Equilibrium MCQs with Answers – Part 9 (NEET Chemistry)

Explanation: \(\textbf{Identify the two hydrogen-ion sources:}\) The weak acid undergoes \[ \mathrm{HA\rightleftharpoons H^++A^-} \] Water also undergoes autoionization: \[ \mathrm{2H_2O\rightleftharpoons H_3O^++OH^-} \] At the specified analytical acid concentration, the hydrogen-ion contribution from water cannot be neglected. \(\textbf{Write the acid-derived material balance:}\) \[ C=[\mathrm{HA}]+[\mathrm{A^-}] \] Using the acid-ionization expression, \[ K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \] we obtain \[ [\mathrm{A^-}] =\frac{K_aC}{[\mathrm{H^+}]+K_a} \] \(\textbf{Apply electrical neutrality:}\) The only appreciable ions are H⁺, OH⁻ and A⁻. Therefore, \[ [\mathrm{H^+}]=[\mathrm{OH^-}]+[\mathrm{A^-}] \] Let \[ h=[\mathrm{H^+}] \] Using \[ [\mathrm{OH^-}]=\frac{K_w}{h} \] gives \[ h=\frac{K_w}{h}+\frac{K_aC}{h+K_a} \] \(\textbf{Identify a useful approximation:}\) The stated acid-ionization constant is \[ K_a=1.00\times10^{-5} \] The expected hydrogen-ion concentration is near \(10^{-7}\,\mathrm{M}\). Thus, \[ h\ll K_a \] and the acid is expected to be almost completely ionized despite being classified as a weak acid at ordinary concentrations. Under this approximation, \[ [\mathrm{A^-}]\approx C \] \(\textbf{Construct the approximate hydrogen-ion equation:}\) \[ h\approx\frac{K_w}{h}+C \] Multiplying by h, \[ h^2-Ch-K_w\approx0 \] \(\textbf{Solve for the physically acceptable root:}\) \[ h\approx\frac{C+\sqrt{C^2+4K_w}}2 \] Substituting, \[ h\approx\frac{10^{-7}+\sqrt{10^{-14}+4(10^{-14})}}2 \] \[ h\approx\frac{10^{-7}+\sqrt5\times10^{-7}}2 \] \[ h\approx1.618\times10^{-7}\,\mathrm{M} \] \(\textbf{Refine the degree of ionization:}\) The fraction ionized is \[ \alpha=\frac{[\mathrm{A^-}]}C \] Using the exact acid-speciation relation, \[ \alpha=\frac{K_a}{K_a+h} \] Substituting the estimated h, \[ \alpha\approx\frac{1.00\times10^{-5}}{1.00\times10^{-5}+1.618\times10^{-7}} \] \[ \alpha\approx0.9841 \] Thus, \[ \alpha\approx98.4\% \] \(\textbf{Check the refined charge balance:}\) The refined A⁻ concentration is \[ [\mathrm{A^-}]\approx0.9841(10^{-7}) \] \[ [\mathrm{A^-}]\approx9.841\times10^{-8}\,\mathrm{M} \] The hydrogen-ion concentration satisfying the charge balance is consequently close to \[ h\approx1.61\times10^{-7}\,\mathrm{M} \] The correction is small at the precision requested. \(\textbf{Determine the pH:}\) \[ pH=-\log_{10}h \] Using the refined value, \[ pH\approx6.79 \] \(\textbf{Interpret almost complete ionization of a weak acid:}\) An acid's classification as weak refers to its ionization equilibrium constant, not to an immutable percentage ionized at every concentration. Extreme dilution can produce a very high degree of ionization. Nevertheless, the total analytical amount of acid remains extremely small. Water autoionization therefore contributes appreciably to the total hydrogen-ion concentration. Treating the acid as almost completely ionized but ignoring water would still produce an inaccurate pH. The problem distinguishes acid strength, fractional ionization and the concentration-dependent importance of water autoionization. \(\textbf{Final answer:}\) \[ \boxed{pH\approx6.79,\qquad\alpha\approx98.4\%} \]

Explanation: \(\textbf{Identify the two simultaneous equilibrium requirements:}\) If solid iodine remains, the molecular iodine vapour pressure is \[ p_{\mathrm{I_2}}=0.200\,\mathrm{bar} \] The gaseous dissociation equilibrium additionally requires \[ \frac{p_{\mathrm I}^2}{p_{\mathrm{I_2}}}=0.800\,\mathrm{bar} \] These conditions must be satisfied together while the solid and both gaseous species coexist. \(\textbf{Determine the atomic iodine pressure under solid coexistence:}\) Substituting the saturated molecular iodine pressure, \[ p_{\mathrm I}^2=(0.800)(0.200) \] \[ p_{\mathrm I}^2=0.160\,\mathrm{bar^2} \] Therefore, \[ p_{\mathrm I}=0.400\,\mathrm{bar} \] \(\textbf{Calculate the corresponding total gas pressure:}\) \[ P_{\mathrm{gas}}=p_{\mathrm{I_2}}+p_{\mathrm I} \] \[ P_{\mathrm{gas}}=0.200+0.400 \] \[ P_{\mathrm{gas}}=0.600\,\mathrm{bar} \] \(\textbf{Determine the iodine inventory required for this coexistence state:}\) Because \[ RT/V=1.00\,\mathrm{bar\,mol^{-1}} \] the amounts of gaseous species are \[ n_{\mathrm{I_2(g)}}=0.200\,\mathrm{mol} \] and \[ n_{\mathrm{I(g)}}=0.400\,\mathrm{mol} \] Each mole of molecular I₂ contains two iodine atoms. Therefore, the gaseous inventory expressed in moles of original I₂ formula units is \[ n_{\mathrm{I_2,equiv}}=n_{\mathrm{I_2(g)}}+\frac12n_{\mathrm{I(g)}} \] \[ n_{\mathrm{I_2,equiv}}=0.200+\frac12(0.400) \] \[ n_{\mathrm{I_2,equiv}}=0.400\,\mathrm{mol} \] \(\textbf{Test the available solid in P:}\) P initially contains \(0.500\,\mathrm{mol}\) I₂. The coexistence state requires \(0.400\,\mathrm{mol}\) of I₂-equivalent material in the gas phase. Therefore, \[ n_{\mathrm{I_2(s)},P,f}=0.500-0.400 \] \[ n_{\mathrm{I_2(s)},P,f}=0.100\,\mathrm{mol} \] Solid remains present. Thus, the calculated coexistence pressures apply to P: \[ P_P=0.600\,\mathrm{bar} \] \(\textbf{Test the available iodine inventory in Q:}\) Q initially contains only \(0.150\,\mathrm{mol}\) I₂. This is insufficient to establish the solid–gas coexistence composition requiring \(0.400\,\mathrm{mol}\) I₂-equivalent material in the gas phase. Therefore, all the solid in Q disappears. The final gases must still satisfy the gaseous dissociation equilibrium. \(\textbf{Construct Q's iodine-atom balance:}\) Let \[ x=n_{\mathrm{I_2(g)}} \] and \[ y=n_{\mathrm{I(g)}} \] Conservation of original I₂ formula units gives \[ x+\frac y2=0.150 \] Therefore, \[ x=0.150-\frac y2 \] \(\textbf{Apply the gaseous dissociation equilibrium in Q:}\) Because the pressure conversion factor is \(1.00\,\mathrm{bar\,mol^{-1}}\), the numerical partial pressures in bar equal the corresponding gas amounts in moles. Thus, \[ 0.800=\frac{y^2}{x} \] Substituting the material balance, \[ y^2=0.800\left(0.150-\frac y2\right) \] \[ y^2+0.400y-0.120=0 \] \(\textbf{Solve for the physically acceptable gas composition:}\) The positive root is \[ y=\frac{-0.400+\sqrt{0.160+0.480}}2 \] \[ y=\frac{-0.400+0.800}{2} \] \[ y=0.200\,\mathrm{mol} \] Therefore, \[ x=0.150-\frac{0.200}{2} \] \[ x=0.0500\,\mathrm{mol} \] \(\textbf{Calculate Q's final gas pressure:}\) \[ p_{\mathrm I}=0.200\,\mathrm{bar} \] \[ p_{\mathrm{I_2}}=0.0500\,\mathrm{bar} \] Therefore, \[ P_Q=0.200+0.0500 \] \[ P_Q=0.250\,\mathrm{bar} \] \(\textbf{Verify that solid iodine cannot remain in Q:}\) The final molecular iodine pressure is \[ p_{\mathrm{I_2}}=0.0500\,\mathrm{bar} \] This is below its characteristic coexistence pressure of \(0.200\,\mathrm{bar}\). The absence of solid iodine is therefore consistent with the calculated final state. \(\textbf{Interpret the coupled phase and chemical equilibria:}\) P has enough iodine to establish both solid–vapour equilibrium and gaseous molecular dissociation while retaining solid material. Q has insufficient iodine to reach that two-phase condition. Its solid disappears, but the remaining gas mixture still satisfies the molecular dissociation equilibrium. Consequently, the two vessels have different total gas pressures despite identical temperatures and gas-space volumes. This example shows that coupled equilibrium conditions must be applied only when the required phases are actually present. \(\textbf{Final answer:}\) \[ \boxed{P_P=0.600\,\mathrm{bar},\qquad P_Q=0.250\,\mathrm{bar}} \]

