Equilibrium MCQs with Answers – Part 8 (NEET Chemistry)

Explanation: Statement I is correct. Hydrogencarbonate ions can accept hydrogen ions to form the dissolved carbon dioxide–carbonic-acid reservoir. That reservoir is connected to gaseous CO₂ through a reversible physical equilibrium. When gaseous CO₂ escapes from an open vessel, the system continuously loses some of the product of the gas-evolution process. Such removal can favour further conversion of dissolved carbon species into CO₂. The observed evolution of gas therefore reflects both acid-base chemistry and gas transfer. Statement II is incorrect. Equal initial HCl additions do not guarantee equal equilibrium CO₂ partial pressures. In P, gas can escape into the surrounding atmosphere. In Q, gas released from the liquid is retained in the finite headspace. Its partial pressure can therefore build up. The equilibrium gas pressure depends on the vessel conditions and the distribution of carbon-derived material, not merely on the initial HCl amount. Statement III is correct. In the sealed vessel, dissolved CO₂ and gaseous CO₂ can approach equilibrium. Their distribution depends on the temperature, gas pressure and the relevant dissolution relationship. Accumulated gaseous CO₂ can therefore affect the amount of carbon dioxide remaining dissolved. This physical equilibrium is coupled to the acid-base behaviour of the hydrogencarbonate system. Statement IV is incorrect. The proton-accepting species in the displayed acid-base reaction is HCO₃⁻. Sodium ions originate from the dissociation of sodium hydrogencarbonate but do not directly accept the proton in that reaction. The distinction between an active conjugate-base ion and a spectator counterion is essential when writing ionic equations. \(\textbf{Interpret the role of the system boundary:}\) A closed vessel can retain the gas formed during acid-base reaction, permitting an equilibrium pressure to develop. An open vessel allows continuous gas exchange with the surroundings. Consequently, the two vessels need not exhibit identical final gas and solution compositions. The comparison integrates Brønsted proton transfer, gas–liquid equilibrium and the effect of an open or closed system on an observed chemical process. Statements I and III are correct.

Explanation: \(\textbf{Express the composition at the specified conversion:}\) The balanced reaction consumes one mole of N₂ and three moles of H₂ while producing two moles of NH₃. A \(20.0\%\) conversion of \(1.00\,\mathrm{mol}\) N₂ corresponds to a reaction extent of \[ \xi=0.200\,\mathrm{mol} \] Therefore, both mixtures have consumed \[ n_{\mathrm{N_2,consumed}}=0.200\,\mathrm{mol} \] \[ n_{\mathrm{H_2,consumed}}=0.600\,\mathrm{mol} \] and formed \[ n_{\mathrm{NH_3,formed}}=0.400\,\mathrm{mol} \] \(\textbf{Determine the reacting-gas amounts in P:}\) \[ n_{\mathrm{N_2},P}=1.00-0.200=0.800\,\mathrm{mol} \] \[ n_{\mathrm{H_2},P}=3.00-0.600=2.40\,\mathrm{mol} \] \[ n_{\mathrm{NH_3},P}=0.400\,\mathrm{mol} \] Hence, \[ n_{\mathrm{total},P}=0.800+2.40+0.400 \] \[ n_{\mathrm{total},P}=3.60\,\mathrm{mol} \] \(\textbf{Calculate the partial pressures in P:}\) Using \[ p_i=\frac{n_i}{n_{\mathrm{total}}}P_{\mathrm{total}} \] we obtain \[ p_{\mathrm{N_2},P}=\frac{0.800}{3.60}(4.00)=\frac89\,\mathrm{bar} \] \[ p_{\mathrm{H_2},P}=\frac{2.40}{3.60}(4.00)=\frac83\,\mathrm{bar} \] \[ p_{\mathrm{NH_3},P}=\frac{0.400}{3.60}(4.00)=\frac49\,\mathrm{bar} \] \(\textbf{Test P against the equilibrium