \(\textbf{Combine the species distributions:}\) Let \[ [\mathrm{H_2A}]=x \] Then, \[ [\mathrm{HA^-}]=10x \] \[ [\mathrm{A^{2-}}]=x \] The analytical acid concentration is conserved: \[ [\mathrm{H_2A}]+[\mathrm{HA^-}]+[\mathrm{A^{2-}}]=0.120 \] Thus, \[ x+10x+x=0.120 \] \[ 12x=0.120 \] \[ x=0.0100\,\mathrm{M} \] Consequently, \[ [\mathrm{H_2A}]=0.0100\,\mathrm{M} \] \[ [\mathrm{HA^-}]=0.100\,\mathrm{M} \] \[ [\mathrm{A^{2-}}]=0.0100\,\mathrm{M} \] \(\textbf{Apply electrical neutrality:}\) The added NaOH provides sodium ions. At equilibrium, \[ [\mathrm{Na^+}]+[\mathrm{H^+}] = [\mathrm{HA^-}]+2[\mathrm{A^{2-}}]+[\mathrm{OH^-}] \] At the stated pH, \[ [\mathrm{OH^-}]=\frac{10^{-14}}{10^{-4}} \] \[ [\mathrm{OH^-}]=10^{-10}\,\mathrm{M} \] Its contribution is negligible at the displayed precision. Therefore, \[ [\mathrm{Na^+}]+0.000100 = 0.100+2(0.0100) \] \[ [\mathrm{Na^+}]=0.120-0.000100 \] \[ [\mathrm{Na^+}]=0.1199\,\mathrm{M} \] \(\textbf{Relate sodium concentration to NaOH addition:}\) The original solution contains no sodium ions. Every mole of added NaOH supplies one mole of sodium ions. Therefore, the analytical concentration of NaOH added equals the final sodium-ion concentration. \(\textbf{Interpret the neutralization requirement:}\) The final solution contains three acid-derived species with different protonation states. A mole of \(\mathrm{HA^-}\) corresponds to removal of one proton from the original acid. A mole of \(\mathrm{A^{2-}}\) corresponds to removal of two protons. The sodium requirement must therefore be determined from charge equivalents rather than merely from the total amount of acid. The small concentration of free hydrogen ions produces the \(0.000100\,\mathrm{M}\) correction. \(\textbf{Final answer:}\) \[ \boxed{C_{\mathrm{NaOH,added}}=0.1199\,\mathrm{mol\,L^{-1}}} \]

Explanation: \(\textbf{Identify the shared hydrogen-ion concentration:}\) The acids undergo separate ionizations: \[ \mathrm{HA\rightleftharpoons H^++A^-} \] \[ \mathrm{HB\rightleftharpoons H^++B^-} \] However, both equilibria exist in the same aqueous solution. Therefore, they must be evaluated using one common equilibrium hydrogen-ion concentration. Let \[ [\mathrm{H^+}]=h \] \(\textbf{Express the ionized concentration of HA:}\) The acid-ionization constant is \[ K_{a,\mathrm{HA}}=\frac{h[\mathrm{A^-}]}{[\mathrm{HA}]} \] The HA material balance gives \[ [\mathrm{HA}]+[\mathrm{A^-}]=0.0200 \] Rearranging the equilibrium relationship, \[ [\mathrm{A^-}] =\frac{C_{\mathrm{HA}}K_{a,\mathrm{HA}}}{h+K_{a,\mathrm{HA}}} \] Therefore, \[ [\mathrm{A^-}] =\frac{0.0200(0.0200)}{h+0.0200} \] \(\textbf{Express the ionized concentration of HB:}\) Similarly, \[ [\mathrm{B^-}] =\frac{C_{\mathrm{HB}}K_{a,\mathrm{HB}}}{h+K_{a,\mathrm{HB}}} \] Thus, \[ [\mathrm{B^-}] =\frac{0.0300(0.0100)}{h+0.0100} \] \(\textbf{Apply electrical neutrality:}\) No other significant ions are present. Therefore, \[ h=[\mathrm{A^-}]+[\mathrm{B^-}] \] Substituting, \[ h=\frac{0.000400}{h+0.0200} +\frac{0.000300}{h+0.0100} \] \(\textbf{Test the proposed equilibrium value:}\) For \[ h=0.0200\,\mathrm{M} \] the first contribution becomes \[ [\mathrm{A^-}] =\frac{0.000400}{0.0400} \] \[ [\mathrm{A^-}]=0.0100\,\mathrm{M} \] The second contribution becomes \[ [\mathrm{B^-}] =\frac{0.000300}{0.0300} \] \[ [\mathrm{B^-}]=0.0100\,\mathrm{M} \] Their sum is \[ [\mathrm{A^-}]+[\mathrm{B^-}] =0.0100+0.0100 \] \[ =0.0200\,\mathrm{M} \] This agrees with the assumed value of h and satisfies charge balance. \(\textbf{Verify both acid equilibria:}\) The remaining unionized concentrations are \[ [\mathrm{HA}]=0.0100\,\mathrm{M} \] \[ [\mathrm{HB}]=0.0200\,\mathrm{M} \] For HA, \[ K_a=\frac{0.0200(0.0100)}{0.0100} \] \[ K_a=0.0200\,\mathrm{M} \] For HB, \[ K_a=\frac{0.0200(0.0100)}{0.0200} \] \[ K_a=0.0100\,\mathrm{M} \] Both equilibrium conditions are satisfied. \(\textbf{Interpret the coupled ionizations:}\) The acids do not ionize independently of one another after being placed in the same solution. Hydrogen ions produced by either acid influence the ionization equilibrium of both. The stronger acid HA has the larger fractional ionization, but the more concentrated HB supplies an equal absolute amount of hydrogen ions in this particular mixture. Calculating the hydrogen-ion concentration of each acid as though it were alone and then adding those separate concentrations would be incorrect. The shared hydrogen-ion concentration, both material balances and electrical neutrality must be satisfied simultaneously. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{H^+}]=0.0200\,\mathrm{M},\quad [\mathrm{A^-}]=[\mathrm{B^-}]=0.0100\,\mathrm{M}} \]

Explanation: \(\textbf{Identify the first species intersection:}\) At point P, the two intersecting species have equal concentrations: \[ [\mathrm{H_2A}]=[\mathrm{HA^-}] \] The first ionization constant is \[ K_{a1}=\frac{[\mathrm{H^+}][\mathrm{HA^-}]}{[\mathrm{H_2A}]} \] Because the two acid-derived concentrations are equal, \[ [\mathrm{H^+}]_P=K_{a1} \] \[ [\mathrm{H^+}]_P=10^{-3}\,\mathrm{M} \] Therefore, \[ pH_P=3.00 \] \(\textbf{Find the minor species at P:}\) From the second ionization, \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]} =\frac{K_{a2}}{[\mathrm{H^+}]} \] At P, \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]} =\frac{10^{-7}}{10^{-3}} \] \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]}=10^{-4} \] Thus, the completely deprotonated species is present at a much smaller concentration than the two intersecting species. \(\textbf{Identify the second intersection:}\) At Q, \[ [\mathrm{HA^-}]=[\mathrm{A^{2-}}] \] The second ionization expression gives \[ [\mathrm{H^+}]_Q=K_{a2} \] \[ [\mathrm{H^+}]_Q=10^{-7}\,\mathrm{M} \] Therefore, \[ pH_Q=7.00 \] \(\textbf{Find the minor species at Q:}\) Rearranging the first ionization expression, \[ \frac{[\mathrm{H_2A}]}{[\mathrm{HA^-}]} =\frac{[\mathrm{H^+}]}{K_{a1}} \] At Q, \[ \frac{[\mathrm{H_2A}]}{[\mathrm{HA^-}]} =\frac{10^{-7}}{10^{-3}} \] \[ \frac{[\mathrm{H_2A}]}{[\mathrm{HA^-}]}=10^{-4} \] \(\textbf{Interpret the distribution graph:}\) Each intersection corresponds to equality of two adjacent protonation states. The first occurs at \(pH=pK_{a1}\). The second occurs at \(pH=pK_{a2}\). An intersection between two species does not mean that every acid-derived species has the same concentration. The third species can be present in a much smaller amount. The separation between the two pKa values explains why the intermediate \(\mathrm{HA^-}\) can predominate over a broad pH range. \(\textbf{Final answer:}\) \[ \boxed{pH_P=3.00,\quad pH_Q=7.00,\quad \frac{[\mathrm{A^{2-}}]_P}{[\mathrm{HA^-}]_P} =\frac{[\mathrm{H_2A}]_Q}{[\mathrm{HA^-}]_Q}=10^{-4}} \]

Explanation: Statement I is correct because ammonia possesses a lone pair on nitrogen. The nitrogen atom donates an electron pair towards the silver(I) centre. Ammonia therefore acts as a Lewis base during formation of the coordination complex. The silver(I) centre functions as the electron-pair acceptor. Statement II is correct because nitric acid supplies hydronium ions in aqueous solution. Free ammonia accepts protons and forms ammonium ions. Ammonium ion does not possess the same available nitrogen lone pair as molecular ammonia. Therefore, protonation reduces the concentration of free ammonia available for coordination to silver(I). The complex-formation equilibrium contains the concentration of free ammonia as a reactant factor. Its formation quotient may be represented as \[ Q_f=\frac{[\mathrm{Ag(NH_3)_2^+}]}{[\mathrm{Ag^+}][\mathrm{NH_3}]^2} \] Reducing