Explanation: \(\textbf{Identify the hydrolysing species:}\) NaA dissociates essentially completely: \[ \mathrm{NaA\rightarrow Na^++A^-} \] The anion A⁻ is the conjugate base of weak acid HA. It can react with water: \[ \mathrm{A^-+H_2O\rightleftharpoons HA+OH^-} \] Therefore, the solution is expected to be basic rather than acidic. \(\textbf{Calculate the conjugate-base constant:}\) \[ K_b(\mathrm{A^-})=\frac{K_w}{K_a(\mathrm{HA})} \] Substituting, \[ K_b=\frac{1.00\times10^{-14}}{1.00\times10^{-5}} \] \[ K_b=1.00\times10^{-9} \] \(\textbf{Examine the usual small-hydrolysis estimate:}\) The analytical salt concentration is \[ C=1.00\times10^{-5}\,\mathrm{M} \] The customary approximation gives \[ [\mathrm{OH^-}]_{\mathrm{approx}}\approx\sqrt{K_bC} \] \[ [\mathrm{OH^-}]_{\mathrm{approx}} \approx\sqrt{(10^{-9})(10^{-5})} \] \[ [\mathrm{OH^-}]_{\mathrm{approx}}=1.00\times10^{-7}\,\mathrm{M} \] \(\textbf{Identify the approximation failure:}\) Pure water already contains hydroxide ions at a concentration near \(10^{-7}\,\mathrm{M}\) at 298 K. The estimated hydroxide concentration due to salt hydrolysis is therefore not large compared with the water contribution. Neglecting water autoionization is unjustified. \(\textbf{Construct the analytical and charge balances:}\) Let \[ h=[\mathrm{H^+}] \] The sodium concentration is \[ [\mathrm{Na^+}]=C \] The total A-derived concentration is \[ C=[\mathrm{HA}]+[\mathrm{A^-}] \] Electrical neutrality gives \[ h+C=[\mathrm{A^-}]+[\mathrm{OH^-}] \] Using the analytical balance, \[ [\mathrm{OH^-}]=h+[\mathrm{HA}] \] \(\textbf{Express HA using the acid-ionization equilibrium:}\) The acid constant gives \[ K_a=\frac{h[\mathrm{A^-}]}{[\mathrm{HA}]} \] Therefore, \[ [\mathrm{HA}]=\frac{Ch}{K_a+h} \] Substituting into the charge-balance result, \[ [\mathrm{OH^-}]=h+\frac{Ch}{K_a+h} \] \(\textbf{Combine the result with water autoionization:}\) \[ K_w=h[\mathrm{OH^-}] \] Thus, \[ K_w=h^2\left(1+\frac{C}{K_a+h}\right) \] \(\textbf{Apply the appropriate small-h approximation:}\) The expected hydrogen-ion concentration is close to \(10^{-7}\,\mathrm{M}\), much smaller than \[ K_a=1.00\times10^{-5} \] Therefore, \[ K_a+h\approx K_a \] and \[ K_w\approx h^2\left(1+\frac C{K_a}\right) \] Because \[ C=K_a \] we obtain \[ K_w\approx2h^2 \] \(\textbf{Calculate the hydrogen-ion concentration:}\) \[ h\approx\sqrt{\frac{K_w}{2}} \] \[ h\approx\sqrt{\frac{1.00\times10^{-14}}2} \] \[ h\approx7.07\times10^{-8}\,\mathrm{M} \] \(\textbf{Determine the pH:}\) \[ pH=-\log_{10}(7.07\times10^{-8}) \] \[ pH\approx7.15 \] \(\textbf{Verify the expected basic character:}\) The corresponding hydroxide concentration is \[ [\mathrm{OH^-}]=\frac{1.00\times10^{-14}}{7.07\times10^{-8}} \] \[ [\mathrm{OH^-}]\approx1.41\times10^{-7}\,\mathrm{M} \] Thus, \[ [\mathrm{OH^-}]\gt [\mathrm{H^+}] \] The solution is slightly basic. \(\textbf{Interpret the importance of the validity check:}\) The usual hydrolysis formula assumes the hydroxide ions produced by hydrolysis dominate those supplied by water. That assumption becomes questionable at very low salt concentrations. The original estimate fails to distinguish the hydroxide generated through salt hydrolysis from the total equilibrium hydroxide concentration. Using electrical neutrality and Kw avoids this problem. The final result also provides a chemical consistency check: a sodium salt of a weak acid should not be classified as acidic merely because an approximation produces a misleading numerical concentration. The problem illustrates why a mathematically familiar formula must be checked against its underlying physical assumptions. \(\textbf{Final answer:}\) \[ \boxed{pH\approx7.15,\qquad[\mathrm{OH^-}]\approx1.41\times10^{-7}\,\mathrm{M}} \]

Explanation: \(\textbf{Represent the dissociation extent in P:}\) Let αP be the fraction of the original A that dissociates. The equilibrium amounts are \[ n_A=1-\alpha_P \] \[ n_B=2\alpha_P \] Therefore, \[ n_{\mathrm{total},P}=1+\alpha_P \] \(\textbf{Express the equilibrium partial pressures in P:}\) Using Dalton's law, \[ p_A=\frac{1-\alpha_P}{1+\alpha_P}P_{\mathrm{total}} \] \[ p_B=\frac{2\alpha_P}{1+\alpha_P}P_{\mathrm{total}} \] \(\textbf{Construct the pressure equilibrium expression:}\) \[ K_p=\frac{p_B^2}{p_A} \] Substituting the mole-fraction expressions, \[ K_p=\frac{4\alpha_P^2P_{\mathrm{total}}}{1-\alpha_P^2} \] \(\textbf{Determine the degree of dissociation in P:}\) Using the supplied pressure and constant, \[ 6.00=\frac{4\alpha_P^2(3.00)}{1-\alpha_P^2} \] \[ 6(1-\alpha_P^2)=12\alpha_P^2 \] \[ 6=18\alpha_P^2 \] \[ \alpha_P^2=\frac13 \] Therefore, \[ \alpha_P=\frac1{\sqrt3} \] \[ \alpha_P\approx0.577 \] \(\textbf{Represent the dissociation extent in Q:}\) Q initially contains one mole of inert gas in addition to one mole of A. The inert gas does not participate chemically. At equilibrium, \[ n_A=1-\alpha_Q \] \[ n_B=2\alpha_Q \] \[ n_{\mathrm{inert}}=1.00 \] Therefore, \[ n_{\mathrm{total},Q}=2+\alpha_Q \] \(\textbf{Express the equilibrium partial pressures in Q:}\) \[ p_A=\frac{1-\alpha_Q}{2+\alpha_Q}P_{\mathrm{total}} \] \[ p_B=\frac{2\alpha_Q}{2+\alpha_Q}P_{\mathrm{total}} \] \(\textbf{Construct Q's equilibrium expression:}\) \[ K_p=\frac{p_B^2}{p_A} \] Substituting, \[ K_p=\frac{4\alpha_Q^2P_{\mathrm{total}}}{(1-\alpha_Q)(2+\alpha_Q)} \] \(\textbf{Solve for the degree of dissociation in Q:}\) \[ 6.00=\frac{4\alpha_Q^2(3.00)}{(1-\alpha_Q)(2+\alpha_Q)} \] Rearranging, \[ 6(1-\alpha_Q)(2+\alpha_Q)=12\alpha_Q^2 \] Dividing by six, \[ (1-\alpha_Q)(2+\alpha_Q)=2\alpha_Q^2 \] Expanding, \[ 2-\alpha_Q-\alpha_Q^2=2\alpha_Q^2 \] \[ 3\alpha_Q^2+\alpha_Q-2=0 \] \(\textbf{Select the physically acceptable root:}\) Factoring, \[ (3\alpha_Q-2)(\alpha_Q+1)=0 \] The root \(\alpha_Q=-1\) is physically impossible. Therefore, \[ \alpha_Q=\frac23 \] \[ \alpha_Q\approx0.667 \] \(\textbf{Compare the two equilibrium dissociation fractions:}\) \[ \alpha_P\approx0.577 \] \[ \alpha_Q\approx0.667 \] Thus, dissociation is greater in Q. \(\textbf{Interpret the inert-gas effect under constant pressure:}\) At fixed total pressure, adding inert gas increases the total gas amount and allows the piston to occupy a larger volume. This lowers the partial pressures of the reacting gases relative to a mixture without inert gas at the same total pressure. The dissociation reaction produces more gaseous molecules than it consumes. The greater expansion associated with inert gas therefore favours dissociation. The effect is different from adding inert gas at fixed temperature and fixed volume, where the reacting-gas partial pressures would remain unchanged immediately. The equilibrium constant itself is identical in P and Q because the temperature is the same. The difference in α arises from the gas composition and pressure constraint rather than a change in Kp. \(\textbf{Final answer:}\) \[ \boxed{\alpha_P=\frac1{\sqrt3}\approx0.577,\qquad \alpha_Q=\frac23\approx0.667} \]