constant:}\) \[ Q_{p,P}=\frac{p_{\mathrm{NH_3}}^2} {p_{\mathrm{N_2}}p_{\mathrm{H_2}}^3} \] Substituting, \[ Q_{p,P}=\frac{(4/9)^2}{(8/9)(8/3)^3} \] \[ Q_{p,P}=\frac3{256} \] \[ Q_{p,P}\approx0.0117\,\mathrm{bar^{-2}} \] Because \[ Q_{p,P}\lt 0.100\,\mathrm{bar^{-2}} \] P favours the forward reaction. \(\textbf{Determine the composition of Q:}\) Q begins with less hydrogen. After the same reaction extent, \[ n_{\mathrm{N_2},Q}=0.800\,\mathrm{mol} \] \[ n_{\mathrm{H_2},Q}=1.00-0.600=0.400\,\mathrm{mol} \] \[ n_{\mathrm{NH_3},Q}=0.400\,\mathrm{mol} \] The total is \[ n_{\mathrm{total},Q}=1.60\,\mathrm{mol} \] \(\textbf{Calculate Q's partial pressures:}\) \[ p_{\mathrm{N_2},Q}=\frac{0.800}{1.60}(4.00)=2.00\,\mathrm{bar} \] \[ p_{\mathrm{H_2},Q}=\frac{0.400}{1.60}(4.00)=1.00\,\mathrm{bar} \] \[ p_{\mathrm{NH_3},Q}=\frac{0.400}{1.60}(4.00)=1.00\,\mathrm{bar} \] \(\textbf{Test Q against the same equilibrium constant:}\) \[ Q_{p,Q}=\frac{(1.00)^2}{(2.00)(1.00)^3} \] \[ Q_{p,Q}=0.500\,\mathrm{bar^{-2}} \] Since \[ Q_{p,Q}\gt K_p \] Q favours the reverse reaction. \(\textbf{Interpret the feed-composition effect:}\) The two mixtures have the same temperature, total pressure and fractional conversion of their initial N₂. Nevertheless, they contain substantially different amounts of hydrogen. Their partial pressures and reaction quotients therefore differ. In P, the relatively large hydrogen partial pressure makes further ammonia formation thermodynamically favoured. In Q, hydrogen is comparatively depleted, making the existing ammonia amount excessive relative to equilibrium. Fractional conversion alone does not determine the reaction direction unless the full reacting-gas composition is also known. The comparison demonstrates why feed composition must be incorporated explicitly into equilibrium-yield calculations.

Explanation: \(\textbf{Separate the wet and dry gas pressures:}\) The total pressure includes the chemically nonparticipating water vapour. Dalton's law gives \[ P_{\mathrm{dry}}=P_{\mathrm{wet}}-p_{\mathrm{H_2O}} \] Substituting, \[ P_{\mathrm{dry}}=2.50-0.500 \] \[ P_{\mathrm{dry}}=2.00\,\mathrm{bar} \] \(\textbf{Recover the individual reacting-gas partial pressures:}\) The supplied mole fractions refer only to the dry A/B mixture. Therefore, \[ p_A=y_{A,\mathrm{dry}}P_{\mathrm{dry}} \] \[ p_A=(0.800)(2.00) \] \[ p_A=1.60\,\mathrm{bar} \] Similarly, \[ p_B=(0.200)(2.00) \] \[ p_B=0.400\,\mathrm{bar} \] \(\textbf{Construct the pressure equilibrium expression:}\) For \[ \mathrm{A(g)\rightleftharpoons2B(g)} \] the expression is \[ K_p=\frac{p_B^2}{p_A} \] The nonreacting water vapour does not appear in this expression. \(\textbf{Calculate Kp from the original equilibrium composition:}\) \[ K_p=\frac{(0.400)^2}{1.60} \] \[ K_p=\frac{0.160}{1.60} \] \[ K_p=0.100\,\mathrm{bar} \] \(\textbf{Evaluate selective water-vapour removal:}\) The temperature and vessel volume remain unchanged. The mole amounts of A and B are also unchanged by the drying process. For each reacting gas, \[ p_i=\frac{n_iRT}{V} \] Therefore, \[ p_A'=1.60\,\mathrm{bar} \] \[ p_B'=0.400\,\mathrm{bar} \] \(\textbf{Check the reaction quotient after drying:}\) \[ Q_p'=\frac{(0.400)^2}{1.60} \] \[ Q_p'=0.100\,\mathrm{bar} \] Thus, \[ Q_p'=K_p \] No net chemical reaction is favoured. \(\textbf{Determine the final total pressure:}\) With water vapour removed, \[ P_f=p_A'+p_B' \] \[ P_f=1.60+0.400 \] \[ P_f=2.00\,\mathrm{bar} \] \(\textbf{Interpret the gas-mixture