free ammonia initially increases this quotient relative to the unchanged formation constant. The complex-dissociation direction is consequently favoured. Statement III is therefore correct. As some complex dissociates, free silver(I) ions and ammonia are released. The liberated ammonia can undergo further protonation, maintaining the tendency towards complex dissociation. The free silver-ion concentration increases under the specified conditions. Statement IV is incorrect because adding acid does not change the intrinsic complex-formation equilibrium constant at fixed temperature. Instead, it changes the chemical availability of a coordinating ligand. The resulting shift arises from coupled protonation and coordination equilibria. The example connects Brønsted–Lowry basicity of ammonia with its Lewis basicity in metal-ion coordination. The same molecule can participate in both processes through its nitrogen lone pair. The decisive distinction is between changing ligand concentration and changing the equilibrium constant. Thus, Statements I, II and III are correct.

Explanation: \(\textbf{Identify the assumptions behind the square-root expression:}\) For a simple weak monobasic acid, \[ K_a=\frac{C\alpha^2}{1-\alpha} \] When the ionized fraction is small and hydrogen ions from water are negligible, \[ [\mathrm{H^+}]\approx\sqrt{K_aC} \] Both assumptions must be assessed. \(\textbf{Estimate the hydrogen-ion concentration in P:}\) For solution P, \[ C_P=0.100\,\mathrm{M} \] Therefore, \[ [\mathrm{H^+}]_P\approx\sqrt{(10^{-9})(0.100)} \] \[ [\mathrm{H^+}]_P\approx\sqrt{10^{-10}} \] \[ [\mathrm{H^+}]_P\approx10^{-5}\,\mathrm{M} \] \(\textbf{Check the assumptions for P:}\) The corresponding approximate degree of ionization is \[ \alpha_P\approx\frac{10^{-5}}{0.100} \] \[ \alpha_P\approx10^{-4} \] This is much smaller than unity. Also, the estimated hydrogen-ion concentration is much larger than the \(10^{-7}\,\mathrm{M}\) characteristic of pure water at the specified temperature. Neglecting water ionization is therefore reasonable for P. \(\textbf{Apply the student's calculation to Q:}\) For the extremely dilute solution, \[ C_Q=10^{-10}\,\mathrm{M} \] The proposed approximation would give \[ [\mathrm{H^+}]_{\mathrm{approx}} =\sqrt{(10^{-9})(10^{-10})} \] \[ [\mathrm{H^+}]_{\mathrm{approx}} =\sqrt{10^{-19}} \] \[ [\mathrm{H^+}]_{\mathrm{approx}}\approx3.16\times10^{-10}\,\mathrm{M} \] \(\textbf{Identify the contradiction:}\) Pure water at \(298\,\mathrm{K}\) already contains approximately \[ [\mathrm{H^+}]=10^{-7}\,\mathrm{M} \] The estimated acid-only concentration is much smaller than this. It is therefore unreasonable to neglect water autoionization. Furthermore, the total amount of HA present is only \(10^{-10}\,\mathrm{mol\,L^{-1}}\). Even complete ionization of all HA could supply no more than this concentration of additional hydrogen-ion equivalents. \(\textbf{Interpret the extremely dilute limit:}\) Solution Q remains very close to the acidity of pure water. Its hydrogen-ion concentration is controlled predominantly by water autoionization, with a very small additional acidic contribution. The simple square-root estimate does not represent the actual hydrogen-ion balance. The failure is not caused merely by an inaccurate value of the equilibrium constant. It results from neglecting a chemically important source of hydrogen ions. \(\textbf{Conclude the approximation test:}\) A weak-acid approximation must be justified for the actual concentration range. An approximation that works for a moderately concentrated weak-acid solution can fail when the same acid is diluted to an extremely low concentration. At such dilution, water ionization and the complete material balance become essential.