Explanation: \(\textbf{Identify the two transformations of the reaction equation:}\) The original reaction is first reversed. Its stoichiometric coefficients are then divided by two. Both operations affect the numerical equilibrium constant. They also affect the reaction Gibbs-energy change because that quantity refers to a specified stoichiometric reaction. \(\textbf{Apply the reaction-reversal rule:}\) For the complete reverse reaction, \[ \mathrm{2NH_3\rightleftharpoons N_2+3H_2} \] the equilibrium constant is \[ K_{p,\mathrm{reverse}}=\frac1{K_{p,\mathrm{forward}}} \] Therefore, \[ K_{p,\mathrm{reverse}}=\frac1{4.00} \] \[ K_{p,\mathrm{reverse}}=0.250\,\mathrm{bar^2} \] \(\textbf{Account for halving the stoichiometric coefficients:}\) Dividing all coefficients by two takes the square root of the equilibrium constant. Hence, \[ K_p'=\left(K_{p,\mathrm{reverse}}\right)^{1/2} \] \[ K_p'=\sqrt{0.250} \] \[ K_p'=0.500\,\mathrm{bar} \] The pressure-unit exponent is also halved. \(\textbf{Determine the quotient for the original forward reaction:}\) \[ Q_{p,f}=\frac{p_{\mathrm{NH_3}}^2} {p_{\mathrm{N_2}}p_{\mathrm{H_2}}^3} \] Substituting, \[ Q_{p,f}=\frac{(1.00)^2}{(2.00)(1.00)^3} \] \[ Q_{p,f}=0.500\,\mathrm{bar^{-2}} \] \(\textbf{Calculate the actual forward reaction Gibbs energy:}\) The governing equation is \[ \Delta_rG_f=RT\ln\frac{Q_{p,f}}{K_{p,f}} \] Substituting, \[ \Delta_rG_f=(2.50)\ln\frac{0.500}{4.00} \] \[ \Delta_rG_f=(2.50)\ln\frac18 \] Since \[ \ln\frac18=-3\ln2 \] we obtain \[ \Delta_rG_f=-(2.50)(3)(0.6931) \] \[ \Delta_rG_f\approx-5.20\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Transform the Gibbs energy for the new reaction:}\) Reversing a reaction changes the sign of its reaction Gibbs energy. Halving the stoichiometric coefficients halves its magnitude. Therefore, \[ \Delta_rG'=-\frac12\Delta_rG_f \] \[ \Delta_rG'=-\frac12(-5.20) \] \[ \Delta_rG'\approx+2.60\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Verify the result using the new reaction quotient:}\) For the newly written reaction, \[ Q_p'=\frac{p_{\mathrm{N_2}}^{1/2}p_{\mathrm{H_2}}^{3/2}} {p_{\mathrm{NH_3}}} \] Substituting, \[ Q_p'=\frac{(2.00)^{1/2}(1.00)^{3/2}}{1.00} \] \[ Q_p'=\sqrt2\,\mathrm{bar} \] Thus, \[ \frac{Q_p'}{K_p'}=\frac{\sqrt2}{0.500} \] \[ \frac{Q_p'}{K_p'}=2\sqrt2 \] Taking the logarithm gives \[ \ln(2\sqrt2)=\frac32\ln2 \] Therefore, \[ \Delta_rG'=(2.50)\left(\frac32\right)(0.6931) \] \[ \Delta_rG'\approx+2.60\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the thermodynamic direction:}\) The actual mixture favours ammonia formation according to the original forward equation. Consequently, the newly written reverse reaction is thermodynamically unfavourable. Its positive Gibbs-energy change is consistent with that conclusion. The equilibrium constant and the reaction Gibbs-energy change must both be transformed when the balanced equation is reversed or rescaled. A negative forward reaction Gibbs energy does not imply that every differently written representation has the same numerical Gibbs-energy change. The problem integrates reaction-equation manipulation, equilibrium-constant algebra and composition-dependent Gibbs energy. \(\textbf{Final answer:}\) \[ \boxed{K_p'=0.500\,\mathrm{bar},\qquad \Delta_rG'\approx+2.60\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Identify the three connected equilibria:}\) The ammonia gas pressure determines the dissolved molecular ammonia concentration. Dissolved ammonia establishes the basic equilibrium \[ \mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-} \] The resulting hydroxide concentration determines whether \[ \mathrm{M(OH)_2(s)\rightleftharpoons M^{2+}+2OH^-} \] can remain entirely dissolved or produces a precipitate. \(\textbf{Determine the hydroxide threshold for precipitation:}\) For M(OH)₂, \[ K_{sp}=[\mathrm{M^{2+}}][\mathrm{OH^-}]^2 \] Before appreciable precipitation, \[ [\mathrm{M^{2+}}]\approx0.0100\,\mathrm{M} \] At the onset of precipitation, \[ [\mathrm{OH^-}]_{\mathrm{critical}} =\sqrt{\frac{K_{sp}}{[\mathrm{M^{2+}}]}} \] Substituting, \[ [\mathrm{OH^-}]_{\mathrm{critical}} =\sqrt{\frac{4.00\times10^{-12}}{0.0100}} \] \[ [\mathrm{OH^-}]_{\mathrm{critical}} =2.00\times10^{-5}\,\mathrm{M} \] \(\textbf{Connect the critical hydroxide concentration to ammonia:}\) The weak-base equilibrium gives \[ K_b=\frac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]} \] Therefore, \[ [\mathrm{NH_3}]_{\mathrm{critical}} =\frac{[\mathrm{NH_4^+}][\mathrm{OH^-}]_{\mathrm{critical}}}{K_b} \] Substituting, \[ [\mathrm{NH_3}]_{\mathrm{critical}} =\frac{(0.0100)(2.00\times10^{-5})}{1.00\times10^{-5}} \] \[ [\mathrm{NH_3}]_{\mathrm{critical}} =0.0200\,\mathrm{M} \] \(\textbf{Recover the critical gas pressure:}\) Using the gas-dissolution relation, \[ p_{\mathrm{critical}}=\frac{[\mathrm{NH_3}]_{\mathrm{critical}}}{k} \] \[ p_{\mathrm{critical}}=\frac{0.0200}{0.100} \] \[ p_{\mathrm{critical}}=0.200\,\mathrm{bar} \] \(\textbf{Evaluate the final dissolved ammonia concentration:}\) At \[ p_{\mathrm{NH_3}}=0.400\,\mathrm{bar} \] Henry's-law-type dissolution gives \[ [\mathrm{NH_3}]=(0.100)(0.400) \] \[ [\mathrm{NH_3}]=0.0400\,\mathrm{M} \] \(\textbf{Calculate the final hydroxide concentration:}\) From the ammonia equilibrium, \[ [\mathrm{OH^-}] =K_b\frac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} \] \[ [\mathrm{OH^-}] =(1.00\times10^{-5})\frac{0.0400}{0.0100} \] \[ [\mathrm{OH^-}]=4.00\times10^{-5}\,\mathrm{M} \] \(\textbf{Determine the free metal-ion concentration after precipitation:}\) Once M(OH)₂ solid is present, \[ [\mathrm{M^{2+}}]_f =\frac{K_{sp}}{[\mathrm{OH^-}]^2} \] Substituting, \[ [\mathrm{M^{2+}}]_f =\frac{4.00\times10^{-12}}{(4.00\times10^{-5})^2} \] \[ [\mathrm{M^{2+}}]_f =\frac{4.00\times10^{-12}}{1.60\times10^{-9}} \] \[ [\mathrm{M^{2+}}]_f=0.00250\,\mathrm{M} \] \(\textbf{Calculate the amount of metal removed:}\) The original metal concentration was \(0.0100\,\mathrm{M}\). Therefore, the precipitated amount in \(1.00\,\mathrm{L}\) is \[ n_{\mathrm{precipitated}} =(0.0100-0.00250)(1.00) \] \[ n_{\mathrm{precipitated}}=0.00750\,\mathrm{mol} \] The original metal inventory was \(0.0100\,\mathrm{mol}\). Hence, \[ f_{\mathrm{precipitated}} =\frac{0.00750}{0.0100} \] \[ f_{\mathrm{precipitated}}=0.750 \] \(\textbf{Check the initial state:}\) At \(0.100\,\mathrm{bar}\), \[ [\mathrm{NH_3}]=0.0100\,\mathrm{M} \] and \[ [\mathrm{OH^-}]=1.00\times10^{-5}\,\mathrm{M} \] The initial metal-hydroxide ionic product is \[ Q_{sp}=(0.0100)(1.00\times10^{-5})^2 \] \[ Q_{sp}=1.00\times10^{-12}\lt K_{sp} \] Thus, precipitation is not favoured initially. \(\textbf{Interpret gas-pressure-controlled precipitation:}\) Increasing ammonia pressure raises the concentration of dissolved NH₃. Because the ammonium concentration is externally maintained, the hydroxide concentration also increases. This can drive a sparingly soluble metal hydroxide beyond its saturation threshold. Once precipitation occurs, the free metal-ion concentration decreases until its ionic product again equals Ksp. The gas pressure, weak-base equilibrium and heterogeneous precipitation equilibrium must therefore be satisfied together. This is not a direct conversion of gas pressure into precipitated mass. The calculation requires three connected chemical relationships and a final metal-ion material balance. \(\textbf{Final answer:}\) \[ \boxed{p_{\mathrm{critical}}=0.200\,\mathrm{bar},\qquad f_{\mathrm{precipitated}}=75.0\%} \]