distinction:}\) Removing water vapour decreases the total pressure. However, it does not decrease the individual partial pressures of A and B at constant temperature and volume. The pressure equilibrium quotient consequently remains unchanged. The system therefore stays at the same chemical equilibrium composition. A change in total pressure is not, by itself, sufficient evidence that a gaseous chemical equilibrium has been disturbed. The chemically relevant partial pressures must be calculated from the correct gas composition, excluding the nonparticipating vapour when determining dry-gas mole fractions. The problem integrates Dalton's law with a pressure equilibrium expression and a selective component-removal experiment. \(\textbf{Final answer:}\) \[ \boxed{K_p=0.100\,\mathrm{bar},\qquad P_f=2.00\,\mathrm{bar}} \]

Explanation: Statement I is correct. For the reaction, \[ K_p=\frac{p_{\mathrm{HI}}^2}{p_{\mathrm{H_2}}p_{\mathrm{I_2}}} \] The initial equilibrium constant at \(T_1\) is \[ K_p(T_1)=\frac{(2.00)^2}{(1.00)(1.00)} \] \[ K_p(T_1)=4.00 \] Halving the vessel volume at unchanged temperature doubles each reacting-gas partial pressure: \[ p_{\mathrm{H_2}}'=2.00\,\mathrm{bar} \] \[ p_{\mathrm{I_2}}'=2.00\,\mathrm{bar} \] \[ p_{\mathrm{HI}}'=4.00\,\mathrm{bar} \] The new quotient is \[ Q_{p,P}=\frac{(4.00)^2}{(2.00)(2.00)} \] \[ Q_{p,P}=4.00=K_p(T_1) \] Thus, there is no net chemical shift. Statement II is correct. In Q, the temperature doubles at constant vessel volume. Before reaction occurs, each partial pressure doubles because \[ p_i=\frac{n_iRT}{V} \] The instantaneous pressures are therefore the same numerical values as those just calculated for P. Consequently, \[ Q_{p,Q}=4.00 \] However, the equilibrium constant appropriate to the new temperature is \[ K_p(T_2)=1.00 \] Therefore, \[ Q_{p,Q}\gt K_p(T_2) \] and the reverse reaction is favoured. This is consistent with the forward reaction being exothermic and the higher temperature favouring the reverse direction. Statement III is incorrect. A catalyst can accelerate the opposing reaction rates and reduce the time required to attain equilibrium. However, it does not change the equilibrium constant or the equilibrium composition at a fixed temperature. Because R begins at equilibrium, catalyst addition does not produce a greater equilibrium HI mole fraction. Statement IV is correct. In P, all reacting-gas partial pressures double during compression. Nevertheless, their ratios and gas-phase mole fractions remain the same immediately after compression. Because the reaction has \[ \Delta n_g=2-(1+1)=0 \] the pressure quotient remains unchanged under uniform isothermal compression. The mixture is already at the equilibrium composition for \(T_1\), so no additional chemical redistribution is required. \(\textbf{Interpret the three disturbances:}\) P and Q can produce the same immediate partial pressures, yet their subsequent reaction tendencies differ. The difference arises because Q has a new temperature-dependent equilibrium constant, while P does not. R changes the kinetics of reaching equilibrium but not the equilibrium position. The example demonstrates why pressure changes, temperature changes and catalytic changes must be treated through their distinct effects on Q, K and reaction rates. A single observation such as doubled gas pressure does not determine the chemical response without knowing how that state was produced. Statements I, II and IV are correct.