Explanation: \(\textbf{Identify the two successive protonation stages:}\) Carbonate ion can accept two protons successively. The first proton produces bicarbonate ion. The second proton converts bicarbonate into carbonic acid. Under the specified conditions, carbonic acid produces escaping carbon dioxide. \(\textbf{Determine the initial carbonate amount:}\) Complete dissociation of sodium carbonate supplies \[ n_{\mathrm{CO_3^{2-}}}=0.00500\,\mathrm{mol} \] The added acid supplies \[ n_{\mathrm{H^+}}=0.00700\,\mathrm{mol} \] \(\textbf{Complete the first protonation stage:}\) Every mole of carbonate requires one mole of hydrogen ions to become bicarbonate. Therefore, \[ n_{\mathrm{H^+,stage\,1}}=0.00500\,\mathrm{mol} \] This produces \[ n_{\mathrm{HCO_3^-,formed}}=0.00500\,\mathrm{mol} \] \(\textbf{Determine the remaining hydrogen ions:}\) \[ n_{\mathrm{H^+,remaining}} =0.00700-0.00500 \] \[ n_{\mathrm{H^+,remaining}}=0.00200\,\mathrm{mol} \] \(\textbf{Apply the second protonation stage:}\) The remaining hydrogen ions convert bicarbonate into carbonic acid in a \(1:1\) mole ratio. Therefore, \[ n_{\mathrm{HCO_3^-,consumed}}=0.00200\,\mathrm{mol} \] The amount of bicarbonate remaining is \[ n_{\mathrm{HCO_3^-,final}} =0.00500-0.00200 \] \[ n_{\mathrm{HCO_3^-,final}}=0.00300\,\mathrm{mol} \] \(\textbf{Calculate the amount of carbon dioxide evolved:}\) Every mole of carbonic acid formed gives one mole of carbon dioxide. Therefore, \[ n_{\mathrm{CO_2}}=0.00200\,\mathrm{mol} \] \(\textbf{Verify carbon conservation:}\) The initial amount of carbon-containing carbonate units is \[ 0.00500\,\mathrm{mol} \] The final carbon-containing amounts are \[ 0.00300+0.00200 \] \[ =0.00500\,\mathrm{mol} \] Thus, the carbon balance is satisfied. \(\textbf{Interpret the acid requirement:}\) Complete conversion of all carbonate into carbon dioxide would require two moles of hydrogen ions per mole of carbonate. For the initial sample, that would require \[ 2(0.00500)=0.0100\,\mathrm{mol} \] Only \(0.00700\,\mathrm{mol}\) is supplied. Therefore, enough acid is available to complete the first protonation but not the second. This explains why both bicarbonate and evolved carbon dioxide are obtained. The calculation demonstrates the importance of treating successive protonation stages separately. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{HCO_3^-,final}}=0.00300\,\mathrm{mol},\quad n_{\mathrm{CO_2}}=0.00200\,\mathrm{mol}} \]