Explanation: \(\textbf{Identify the acid-base source of sulfide ions:}\) Hydrogen sulfide ionizes successively: \[ \mathrm{H_2S\rightleftharpoons H^++HS^-} \] \[ \mathrm{HS^-\rightleftharpoons H^++S^{2-}} \] The sulfide-ion concentration depends strongly on the maintained hydrogen-ion concentration. \(\textbf{Combine the successive ionization constants:}\) For the first stage, \[ K_{a1}=\frac{[\mathrm{H^+}][\mathrm{HS^-}]}{[\mathrm{H_2S}]} \] For the second stage, \[ K_{a2}=\frac{[\mathrm{H^+}][\mathrm{S^{2-}}]}{[\mathrm{HS^-}]} \] Multiplying, \[ K_{a1}K_{a2} =\frac{[\mathrm{H^+}]^2[\mathrm{S^{2-}}]}{[\mathrm{H_2S}]} \] Therefore, \[ [\mathrm{S^{2-}}] =\frac{K_{a1}K_{a2}[\mathrm{H_2S}]}{[\mathrm{H^+}]^2} \] \(\textbf{Insert the specified acid constants:}\) \[ K_{a1}K_{a2}[\mathrm{H_2S}] =(10^{-7})(10^{-14})(0.100) \] \[ K_{a1}K_{a2}[\mathrm{H_2S}]=1.00\times10^{-22} \] Hence, \[ [\mathrm{S^{2-}}] =\frac{1.00\times10^{-22}}{[\mathrm{H^+}]^2} \] \(\textbf{Determine the sulfide concentration needed for MS precipitation:}\) The MS saturation condition is \[ K_{sp}(\mathrm{MS})=[\mathrm{M^{2+}}][\mathrm{S^{2-}}] \] At the onset, \[ [\mathrm{S^{2-}}]_{\mathrm{M,critical}} =\frac{1.00\times10^{-15}}{0.0100} \] \[ [\mathrm{S^{2-}}]_{\mathrm{M,critical}} =1.00\times10^{-13}\,\mathrm{M} \] \(\textbf{Recover the hydrogen-ion concentration at the first threshold:}\) \[ 1.00\times10^{-13} =\frac{1.00\times10^{-22}}{[\mathrm{H^+}]^2} \] Therefore, \[ [\mathrm{H^+}]^2=1.00\times10^{-9} \] \[ [\mathrm{H^+}]=3.162\times10^{-5}\,\mathrm{M} \] Thus, \[ pH_{\mathrm{M,critical}}=4.50 \] \(\textbf{Determine the sulfide concentration needed for NS precipitation:}\) For NS, \[ K_{sp}(\mathrm{NS})=[\mathrm{N^{2+}}][\mathrm{S^{2-}}] \] At its onset, \[ [\mathrm{S^{2-}}]_{\mathrm{N,critical}} =\frac{1.00\times10^{-12}}{0.0100} \] \[ [\mathrm{S^{2-}}]_{\mathrm{N,critical}} =1.00\times10^{-10}\,\mathrm{M} \] \(\textbf{Recover the second pH threshold:}\) \[ 1.00\times10^{-10} =\frac{1.00\times10^{-22}}{[\mathrm{H^+}]^2} \] Therefore, \[ [\mathrm{H^+}]^2=1.00\times10^{-12} \] \[ [\mathrm{H^+}]=1.00\times10^{-6}\,\mathrm{M} \] Thus, \[ pH_{\mathrm{N,critical}}=6.00 \] \(\textbf{Identify the selective-precipitation region:}\) MS reaches its saturation threshold at pH 4.50. NS reaches its threshold only at pH 6.00. Therefore, between these values, MS precipitation is favoured while NS remains below its own precipitation threshold under the stated assumptions. \(\textbf{Interpret the role of acidity:}\) Raising pH decreases the hydrogen-ion concentration. The combined hydrogen sulfide ionization relation then increases free sulfide concentration through an inverse-square dependence on hydrogen-ion concentration. The two metal sulfides respond at different pH thresholds because their solubility products differ. This provides a chemical basis for selective precipitation under controlled acidity. The threshold calculations concern the BEGINNING of precipitation, not the assumption that either metal is completely precipitated at its threshold. As M²⁺ is removed, its free concentration decreases and must continue to satisfy the MS solubility-product condition. The result connects multistage acid ionization with two competing heterogeneous precipitation equilibria. \(\textbf{Final answer:}\) \[ \boxed{4.50\lt pH\lt 6.00} \]

Explanation: \(\textbf{Verify the original equilibrium composition:}\) For the reaction as written, \[ K_p=\frac{p_{\mathrm{CO_2}}p_{\mathrm{H_2}}} {p_{\mathrm{CO}}p_{\mathrm{H_2O}}} \] The number of gaseous molecules is the same on both sides. Therefore, the common ideal-gas factor \(RT/V\) cancels. The equilibrium constant can be evaluated directly from gas mole amounts: \[ K_p=\frac{n_{\mathrm{CO_2}}n_{\mathrm{H_2}}} {n_{\mathrm{CO}}n_{\mathrm{H_2O}}} \] Substituting, \[ K_p=\frac{(2.00)(2.00)}{(1.00)(1.00)} \] \[ K_p=4.00 \] \(\textbf{Determine the required additional reaction extent:}\) The initial CO amount is \(1.00\,\mathrm{mol}\). The target final amount is \(0.750\,\mathrm{mol}\). Therefore, the required additional forward reaction extent is \[ \xi=1.00-0.750 \] \[ \xi=0.250\,\mathrm{mol} \] \(\textbf{Represent the selective hydrogen removal:}\) Let x moles of H₂ be removed. Immediately after removal, \[ n_{\mathrm{CO}}=1.00 \] \[ n_{\mathrm{H_2O}}=1.00 \] \[ n_{\mathrm{CO_2}}=2.00 \] \[ n_{\mathrm{H_2}}=2.00-x \] \(\textbf{Construct the required final composition:}\) The desired forward extent is \(0.250\,\mathrm{mol}\). Therefore, \[ n_{\mathrm{CO},f}=1.00-0.250 \] \[ n_{\mathrm{CO},f}=0.750\,\mathrm{mol} \] Similarly, \[ n_{\mathrm{H_2O},f}=0.750\,\mathrm{mol} \] The carbon dioxide amount becomes \[ n_{\mathrm{CO_2},f}=2.00+0.250 \] \[ n_{\mathrm{CO_2},f}=2.250\,\mathrm{mol} \] The final hydrogen amount is \[ n_{\mathrm{H_2},f}=2.00-x+0.250 \] \[ n_{\mathrm{H_2},f}=2.250-x \] \(\textbf{Impose the equilibrium condition at the target conversion:}\) Because temperature is unchanged, \[ K_p=4.00 \] Therefore, \[ 4.00=\frac{(2.250)(2.250-x)}{(0.750)(0.750)} \] \(\textbf{Solve for the hydrogen amount removed:}\) \[ 4.00(0.5625)=2.250(2.250-x) \] \[ 2.250=2.250(2.250-x) \] Dividing by \(2.250\), \[ 1.00=2.250-x \] Hence, \[ x=1.250\,\mathrm{mol} \] \(\textbf{Check the immediate direction after removal:}\) Immediately after removing hydrogen, \[ n_{\mathrm{H_2}}'=2.00-1.25 \] \[ n_{\mathrm{H_2}}'=0.750\,\mathrm{mol} \] The immediate quotient becomes \[ Q_p'=\frac{(2.00)(0.750)}{(1.00)(1.00)} \] \[ Q_p'=1.50 \] Since \[ Q_p'\lt K_p \] the additional forward conversion required by the experimental objective is favoured. \(\textbf{Calculate the initial total gas amount:}\) \[ n_{\mathrm{total},i}=1.00+1.00+2.00+2.00 \] \[ n_{\mathrm{total},i}=6.00\,\mathrm{mol} \] \(\textbf{Calculate the final total gas amount:}\) At the new equilibrium, \[ n_{\mathrm{total},f}=0.750+0.750+2.250+1.000 \] \[ n_{\mathrm{total},f}=4.750\,\mathrm{mol} \] \(\textbf{Determine the total-pressure ratio:}\) The vessel temperature and volume are fixed. Therefore, \[ P_{\mathrm{total}}\propto n_{\mathrm{total}} \] Hence, \[ \frac{P_f}{P_i}=\frac{4.750}{6.00} \] \[ \frac{P_f}{P_i}\approx0.792 \] \(\textbf{Interpret the material and pressure changes:}\) Selective hydrogen removal decreases the reaction quotient and favours additional hydrogen formation. The required removal amount must be determined from the target equilibrium composition rather than from the stoichiometric extent alone. Although the subsequent forward chemical reaction proceeds, it has zero change in total gaseous stoichiometric mole number. Consequently, chemical re-equilibration itself does not alter the total gas-mole inventory in this reaction. The net total-pressure reduction is caused by selective hydrogen removal. The result combines an inverse equilibrium-design calculation with a separate gas-mole conservation argument. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{H_2,removed}}=1.250\,\mathrm{mol},\qquad \frac{P_f}{P_i}\approx0.792} \]