Explanation: \(\textbf{Account for the ionic stoichiometry of barium hydroxide:}\) Barium hydroxide dissociates as \[ \mathrm{Ba(OH)_2\rightarrow Ba^{2+}+2OH^-} \] Each mole supplies one mole of Ba²⁺ and two moles of OH⁻. The original amount of Ba(OH)₂ is \[ n_{\mathrm{Ba(OH)_2}}=(0.100)(0.1000) \] \[ n_{\mathrm{Ba(OH)_2}}=0.0100\,\mathrm{mol} \] Therefore, \[ n_{\mathrm{Ba^{2+}}}=0.0100\,\mathrm{mol} \] \[ n_{\mathrm{OH^-},0}=0.0200\,\mathrm{mol} \] \(\textbf{Determine the strong-acid inventory:}\) HCl dissociates essentially completely: \[ \mathrm{HCl\rightarrow H^++Cl^-} \] The amount added is \[ n_{\mathrm{HCl}}=(0.150)(0.1000) \] \[ n_{\mathrm{HCl}}=0.0150\,\mathrm{mol} \] Thus, \(0.0150\,\mathrm{mol}\) H⁺ and the same amount of Cl⁻ enter the mixture. \(\textbf{Identify the limiting neutralization component:}\) The ionic neutralization is \[ \mathrm{H^++OH^-\rightarrow H_2O} \] The available hydroxide amount exceeds the hydrogen-ion amount. Hence, H⁺ is consumed, leaving \[ n_{\mathrm{OH^-},f}=0.0200-0.0150 \] \[ n_{\mathrm{OH^-},f}=0.00500\,\mathrm{mol} \] \(\textbf{Use the correct mixed volume:}\) \[ V_f=0.1000+0.1000 \] \[ V_f=0.2000\,\mathrm{L} \] Therefore, \[ [\mathrm{OH^-}]_f=\frac{0.00500}{0.2000} \] \[ [\mathrm{OH^-}]_f=0.0250\,\mathrm{M} \] \(\textbf{Calculate the final pH:}\) \[ pOH=-\log_{10}(0.0250) \] \[ pOH\approx1.602 \] At \(298\,\mathrm{K}\), \[ pH=14.000-1.602 \] \[ pH\approx12.398 \] \(\textbf{Recover the spectator-ion concentrations:}\) Barium and chloride ions remain in solution under the stated assumptions: \[ [\mathrm{Ba^{2+}}]=\frac{0.0100}{0.2000} \] \[ [\mathrm{Ba^{2+}}]=0.0500\,\mathrm{M} \] \[ [\mathrm{Cl^-}]=\frac{0.0150}{0.2000} \] \[ [\mathrm{Cl^-}]=0.0750\,\mathrm{M} \] \(\textbf{Verify electrical neutrality:}\) Total positive charge concentration is \[ 2[\mathrm{Ba^{2+}}]=2(0.0500)=0.100\,\mathrm{M} \] Total negative charge concentration is \[ [\mathrm{Cl^-}]+[\mathrm{OH^-}] =0.0750+0.0250 \] \[ =0.100\,\mathrm{M} \] The ionic composition satisfies charge balance. \(\textbf{Interpret the electrolyte calculation:}\) Barium hydroxide is a strong electrolyte, but its formula produces twice as many hydroxide ions as barium ions. Counting Ba(OH)₂ formula units as hydroxide ions would give the wrong limiting reagent and pH. The spectator ions do not disappear during neutralization. Their final concentrations depend on their original mole inventories and the combined solution volume. Electrical neutrality provides an independent check on the completed calculation. The result integrates dissociation stoichiometry, strong-acid neutralization, pH and ionic charge conservation. \(\textbf{Final answer:}\) \[ \boxed{pH\approx12.40,\quad[\mathrm{Ba^{2+}}]=0.0500\,\mathrm{M},\quad[\mathrm{Cl^-}]=0.0750\,\mathrm{M}} \]

Explanation: \(\textbf{Use the exact ionization relationship:}\) For the equilibrium \[ \mathrm{HA\rightleftharpoons H^++A^-} \] a solution with analytical concentration C and degree of ionization α has \[ [\mathrm{HA}]=C(1-\alpha) \] \[ [\mathrm{H^+}]=[\mathrm{A^-}]=C\alpha \] Therefore, \[ K_a=\frac{C\alpha^2}{1-\alpha} \] \(\textbf{Infer Ka from solution P:}\) The measured ionization degree is \(0.200\), which is not sufficiently small to neglect in the denominator without significant error. Using the exact expression, \[ K_a=\frac{(0.100)(0.200)^2}{1-0.200} \] \[ K_a=\frac{0.00400}{0.800} \] \[ K_a=0.00500 \] \(\textbf{Set up the equilibrium after dilution:}\) The temperature is unchanged, so Ka remains \(0.00500\). For Q, \[ K_a=\frac{C_Q\alpha_Q^2}{1-\alpha_Q} \] Substituting, \[ 0.00500=\frac{0.0300\alpha_Q^2}{1-\alpha_Q} \] Rearranging, \[ 0.00500-0.00500\alpha_Q=0.0300\alpha_Q^2 \] Dividing by \(0.00500\), \[ 6\alpha_Q^2+\alpha_Q-1=0 \] \(\textbf{Select the physically