Explanation: \(\textbf{Verify the original complex-formation equilibrium:}\) The equilibrium constant is \[ K_c=\frac{[\mathrm{FeSCN^{2+}}]} {[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]} \] Substituting the initial values, \[ K_c=\frac{0.0200}{(0.0200)(0.0100)} \] \[ K_c=100\,\mathrm{L\,mol^{-1}} \] The original mixture is therefore consistent with equilibrium. \(\textbf{Determine the immediate effect of dilution:}\) The solution volume doubles from \(1.00\) to \(2.00\,\mathrm{L}\). Before further reaction occurs, each reacting-species concentration is halved: \[ [\mathrm{Fe^{3+}}]'=0.0100\,\mathrm{M} \] \[ [\mathrm{SCN^-}]'=0.00500\,\mathrm{M} \] \[ [\mathrm{FeSCN^{2+}}]'=0.0100\,\mathrm{M} \] \(\textbf{Evaluate the reaction quotient after dilution:}\) \[ Q_c'=\frac{0.0100}{(0.0100)(0.00500)} \] \[ Q_c'=200\,\mathrm{L\,mol^{-1}} \] Since \[ Q_c'\gt K_c \] the reverse reaction would be favoured if no corrective reagent were introduced. \(\textbf{Analyze procedure P using additional Fe³⁺:}\) Only the free Fe³⁺ concentration is changed directly. To restore equilibrium immediately, \[ K_c=\frac{[\mathrm{FeSCN^{2+}}]'} {[\mathrm{Fe^{3+}}]_{\mathrm{required}}[\mathrm{SCN^-}]'} \] Substituting, \[ 100=\frac{0.0100}{[\mathrm{Fe^{3+}}]_{\mathrm{required}}(0.00500)} \] \(\textbf{Calculate the required iron concentration:}\) \[ [\mathrm{Fe^{3+}}]_{\mathrm{required}} =\frac{0.0100}{(100)(0.00500)} \] \[ [\mathrm{Fe^{3+}}]_{\mathrm{required}}=0.0200\,\mathrm{M} \] The concentration before addition was \(0.0100\,\mathrm{M}\). Therefore, \[ \Delta[\mathrm{Fe^{3+}}]=0.0100\,\mathrm{M} \] \(\textbf{Convert the required concentration increase into moles:}\) The diluted solution volume is \(2.00\,\mathrm{L}\). Thus, \[ n_{\mathrm{Fe^{3+},added}} =(0.0100)(2.00) \] \[ n_{\mathrm{Fe^{3+},added}}=0.0200\,\mathrm{mol} \] \(\textbf{Analyze procedure Q using additional SCN⁻:}\) Now only the free thiocyanate concentration is directly increased. The equilibrium requirement is \[ 100=\frac{0.0100}{(0.0100)[\mathrm{SCN^-}]_{\mathrm{required}}} \] Therefore, \[ [\mathrm{SCN^-}]_{\mathrm{required}} =\frac{0.0100}{(100)(0.0100)} \] \[ [\mathrm{SCN^-}]_{\mathrm{required}}=0.0100\,\mathrm{M} \] \(\textbf{Determine the required thiocyanate addition:}\) The concentration must increase from \(0.00500\) to \(0.0100\,\mathrm{M}\). Hence, \[ \Delta[\mathrm{SCN^-}]=0.00500\,\mathrm{M} \] In the \(2.00\,\mathrm{L}\) solution, \[ n_{\mathrm{SCN^-,added}} =(0.00500)(2.00) \] \[ n_{\mathrm{SCN^-,added}}=0.0100\,\mathrm{mol} \] \(\textbf{Verify both proposed compensations:}\) For P, \[ Q_{c,P}=\frac{0.0100}{(0.0200)(0.00500)} \] \[ Q_{c,P}=100\,\mathrm{L\,mol^{-1}} \] For Q, \[ Q_{c,Q}=\frac{0.0100}{(0.0100)(0.0100)} \] \[ Q_{c,Q}=100\,\mathrm{L\,mol^{-1}} \] Both restored quotients equal Kc. \(\textbf{Interpret the unequal reagent amounts:}\) The two procedures achieve the same immediate equilibrium condition by changing different free-reactant concentrations. However, the diluted mixture initially contains unequal concentrations of Fe³⁺ and SCN⁻. The amount required to double each concentration is therefore different. Equal fractional concentration changes do not necessarily require equal amounts of reagent. The calculation also distinguishes restoring equilibrium immediately from allowing the diluted solution to move gradually towards a new equilibrium composition. The equilibrium constant remains unchanged because temperature is fixed. The problem integrates complex-formation equilibrium, volume dilution, inverse reagent design and concentration-to-mole conversion. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{Fe^{3+},added}}=0.0200\,\mathrm{mol},\qquad n_{\mathrm{SCN^-,added}}=0.0100\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the two linked equilibrium stages:}\) The external CO₂ partial pressure controls the concentration of dissolved molecular CO₂. The dissolved gas then participates in an acid-ionization equilibrium producing H⁺ and HCO₃⁻. Thus, a pressure change affects pH indirectly through the dissolved acid-forming species. \(\textbf{Determine the initial dissolved CO₂ concentration:}\) The specified dissolution relation is \[ [\mathrm{CO_2(aq)}]=kp_{\mathrm{CO_2}} \] At the initial pressure, \[ [\mathrm{CO_2}]_1=(0.100)(0.100) \] \[ [\mathrm{CO_2}]_1=0.0100\,\mathrm{M} \] \(\textbf{Construct the aqueous acid-ionization expression:}\) \[ K_a^*=\frac{[\mathrm{H^+}][\mathrm{HCO_3^-}]}{[\mathrm{CO_2(aq)}]} \] With negligible other ions, \[ [\mathrm{H^+}]\approx[\mathrm{HCO_3^-}] \] Therefore, \[ K_a^*\approx\frac{[\mathrm{H^+}]^2}{[\mathrm{CO_2(aq)}]} \] Rearranging, \[ [\mathrm{H^+}]\approx\sqrt{K_a^*[\mathrm{CO_2(aq)}]} \] \(\textbf{Calculate the original hydrogen-ion concentration:}\) \[ [\mathrm{H^+}]_1 \approx\sqrt{(1.00\times10^{-6})(0.0100)} \] \[ [\mathrm{H^+}]_1\approx1.00\times10^{-4}\,\mathrm{M} \] Thus, \[ pH_1\approx4.000 \] \(\textbf{Determine the dissolved CO₂ concentration at higher pressure:}\) At the final pressure, \[ [\mathrm{CO_2}]_2=(0.100)(0.400) \] \[ [\mathrm{CO_2}]_2=0.0400\,\mathrm{M} \] The dissolved molecular CO₂ concentration has increased fourfold. \(\textbf{Calculate the final hydrogen-ion concentration:}\) \[ [\mathrm{H^+}]_2 \approx\sqrt{(1.00\times10^{-6})(0.0400)} \] \[ [\mathrm{H^+}]_2\approx2.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Determine the final pH:}\) \[ pH_2=-\log_{10}(2.00\times10^{-4}) \] \[ pH_2=4.000-\log_{10}2 \] \[ pH_2\approx3.699 \] \(\textbf{Calculate the pH change:}\) \[ \Delta pH=pH_2-pH_1 \] \[ \Delta pH=3.699-4.000 \] \[ \Delta pH=-0.301 \] \(\textbf{Compare the hydrogencarbonate concentrations:}\) Under the specified approximation, \[ [\mathrm{HCO_3^-}]\approx[\mathrm{H^+}] \] Therefore, \[ \frac{[\mathrm{HCO_3^-}]_2}{[\mathrm{HCO_3^-}]_1} \approx\frac{2.00\times10^{-4}}{1.00\times10^{-4}} \] \[ \frac{[\mathrm{HCO_3^-}]_2}{[\mathrm{HCO_3^-}]_1}\approx2.00 \] \(\textbf{Interpret the nonlinear pressure response:}\) Henry's law gives a linear relationship between gas partial pressure and dissolved molecular CO₂ concentration. However, the hydrogen-ion concentration depends approximately on the square root of the dissolved molecular CO₂ concentration. Consequently, a fourfold increase in CO₂ pressure produces approximately a twofold increase in hydrogen-ion concentration. The pH decreases logarithmically by approximately \(0.301\), rather than decreasing fourfold. The hydrogencarbonate concentration also doubles under the stated charge-balance approximation. The problem distinguishes the gas-dissolution response from the subsequent weak-acid ionization response. It demonstrates how an apparently simple pressure change can produce a nonlinear aqueous-equilibrium consequence. \(\textbf{Final answer:}\) \[ \boxed{\Delta pH\approx-0.301,\qquad \frac{[\mathrm{HCO_3^-}]_2}{[\mathrm{HCO_3^-}]_1}\approx2.00} \]

Explanation: \(\textbf{Identify the relevant carbonate buffer pair:}\) The buffer contains the conjugate pair \[ \mathrm{HCO_3^-/CO_3^{2-}} \] The relevant equilibrium is \[ \mathrm{HCO_3^-\rightleftharpoons H^++CO_3^{2-}} \] Its acid-ionization constant is \(K_{a2}\). \(\textbf{Establish the original carbonate concentration in P:}\) P contains \(0.100\,\mathrm{mol}\) Na₂CO₃ in \(1.00\,\mathrm{L}\). Under the stipulated buffer approximation, \[ [\mathrm{CO_3^{2-}}]_P\approx0.100\,\mathrm{M} \] The HCO₃⁻ and CO₃²⁻ inventories are initially equal, so \[ pH_P\approx pK_{a2} \] \[ pH_P\approx10.00 \] \(\textbf{Calculate P's calcium carbonate ionic product:}\) \[ Q_{sp,P}=[\mathrm{Ca^{2+}}][\mathrm{CO_3^{2-}}] \] Substituting, \[ Q_{sp,P}=(6.00\times10^{-8})(0.100) \] \[ Q_{sp,P}=6.00\times10^{-9} \] \(\textbf{Compare P with the saturation threshold:}\) \[ Q_{sp,P}=6.00\times10^{-9} \] whereas \[ K_{sp}=4.00\times10^{-9} \] Therefore, \[ Q_{sp,P}\gt K_{sp} \] CaCO₃ precipitation is favoured in P. \(\textbf{Identify the effect of HCl in Q:}\) Hydrogen ions protonate carbonate ions: \[ \mathrm{CO_3^{2-}+H^+\rightarrow HCO_3^-} \] The added HCl amount is \[ 0.0500\,\mathrm{mol} \] This is smaller than the original carbonate inventory, so both buffer components remain present. \(\textbf{Determine the new Q buffer composition:}\) The carbonate amount becomes \[ n_{\mathrm{CO_3^{2-}},Q}=0.100-0.0500 \] \[ n_{\mathrm{CO_3^{2-}},Q}=0.0500\,\mathrm{mol} \] The hydrogencarbonate amount becomes \[ n_{\mathrm{HCO_3^-},Q}=0.100+0.0500 \] \[ n_{\mathrm{HCO_3^-},Q}=0.150\,\mathrm{mol} \] \(\textbf{Determine Q's approximate pH:}\) \[ pH_Q\approx pK_{a2}+ \log_{10}\frac{n_{\mathrm{CO_3^{2-}}}}{n_{\mathrm{HCO_3^-}}} \] Substituting, \[ pH_Q\approx10.00+\log_{10}\frac{0.0500}{0.150} \] \[ pH_Q\approx10.00+\log_{10}\frac13 \] \[ pH_Q\approx9.523 \] \(\textbf{Calculate the carbonate ionic product in Q:}\) Under the stipulated approximation, \[ [\mathrm{CO_3^{2-}}]_Q\approx0.0500\,\mathrm{M} \] Therefore, \[ Q_{sp,Q}=(6.00\times10^{-8})(0.0500) \] \[ Q_{sp,Q}=3.00\times10^{-9} \] \(\textbf{Compare Q with its saturation threshold:}\) \[ Q_{sp,Q}=3.00\times10^{-9} \] Since \[ Q_{sp,Q}\lt K_{sp} \] CaCO₃ precipitation is not favoured in Q at the specified initial calcium concentration. \(\textbf{Interpret the chemical connection:}\) Both samples receive the same calcium-ion addition. Their different precipitation tendencies arise from their different free carbonate-ion concentrations. HCl addition converts part of the carbonate reserve into hydrogencarbonate. This lowers both the buffer pH and the carbonate concentration available for calcium carbonate precipitation. The solubility product itself remains unchanged because temperature is fixed. The example demonstrates that an acid-base reaction can control a separate precipitation equilibrium by changing the free concentration of one of its constituent ions. The calcium addition is tested through the immediate ionic product; subsequent extensive precipitation would require a renewed material-balance calculation.

Explanation: Statement I is correct. In P, the reaction is \[ \mathrm{A\rightleftharpoons2B} \] The number of gaseous molecules increases in the forward direction. Doubling the volume lowers the reacting-gas partial pressures and favours the side containing more gaseous molecules. Thus, P has an immediate forward tendency. In Q, the reaction is \[ \mathrm{2C\rightleftharpoons D} \] The number of gaseous molecules decreases in the forward direction. Halving the volume raises the reacting-gas partial pressures and favours the side containing fewer gaseous molecules. Thus, Q also has an immediate forward tendency. Statement II is correct. For R, \[ \mathrm{E+F\rightleftharpoons G+H} \] the total gaseous stoichiometric mole number is the same on both sides. A uniform increase in all reacting-gas partial pressures leaves the corresponding reaction quotient unchanged. Since R was initially at equilibrium, its quotient remains equal to Kp immediately after compression. Therefore, a volume change alone does not produce a net reaction tendency in R. Statement III is incorrect. In S, inert gas is added at constant temperature and constant volume. The mole amounts of J, K and L remain unchanged during this addition. Their individual partial pressures therefore remain unchanged. Although the total pressure increases, the reaction quotient \[ Q_p=\frac{p_Kp_L}{p_J} \] is unchanged. Consequently, S does not acquire a net forward tendency merely because inert gas has increased the total pressure. Statement IV is correct. All four disturbances occur at unchanged temperature. Changing volume alters reacting-gas partial pressures and potentially the reaction quotient. Adding inert gas at constant volume changes the total gas pressure but not the reacting-gas partial pressures. Neither operation changes the equilibrium constant for a specified reaction at a fixed temperature. \(\textbf{Interpret the common principle:}\) The four experiments have different changes in total gas pressure or volume. However, the resulting equilibrium responses depend on the stoichiometry and the actual reacting-gas partial pressures. P and Q shift forward for different volume changes because their gaseous stoichiometric mole-number differences have opposite signs. R remains undisturbed because its quotient is invariant under uniform compression. S remains undisturbed because the reacting-gas partial pressures are unchanged by the specified inert-gas addition. The comparison demonstrates why total pressure alone is not enough to determine a chemical equilibrium shift. Statements I, II and IV are correct.

Explanation: \(\textbf{Determine the original equilibrium constant:}\) For the reaction as written, \[ K_p=\frac{p_{\mathrm{N_2O_4}}}{p_{\mathrm{NO_2}}^2} \] Substituting the original equilibrium pressures, \[ K_p=\frac{1.00}{(0.500)^2} \] \[ K_p=4.00\,\mathrm{bar^{-1}} \] \(\textbf{Identify the instantaneous effect of compression:}\) At constant temperature, the partial pressure of each ideal gas is proportional to its concentration. The vessel volume is halved before appreciable chemical reaction occurs. Therefore, both gas partial pressures immediately double. \[ p_{\mathrm{NO_2}}'=1.00\,\mathrm{bar} \] \[ p_{\mathrm{N_2O_4}}'=2.00\,\mathrm{bar} \] The NO₂ concentration consequently increases immediately after compression. \(\textbf{Evaluate the reaction quotient after compression:}\) \[ Q_p'=\frac{2.00}{(1.00)^2} \] \[ Q_p'=2.00\,\mathrm{bar^{-1}} \] Since \[ Q_p'\lt K_p \] the forward dimerization reaction is favoured. The resulting chemical adjustment consumes NO₂ and forms additional N₂O₄. \(\textbf{Represent the subsequent equilibrium adjustment:}\) Let x denote the pressure-equivalent extent of additional dimerization at the new fixed volume. The partial pressures during chemical relaxation can be written as \[ p_{\mathrm{NO_2}}=1.00-2x \] \[ p_{\mathrm{N_2O_4}}=2.00+x \] The factor of two in the NO₂ change follows from the balanced reaction. \(\textbf{Apply the equilibrium condition:}\) \[ 4.00=\frac{2.00+x}{(1.00-2x)^2} \] Rearranging, \[ 2.00+x=4.00(1.00-4x+4x^2) \] \[ 2.00+x=4.00-16x+16x^2 \] Therefore, \[ 16x^2-17x+2=0 \] \(\textbf{Select the physically acceptable solution:}\) \[ x=\frac{17\pm\sqrt{161}}{32} \] Only the smaller root leaves a nonnegative NO₂ pressure. Thus, \[ x=\frac{17-\sqrt{161}}{32} \] \[ x\approx0.1347\,\mathrm{bar} \] \(\textbf{Calculate the final NO₂ partial pressure:}\) \[ p_{\mathrm{NO_2},f}=1.00-2(0.1347) \] \[ p_{\mathrm{NO_2},f}\approx0.7305\,\mathrm{bar} \] \(\textbf{Compare all three stages:}\) Initially, \[ p_{\mathrm{NO_2},i}=0.500\,\mathrm{bar} \] Immediately after compression, \[ p_{\mathrm{NO_2}}'=1.00\,\mathrm{bar} \] After re-equilibration, \[ p_{\mathrm{NO_2},f}\approx0.731\,\mathrm{bar} \] Therefore, \[ p_{\mathrm{NO_2},i}\lt p_{\mathrm{NO_2},f}\lt p_{\mathrm{NO_2}}' \] \(\textbf{Interpret the colour observation:}\) Compression immediately increases the NO₂ concentration and therefore increases its brown colour intensity under the stated optical comparison. Subsequent chemical dimerization reduces the NO₂ concentration relative to that immediate compressed state. However, the reaction does not consume enough NO₂ to restore its original concentration. The final mixture can consequently remain more intensely coloured than the original mixture even though the chemical equilibrium has shifted towards colourless N₂O₄. A shift towards a colourless product does not necessarily imply that the final colour intensity must be lower than its value before compression. The experiment requires separating the instantaneous physical concentration change from the subsequent chemical-equilibrium response. The equilibrium constant remains unchanged throughout because temperature is fixed. \(\textbf{Final answer:}\) \[ \boxed{p_{\mathrm{NO_2},f}\approx0.731\,\mathrm{bar}} \] The NO₂ concentration rises immediately after compression, then falls during re-equilibration but remains above its original value: \(0.500\lt0.731\lt1.00\) (partial pressures in bar).