acceptable root:}\) Factorising, \[ (3\alpha_Q-1)(2\alpha_Q+1)=0 \] The negative root is not a possible ionization fraction. Thus, \[ \alpha_Q=\frac13 \] The degree of ionization increases from \(20.0\%\) to approximately \(33.3\%\). \(\textbf{Compare the actual hydrogen-ion concentrations:}\) In P, \[ [\mathrm{H^+}]_P=C_P\alpha_P \] \[ [\mathrm{H^+}]_P=(0.100)(0.200) \] \[ [\mathrm{H^+}]_P=0.0200\,\mathrm{M} \] In Q, \[ [\mathrm{H^+}]_Q=C_Q\alpha_Q \] \[ [\mathrm{H^+}]_Q=(0.0300)\left(\frac13\right) \] \[ [\mathrm{H^+}]_Q=0.0100\,\mathrm{M} \] Therefore, \[ \frac{[\mathrm{H^+}]_Q}{[\mathrm{H^+}]_P} =\frac{0.0100}{0.0200} \] \[ \frac{[\mathrm{H^+}]_Q}{[\mathrm{H^+}]_P}=\frac12 \] \(\textbf{Interpret fractional versus absolute ionization:}\) Dilution favours a greater fraction of the weak-acid molecules becoming ionized. However, the total analytical concentration falls sufficiently that the actual hydrogen-ion concentration decreases. The percentage of ionization and the number of ions per unit volume therefore move in opposite directions. The common square-root approximation would be unreliable for the measured initial ionization degree. The exact equilibrium relationship accounts for the remaining unionized acid in both solutions. This experiment distinguishes degree of ionization, equilibrium ion concentration and the effect of dilution. \(\textbf{Final answer:}\) \[ \boxed{\alpha_Q=\frac13,\qquad\frac{[\mathrm{H^+}]_Q}{[\mathrm{H^+}]_P}=\frac12} \]

Explanation: Statement I is correct. NaCl is an ionic compound whose dissolved formula units separate into hydrated sodium and chloride ions. HCl is molecular before dissolution but transfers protons essentially completely to water under the usual aqueous strong-acid treatment. Both provide abundant mobile ions in aqueous solution and are classified as strong electrolytes. The classification refers to the extent of ion formation in solution, not merely to whether a substance is ionic or molecular in its undissolved state. Statement II is incorrect. Glucose dissolves as molecules without producing an appreciable concentration of ions under ordinary conditions. It is therefore a nonelectrolyte rather than a weak electrolyte. A weak electrolyte undergoes partial ionization and has an equilibrium involving both ionized and unionized species. Simple molecular dissolution without significant ion formation does not satisfy that description. Statement III is correct. Ionic NaCl contains ions in its crystal structure, and water stabilizes them when the crystal dissolves. Dissociation describes the separation of these pre-existing ions. In contrast, aqueous HCl produces hydronium and chloride through proton transfer: \[ \mathrm{HCl+H_2O\rightarrow H_3O^++Cl^-} \] This illustrates ionization of a molecular electrolyte. The terms are sometimes used broadly in introductory electrolyte discussions, but the microscopic distinction remains chemically useful. Statement IV is correct. Ethanoic acid is a weak acid and establishes \[ \mathrm{CH_3COOH+H_2O\rightleftharpoons CH_3COO^-+H_3O^+} \] Both the unionized acid and its ions are present in equilibrium. Aqueous HCl, by comparison, is treated as essentially completely ionized in ordinary dilute-solution calculations. Its behaviour is not represented by the same appreciable unionized-acid reservoir as a typical weak acid. \(\textbf{Interpret the classification:}\) The ability to dissolve in water is not sufficient to establish electrolyte character. The presence and extent of mobile-ion formation distinguish strong electrolytes, weak electrolytes and nonelectrolytes. A molecular substance can be a strong electrolyte after reaction with water, while another molecular substance can remain a nonelectrolyte. The correct statements are I, III and IV.