Explanation: \(\textbf{Distinguish initial salt dissociation from further ionization:}\) NaHSO₄ is treated as a completely dissociated soluble sodium salt. However, HSO₄⁻ is not assumed to release its remaining proton completely. Its further ionization must be calculated using the supplied Ka. The two solutions contain the same analytical amount of HSO₄⁻ but different initial hydrogen-ion concentrations. \(\textbf{Construct the equilibrium in P:}\) Initially, \[ [\mathrm{HSO_4^-}]_0=0.0100\,\mathrm{M} \] Let x represent the concentration of HSO₄⁻ that ionizes. Then, \[ [\mathrm{SO_4^{2-}}]_P=x \] \[ [\mathrm{H^+}]_P=x \] \[ [\mathrm{HSO_4^-}]_P=0.0100-x \] \(\textbf{Apply the acid-ionization constant in P:}\) \[ K_a=\frac{[\mathrm{H^+}][\mathrm{SO_4^{2-}}]} {[\mathrm{HSO_4^-}]} \] Therefore, \[ 1.00\times10^{-2}=\frac{x^2}{0.0100-x} \] Rearranging, \[ x^2+0.0100x-1.00\times10^{-4}=0 \] \(\textbf{Solve without assuming negligible ionization:}\) The positive root is \[ x=\frac{-0.0100+\sqrt{(0.0100)^2+4(1.00\times10^{-4})}}2 \] \[ x=\frac{-0.0100+\sqrt{5.00\times10^{-4}}}2 \] \[ x\approx6.18\times10^{-3}\,\mathrm{M} \] The ionization fraction is appreciable, so simply replacing the remaining HSO₄⁻ concentration by its initial value would be inappropriate. \(\textbf{Construct the equilibrium in Q:}\) Q initially contains hydrogen ions from HCl: \[ [\mathrm{H^+}]_0=0.100\,\mathrm{M} \] Let y represent the additional ionization of HSO₄⁻. Then, \[ [\mathrm{SO_4^{2-}}]_Q=y \] \[ [\mathrm{H^+}]_Q=0.100+y \] \[ [\mathrm{HSO_4^-}]_Q=0.0100-y \] \(\textbf{Apply the ionization constant in Q:}\) \[ 1.00\times10^{-2} =\frac{(0.100+y)y}{0.0100-y} \] Rearranging, \[ 1.00\times10^{-4}-0.0100y =0.100y+y^2 \] Therefore, \[ y^2+0.110y-1.00\times10^{-4}=0 \] \(\textbf{Select the positive root:}\) \[ y=\frac{-0.110+\sqrt{(0.110)^2+4(1.00\times10^{-4})}}2 \] \[ y\approx9.02\times10^{-4}\,\mathrm{M} \] \(\textbf{Compare the extent of further ionization:}\) \[ \frac{[\mathrm{SO_4^{2-}}]_P}{[\mathrm{SO_4^{2-}}]_Q} =\frac{6.18\times10^{-3}}{9.02\times10^{-4}} \] \[ \frac{[\mathrm{SO_4^{2-}}]_P}{[\mathrm{SO_4^{2-}}]_Q}\approx6.85 \] The sulfate-ion concentration is substantially smaller in Q. \(\textbf{Interpret the common-ion effect:}\) HCl contributes H⁺, which is already a product of hydrogensulfate ionization. Increasing the hydrogen-ion concentration suppresses the additional ionization of HSO₄⁻. The amount of sulfate produced consequently decreases even though both solutions begin with the same amount of NaHSO₄. The equilibrium constant itself remains unchanged. P cannot be treated using a small-ionization approximation because a substantial fraction of HSO₄⁻ ionizes. Q illustrates common-ion suppression caused by an externally supplied strong acid rather than by adding a salt containing sulfate. The calculation connects partial second-stage ionization, an external common ion and the validity of equilibrium approximations. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{SO_4^{2-}}]_P\approx6.18\times10^{-3}\,\mathrm{M},\qquad [\mathrm{SO_4^{2-}}]_Q\approx9.02\times10^{-4}\,\mathrm{M}} \]

Explanation: \(\textbf{Separate free fluoride from total dissolved fluorine:}\) Every mole of CaF₂ that dissolves produces one mole of dissolved calcium and two moles of fluorine-containing species. Let S denote the total molar solubility. Then, \[ [\mathrm{Ca^{2+}}]=S \] and \[ [\mathrm{F^-}]+[\mathrm{HF}]=2S \] \(\textbf{Relate the two fluorine-containing species:}\) For HF, \[ K_a=\frac{[\mathrm{H^+}][\mathrm{F^-}]}{[\mathrm{HF}]} \] Therefore, \[ \frac{[\mathrm{HF}]}{[\mathrm{F^-}]} =\frac{[\mathrm{H^+}]}{K_a} \] \(\textbf{Express the free fluoride fraction:}\) Define f as the fraction of dissolved fluorine-containing species present as F⁻. Then, \[ f=\frac{[\mathrm{F^-}]}{[\mathrm{F^-}]+[\mathrm{HF}]} \] Using the acid-ionization relationship, \[ f=\frac{K_a}{K_a+[\mathrm{H^+}]} \] Consequently, \[ [\mathrm{F^-}]=2Sf \] \(\textbf{Connect the speciation fraction with Ksp:}\) The solubility-product expression is \[ K_{sp}=[\mathrm{Ca^{2+}}][\mathrm{F^-}]^2 \] Substituting, \[ K_{sp}=S(2Sf)^2 \] Therefore, \[ K_{sp}=4S^3f^2 \] Rearranging, \[ S=\left(\frac{K_{sp}}{4f^2}\right)^{1/3} \] \(\textbf{Calculate the fluoride fraction at pH 4.00:}\) For P, \[ [\mathrm{H^+}]_P=1.00\times10^{-4}\,\mathrm{M} \] This equals the HF ionization constant. Thus, \[ f_P=\frac{10^{-4}}{10^{-4}+10^{-4}} \] \[ f_P=\frac12 \] \(\textbf{Calculate the solubility in P:}\) \[ S_P^3=\frac{4.00\times10^{-11}}{4(1/2)^2} \] \[ S_P^3=4.00\times10^{-11} \] Therefore, \[ S_P\approx3.42\times10^{-4}\,\mathrm{M} \] \(\textbf{Calculate the fluoride fraction at pH 5.00:}\) For Q, \[ [\mathrm{H^+}]_Q=1.00\times10^{-5}\,\mathrm{M} \] Therefore, \[ f_Q=\frac{10^{-4}}{10^{-4}+10^{-5}} \] \[ f_Q=\frac{10}{11} \] \(\textbf{Calculate the solubility in Q:}\) \[ S_Q^3=\frac{4.00\times10^{-11}}{4(10/11)^2} \] \[ S_Q^3=(4.00\times10^{-11})\frac{121}{400} \] \[ S_Q^3=1.21\times10^{-11} \] Therefore, \[ S_Q\approx2.30\times10^{-4}\,\mathrm{M} \] \(\textbf{Determine the solubility ratio:}\) The intrinsic Ksp is identical in both experiments. Therefore, \[ \frac{S_P}{S_Q} =\left(\frac{f_Q}{f_P}\right)^{2/3} \] Substituting, \[ \frac{S_P}{S_Q} =\left(\frac{10/11}{1/2}\right)^{2/3} \] \[ \frac{S_P}{S_Q} =\left(\frac{20}{11}\right)^{2/3} \] \[ \frac{S_P}{S_Q}\approx1.49 \] \(\textbf{Interpret the unusual exponent:}\) At lower pH, more dissolved fluoride is converted into molecular HF. This reduces the free F⁻ concentration relative to the total fluorine inventory. Additional CaF₂ can consequently dissolve while maintaining the same solubility product. The relationship is not the same as for a sparingly soluble salt that releases only one anion per formula unit. CaF₂ produces two fluoride ions per calcium ion, causing the free fluoride concentration to appear squared in Ksp. The resulting total solubility depends on the cube root of the governing expression. Ignoring dissolution stoichiometry would therefore produce an incorrect solubility ratio even if the HF ionization equilibrium were handled correctly. \(\textbf{Final answer:}\) \[ \boxed{\frac{S_P}{S_Q}=\left(\frac{20}{11}\right)^{2/3}\approx1.49} \]

Explanation: Statement I is correct. In P, HCO₃⁻ accepts a proton from water and forms H₂CO₃. A species that accepts a proton acts as a Brønsted–Lowry base. Water donates the proton in this process and forms OH⁻. In Q, HCO₃⁻ instead donates a proton to water. This produces CO₃²⁻ and H₃O⁺. HCO₃⁻ therefore acts as a Brønsted–Lowry acid in Q. The same species can behave as an acid or base depending on its reaction partner and the direction of proton transfer. Statement II is correct. In R, BF₃ accepts an electron pair from F⁻. BF₃ is electron-deficient and acts as a Lewis acid. Fluoride supplies an electron pair and acts as a Lewis base. No proton transfer is involved, but the process still satisfies the Lewis acid-base definition. Statement III is correct. In S, NH₃ accepts a proton supplied by HCl. It therefore acts as a Brønsted–Lowry base. The nitrogen lone pair participates in forming the additional N–H bond. NH₃ can consequently also be described as an electron-pair donor, which is the defining behaviour of a Lewis base. Statement IV is incorrect. The ability of HCO₃⁻ to release a proton does not establish that it is a strong acid. Acid strength depends on the extent of ionization and the corresponding acid-ionization constant. Hydrogencarbonate is amphiprotic and can participate in either proton donation or proton acceptance. Its formation of CO₃²⁻ in a suitable reaction is not evidence of complete proton donation in water. \(\textbf{Interpret the context dependence of acid-base classification:}\) Acid-base roles are assigned from what a species actually does in the specified reaction. An amphiprotic species can act as either a proton donor or proton acceptor. A Lewis acid-base reaction can occur without any proton transfer. A Brønsted–Lowry base can simultaneously satisfy the Lewis-base definition when its electron pair accepts a proton. The classification of a species must therefore not be inferred solely from the presence or absence of hydrogen or hydroxyl groups in its formula. Statements I, II and III are correct.

Explanation: \(\textbf{Identify the ionic stoichiometry in pure water:}\) Each dissolved formula unit produces two M³⁺ ions and three X²⁻ ions. Therefore, in pure water, \[ [\mathrm{M^{3+}}]=2S_0 \] \[ [\mathrm{X^{2-}}]=3S_0 \] \(\textbf{Construct the solubility-product expression:}\) For the balanced dissolution reaction, \[ K_{sp}=[\mathrm{M^{3+}}]^2[\mathrm{X^{2-}}]^3 \] Substituting the pure-water concentration relationships, \[ K_{sp}=(2S_0)^2(3S_0)^3 \] \[ K_{sp}=4S_0^2\times27S_0^3 \] \[ K_{sp}=108S_0^5 \] \(\textbf{Determine Ksp from the measured pure-water solubility:}\) \[ K_{sp}=108(1.00\times10^{-3})^5 \] \[ K_{sp}=108\times10^{-15} \] \[ K_{sp}=1.08\times10^{-13} \] \(\textbf{Represent dissolution in the common-ion solution:}\) Let S be the new molar solubility. The dissolved metal concentration is \[ [\mathrm{M^{3+}}]=2S \] The total free X²⁻ concentration is \[ [\mathrm{X^{2-}}]=0.0300+3S \] The added common ion is already present before any M₂X₃ dissolves. \(\textbf{Construct the exact saturation equation:}\) \[ 1.08\times10^{-13} =(2S)^2(0.0300+3S)^3 \] \(\textbf{Apply the common-ion-dominance approximation:}\) If \[ 3S\ll0.0300 \] then \[ 0.0300+3S\approx0.0300 \] The saturation equation becomes \[ 1.08\times10^{-13} \approx4S^2(0.0300)^3 \] \(\textbf{Calculate the approximate solubility:}\) \[ S^2\approx\frac{1.08\times10^{-13}}{4(0.0300)^3} \] Since \[ (0.0300)^3=2.70\times10^{-5} \] we obtain \[ S^2\approx\frac{1.08\times10^{-13}}{1.08\times10^{-4}} \] \[ S^2\approx1.00\times10^{-9} \] Therefore, \[ S\approx3.16\times10^{-5}\,\mathrm{M} \] \(\textbf{Verify the approximation:}\) At the calculated solubility, \[ 3S\approx9.49\times10^{-5}\,\mathrm{M} \] The fractional contribution to the background common-ion concentration is \[ \frac{3S}{0.0300}\approx0.00316 \] Thus, dissolution contributes only about \(0.316\%\) of the externally supplied X²⁻ concentration. The approximation is appropriate at the requested precision. \(\textbf{Interpret the suppression of dissolution:}\) The external X²⁻ ions increase the concentration of a product of the dissolution equilibrium. This suppresses the net dissolution of M₂X₃. The intrinsic solubility product remains unchanged because temperature is fixed. The fifth-power dependence in pure water arises from the combined stoichiometric powers of the metal and anion concentrations. In the common-ion solution, the anion concentration becomes approximately fixed by the external source. Consequently, the unknown solubility appears predominantly through the square of the metal concentration rather than its fifth power. The example illustrates why the relationship between Ksp and molar solubility changes when a common ion is present. Correct solution requires both stoichiometric concentration factors and an explicit validity check of the approximation. \(\textbf{Final answer:}\) \[ \boxed{S\approx3.16\times10^{-5}\,\mathrm{M}} \]

Explanation: Statement I is correct. Adding the two chemical equations eliminates the intermediate B and produces \[ \mathrm{A\rightleftharpoons C} \] The overall equilibrium constant is \[ K_{\mathrm{overall}}=K_1K_2 \] \[ K_{\mathrm{overall}}=(2.00)(3.00)=6.00 \] The actual overall quotient is \[ Q_{\mathrm{overall}}=\frac{[C]}{[A]} \] \[ Q_{\mathrm{overall}}=\frac{0.600}{0.100}=6.00 \] Thus, the overall quotient equals the overall equilibrium constant. Statement II is incorrect. The individual reaction quotients must also be examined. For the first reaction, \[ Q_1=\frac{[B]}{[A]} \] \[ Q_1=\frac{0.100}{0.100}=1.00 \] Since \[ Q_1\lt K_1 \] the first reaction is not at equilibrium. For the second reaction, \[ Q_2=\frac{[C]}{[B]} \] \[ Q_2=\frac{0.600}{0.100}=6.00 \] Since \[ Q_2\gt K_2 \] the second reaction is also not at equilibrium. Equality of the overall quotient with the product of the constants does not guarantee that both coupled equilibria are individually satisfied. Statement III is correct. The first reaction has \(Q_1\lt K_1\), so it initially favours conversion of A into B. The second reaction has \(Q_2\gt K_2\), so it initially favours the reverse conversion of C into B. Both tendencies therefore produce the intermediate B. This makes the original mixture inconsistent with simultaneous equilibrium despite its correct overall A-to-C ratio. Statement IV is correct. The individual equilibrium conditions require \[ [B]=2[A] \] and \[ [C]=3[B] \] Therefore, \[ [A]:[B]:[C]=1:2:6 \] The total analytical concentration is \[ C_T=0.100+0.100+0.600 \] \[ C_T=0.800\,\mathrm{M} \] At simultaneous equilibrium, \[ [B]_{\mathrm{eq}}=\frac29(0.800) \] \[ [B]_{\mathrm{eq}}\approx0.1778\,\mathrm{M} \] This exceeds the original B concentration of \(0.100\,\mathrm{M}\). Thus, B must increase as the coupled system approaches its equilibrium composition. \(\textbf{Interpret the coupled-equilibrium consistency test:}\) The overall reaction eliminates B algebraically, but B remains a physically present intermediate. The overall equilibrium condition is necessary but not sufficient for simultaneous equilibrium of both reactions. Each independently reversible step must satisfy its own equilibrium relationship. Here, opposing deviations of the individual quotients combine to give an apparently correct overall quotient. Nevertheless, both steps initially favour formation of B. The question demonstrates why coupled chemical equilibria cannot always be assessed from a single net reaction equation. Statements I, III and IV are correct.

Explanation: Statement I is correct. For a gas-phase reaction, multiplying every reacting-gas partial pressure by the same factor f changes the reaction quotient according to \[ \frac{Q_p'}{Q_p}=f^{\Delta n_g} \] Here, \(\Delta n_g\) is the sum of gaseous product coefficients minus the sum of gaseous reactant coefficients. The mixture was initially at equilibrium, so \[ Q_p=K_p \] Tripling the volume at constant temperature divides every reacting-gas partial pressure by three: \[ f=\frac13 \] Therefore, \[ \frac{Q_p'}{K_p}=\left(\frac13\right)^{\Delta n_g} \] The measured ratio is \[ 9.00=\left(\frac13\right)^{\Delta n_g} \] Since \[ 9.00=3^2 \] we obtain \[ \Delta n_g=-2 \] Thus, the forward reaction decreases the gaseous stoichiometric mole number by two. Statement II is correct. The measured post-expansion quotient satisfies \[ Q_p'=9K_p \] Therefore, \[ Q_p'\gt K_p \] The reverse reaction is initially favoured. This is also consistent with the gaseous stoichiometry. Expansion favours the direction containing more gaseous molecules, which is the reverse direction when \(\Delta n_g=-2\). Statement III is incorrect. At unchanged temperature and fixed volume, introducing an inert ideal gas does not change the partial pressures of the reacting gases. The reaction quotient consequently remains unchanged during the inert-gas addition. The total pressure increases, but the reacting-gas quotient remains \[ Q_p'=9K_p \] Therefore, the mixture is not restored to equilibrium by this operation. Statement IV is correct. Isothermal compression increases the partial pressures of the reacting gases. For a reaction with \[ \Delta n_g=-2 \] a uniform pressure increase reduces the reaction quotient relative to its original equilibrium value. Thus, immediately after compressing an initially equilibrated mixture, \[ Q_p'\lt K_p \] and the forward reaction is favoured. \(\textbf{Interpret the reverse inference:}\) The measured quotient change reveals information about the balanced reaction even though its individual coefficients were not supplied. The key is that a uniform pressure change affects the quotient through the net gaseous stoichiometric mole number. The inferred sign then predicts the directions favoured by compression and expansion. The inert-gas operation must be considered separately because constant-volume gas addition does not produce the same partial-pressure change as mechanical compression. The problem integrates experimental quotient data, stoichiometric inference and the distinction between total pressure and reacting-gas partial pressures. Statements I, II and IV are correct.

Explanation: Statement I is correct. For the reaction as written, pure liquid water is omitted from the concentration equilibrium expression. Therefore, \[ K_c=\frac{[\mathrm{CoCl_4^{2-}}]} {[\mathrm{Co(H_2O)_6^{2+}}][\mathrm{Cl^-}]^4} \] The solution is initially at equilibrium, so its initial quotient equals Kc. Doubling the volume immediately halves the concentrations of the two cobalt complexes and the free chloride ions. Therefore, \[ Q_c'= \frac{\frac12[\mathrm{CoCl_4^{2-}}]} {\left(\frac12[\mathrm{Co(H_2O)_6^{2+}}]\right) \left(\frac12[\mathrm{Cl^-}]\right)^4} \] The factors associated with the two cobalt-complex concentrations cancel. The chloride factor contributes \[ \left(\frac12\right)^{-4}=16 \] Thus, \[ Q_c'=16K_c \] Statement II is correct. Since \[ Q_c'\gt K_c \] the reverse reaction is favoured. The reverse reaction converts blue tetrachlorocobaltate(II) into the pink hydrated cobalt(II) complex. The colour change is therefore consistent with the reaction-quotient comparison. The direction should not be inferred merely from the statement that water was added; dilution also lowers the free chloride concentration. Statement III is incorrect. The reaction involves water, but water is present as the pure liquid solvent. Its activity is treated as constant in the equilibrium expression. The temperature is unchanged, so Kc does not change. Dilution changes the reaction quotient and the equilibrium composition, not the intrinsic equilibrium constant. Statement IV is correct. Chloride ions are reactants in the forward formation of the blue complex. Increasing their free concentration lowers the reaction quotient for the reaction as written. A sufficiently large chloride addition can consequently favour formation of \(\mathrm{[CoCl_4]^{2-}}\). This can oppose the pink-shifting tendency produced by dilution. \(\textbf{Interpret the experimental observation:}\) The two cobalt complexes are connected by a reversible coordination equilibrium. The concentration of chloride strongly influences their relative proportions because four chloride ions participate in the written reaction. Doubling the solution volume does not merely halve the concentrations and leave their equilibrium relationship unchanged. The fourth power of chloride creates a much larger quotient change. At the same time, the pure liquid solvent does not contribute an ordinary concentration factor to Kc. The example integrates a colour observation, the law of chemical equilibrium and selective reagent addition. Statements I, II and IV are correct.