Explanation: \(\textbf{Identify the first precipitation threshold:}\) For either divalent metal hydroxide, \[ \mathrm{X(OH)_2(s)\rightleftharpoons X^{2+}+2OH^-} \] The ionic product is \[ Q_{sp}=[\mathrm{X^{2+}}][\mathrm{OH^-}]^2 \] Precipitation begins when the ionic product reaches the corresponding Ksp. Before precipitation, the free metal concentration is approximately its initial value. \(\textbf{Find the pH where M(OH)₂ begins forming:}\) \[ [\mathrm{OH^-}]_{\mathrm{M,onset}} =\sqrt{\frac{K_{sp,\mathrm M}}{[\mathrm{M^{2+}}]_0}} \] Substituting, \[ [\mathrm{OH^-}]_{\mathrm{M,onset}} =\sqrt{\frac{10^{-12}}{10^{-2}}} \] \[ [\mathrm{OH^-}]_{\mathrm{M,onset}}=1.00\times10^{-5}\,\mathrm{M} \] Thus, \[ pOH_{\mathrm{M,onset}}=5.00 \] \[ pH_{\mathrm{M,onset}}=9.00 \] \(\textbf{Find the onset condition for N(OH)₂:}\) \[ [\mathrm{OH^-}]_{\mathrm{N,onset}} =\sqrt{\frac{10^{-10}}{10^{-2}}} \] \[ [\mathrm{OH^-}]_{\mathrm{N,onset}}=1.00\times10^{-4}\,\mathrm{M} \] Therefore, \[ pOH_{\mathrm{N,onset}}=4.00 \] \[ pH_{\mathrm{N,onset}}=10.00 \] \(\textbf{Determine the free M²⁺ concentration at pH 10:}\) M(OH)₂ has already begun precipitating at the lower pH. At pH 10.00, its remaining free ions satisfy \[ [\mathrm{M^{2+}}][\mathrm{OH^-}]^2=K_{sp,\mathrm M} \] Therefore, \[ [\mathrm{M^{2+}}]_{\mathrm{free}} =\frac{10^{-12}}{(10^{-4})^2} \] \[ [\mathrm{M^{2+}}]_{\mathrm{free}} =1.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Calculate the fraction of M²⁺ removed:}\) The original analytical concentration was \(0.0100\,\mathrm{M}\). The dissolved fraction remaining is \[ f_{\mathrm{remaining}}=\frac{1.00\times10^{-4}}{0.0100} \] \[ f_{\mathrm{remaining}}=0.0100 \] Therefore, \[ f_{\mathrm{precipitated}}=1-0.0100 \] \[ f_{\mathrm{precipitated}}=0.990 \] Thus, approximately \(99.0\%\) of the original M²⁺ has precipitated. \(\textbf{Check the state of N²⁺:}\) At pH 10.00, \[ Q_{sp,\mathrm N}=(0.0100)(10^{-4})^2 \] \[ Q_{sp,\mathrm N}=1.00\times10^{-10} \] This equals its Ksp. N(OH)₂ is just reaching its precipitation threshold rather than having already lost \(99.0\%\) of its ions. \(\textbf{Interpret selective precipitation:}\) The more sparingly soluble hydroxide reaches saturation first. As pH rises further, precipitation lowers its free metal-ion concentration while the hydroxide ionic product remains at Ksp. The second metal hydroxide does not begin precipitating until a greater hydroxide concentration is reached. By that stage, the first metal can already have been removed extensively. The calculation shows how two Ksp values, the initial metal concentrations and the pH scale together determine the degree of separation. \(\textbf{Final answer:}\) \[ \boxed{pH_{\mathrm{N,onset}}=10.00,\qquad \%\mathrm{M^{2+}\ precipitated}\approx99.0\%} \]

Explanation: \(\textbf{Recognize the extreme dilution of HCl in P:}\) HCl is a strong acid, but its analytical concentration is only \[ C=1.00\times10^{-8}\,\mathrm{M} \] Pure water at \(298\,\mathrm{K}\) already contains approximately \[ 1.00\times10^{-7}\,\mathrm{M} \] hydrogen ions. Water autoionization is therefore significant relative to the added acid. \(\textbf{Apply charge balance in P:}\) Complete HCl dissociation gives \[ [\mathrm{Cl^-}]=C \] Electrical neutrality requires \[ [\mathrm{H^+}]=[\mathrm{Cl^-}]+[\mathrm{OH^-}] \] Let \[ h=[\mathrm{H^+}] \] Using \[ [\mathrm{OH^-}]=\frac{K_w}{h} \] we obtain \[ h=C+\frac{K_w}{h} \] \(\textbf{Solve for the actual hydrogen-ion concentration:}\) \[ h^2-Ch-K_w=0 \] The positive root is \[ h=\frac{C+\sqrt{C^2+4K_w}}2 \] Substituting, \[ h=\frac{10^{-8}+\sqrt{10^{-16}+4.00\times10^{-14}}}{2} \] \[ h\approx1.051\times10^{-7}\,\mathrm{M} \] Therefore, \[ pH_P=-\log_{10}(1.051\times10^{-7}) \] \[ pH_P\approx6.98 \] \(\textbf{Determine the weak-acid equilibrium in Q:}\) HA undergoes partial ionization: \[ \mathrm{HA\rightleftharpoons H^++A^-} \] Let x be the equilibrium concentration of hydrogen ions produced by HA. Then, \[ K_a=\frac{x^2}{C_{\mathrm{HA}}-x} \] Substituting, \[ 1.00\times10^{-5}=\frac{x^2}{0.00100-x} \] \[ x^2+1.00\times10^{-5}x-1.00\times10^{-8}=0 \] \(\textbf{Use the positive equilibrium root:}\) \[ x=\frac{-1.00\times10^{-5} +\sqrt{(1.00\times10^{-5})^2+4.00\times10^{-8}}}{2} \] \[ x\approx9.51\times10^{-5}\,\mathrm{M} \] Water's hydrogen-ion contribution is negligible on this scale. Thus, \[ pH_Q\approx-\log_{10}(9.51\times10^{-5}) \] \[ pH_Q\approx4.02 \] \(\textbf{Compare the actual solution acidities:}\) \[ pH_P\approx6.98 \] \[ pH_Q\approx4.02 \] Both solutions are acidic at \(298\,\mathrm{K}\), but Q has the much greater hydrogen-ion concentration. \(\textbf{Interpret acid strength versus solution concentration:}\) HCl is stronger than HA because it ionizes essentially completely under the specified conditions. However, its extremely small analytical concentration limits the number of hydrogen ions it introduces. The more concentrated weak-acid solution produces substantially more hydrogen ions despite partial ionization. A solution's pH depends on both acid strength and analytical concentration. At extreme dilution, water autoionization adds an additional constraint that cannot be ignored. The example distinguishes intrinsic electrolyte strength from the measured acidity of a particular solution. \(\textbf{Final answer:}\) \[ \boxed{pH_P\approx6.98,\qquad pH_Q\approx4.02} \]

Explanation: \(\textbf{Establish the state after the first ionization:}\) The first stage is stipulated to be essentially complete. Therefore, its contribution to hydrogen ions is \[ [\mathrm{H^+}]_{\mathrm{first}}=0.0300\,\mathrm{M} \] It also produces \[ [\mathrm{HA^-}]_{\mathrm{initial}}=0.0300\,\mathrm{M} \] The second ionization must be calculated in the presence of the hydrogen ions already produced by the first stage. \(\textbf{Represent the second-stage ionization:}\) Let x be the concentration of HA⁻ undergoing the second ionization. Then, \[ [\mathrm{HA^-}]=0.0300-x \] \[ [\mathrm{A^{2-}}]=x \] \[ [\mathrm{H^+}]=0.0300+x \] \(\textbf{Construct the second ionization equilibrium:}\) \[ K_{a2}=\frac{[\mathrm{H^+}][\mathrm{A^{2-}}]}{[\mathrm{HA^-}]} \] Substituting, \[ 0.0200=\frac{(0.0300+x)x}{0.0300-x} \] \(\textbf{Solve the coupled equilibrium expression:}\) Cross-multiplying, \[ 0.0200(0.0300-x)=(0.0300+x)x \] \[ 0.000600-0.0200x=0.0300x+x^2 \] Therefore, \[ x^2+0.0500x-0.000600=0 \] Factorising, \[ (x-0.0100)(x+0.0600)=0 \] The negative root is physically unacceptable. Hence, \[ x=0.0100\,\mathrm{M} \] \(\textbf{Recover the final ionic concentrations:}\) \[ [\mathrm{A^{2-}}]=0.0100\,\mathrm{M} \] \[ [\mathrm{HA^-}]=0.0300-0.0100 \] \[ [\mathrm{HA^-}]=0.0200\,\mathrm{M} \] \[ [\mathrm{H^+}]=0.0300+0.0100 \] \[ [\mathrm{H^+}]=0.0400\,\mathrm{M} \] \(\textbf{Determine the fully ionized fraction:}\) \[ f_{\mathrm{A^{2-}}}=\frac{[\mathrm{A^{2-}}]}{C} \] \[ f_{\mathrm{A^{2-}}}=\frac{0.0100}{0.0300} \] \[ f_{\mathrm{A^{2-}}}=\frac13 \] \(\textbf{Verify charge balance:}\) The negative charge concentration is \[ [\mathrm{HA^-}]+2[\mathrm{A^{2-}}] \] \[ =0.0200+2(0.0100) \] \[ =0.0400\,\mathrm{M} \] This agrees with the hydrogen-ion concentration. \(\textbf{Interpret the two-stage dependency:}\) The second ionization cannot be treated as though it begins in a solution containing no hydrogen ions. The nearly complete first stage has already established a substantial hydrogen-ion concentration. This common hydrogen-ion concentration influences the second equilibrium and limits the additional ionization. Nevertheless, the second-stage contribution is appreciable under the specified conditions. Ignoring it would underestimate the final acidity and miss the doubly charged anions. The calculation combines sequential ionization, conservation of acid-derived material and electrical neutrality. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{H^+}]=0.0400\,\mathrm{M},\qquad f_{\mathrm{A^{2-}}}=\frac13} \]

Explanation: \(\textbf{Determine the equilibrium hydrogen-ion concentration in P:}\) For a weak monobasic acid, \[ K_a=\frac{h^2}{C-h} \] where h is the equilibrium hydrogen-ion concentration and water autoionization is neglected. For P, \[ 1.00\times10^{-5}=\frac{h_P^2}{0.101-h_P} \] Testing \[ h_P=1.00\times10^{-3}\,\mathrm{M} \] gives \[ \frac{(1.00\times10^{-3})^2}{0.101-0.00100} =\frac{10^{-6}}{0.100} \] \[ =1.00\times10^{-5} \] Thus, \[ pH_P=3.00 \] \(\textbf{Determine the equilibrium hydrogen-ion concentration in Q:}\) For HB, \[ 1.00\times10^{-4}=\frac{h_Q^2}{0.0110-h_Q} \] Testing \[ h_Q=1.00\times10^{-3}\,\mathrm{M} \] gives \[ \frac{(1.00\times10^{-3})^2}{0.0110-0.00100} =\frac{10^{-6}}{0.0100} \] \[ =1.00\times10^{-4} \] Therefore, \[ pH_Q=3.00 \] The two acids have different strengths and concentrations but the same equilibrium hydrogen-ion concentration. \(\textbf{Identify the species inventories before mixing:}\) In P, \[ [\mathrm{A^-}]=0.00100\,\mathrm{M} \] \[ [\mathrm{HA}]=0.100\,\mathrm{M} \] In Q, \[ [\mathrm{B^-}]=0.00100\,\mathrm{M} \] \[ [\mathrm{HB}]=0.0100\,\mathrm{M} \] \(\textbf{Construct the equal-volume mixture:}\) Each original solution contributes half the final volume. If the common hydrogen-ion concentration remains \(0.00100\,\mathrm{M}\), the candidate mixed composition is \[ [\mathrm{HA}]=0.0500\,\mathrm{M} \] \[ [\mathrm{A^-}]=0.000500\,\mathrm{M} \] \[ [\mathrm{HB}]=0.00500\,\mathrm{M} \] \[ [\mathrm{B^-}]=0.000500\,\mathrm{M} \] \[ [\mathrm{H^+}]=0.00100\,\mathrm{M} \] \(\textbf{Verify the HA ionization equilibrium:}\) \[ Q_{a,\mathrm{HA}} =\frac{(0.00100)(0.000500)}{0.0500} \] \[ Q_{a,\mathrm{HA}}=1.00\times10^{-5} \] This equals \(K_a(\mathrm{HA})\). \(\textbf{Verify the HB ionization equilibrium:}\) \[ Q_{a,\mathrm{HB}} =\frac{(0.00100)(0.000500)}{0.00500} \] \[ Q_{a,\mathrm{HB}}=1.00\times10^{-4} \] This equals \(K_a(\mathrm{HB})\). \(\textbf{Check electrical neutrality:}\) The combined conjugate-base concentration is \[ [\mathrm{A^-}]+[\mathrm{B^-}] =0.000500+0.000500 \] \[ =0.00100\,\mathrm{M} \] This agrees with the proposed hydrogen-ion concentration. All material balances and both ionization equilibria are therefore satisfied. \(\textbf{Interpret the unexpected mixing result:}\) Each starting solution already has the same hydrogen-ion concentration. Mixing equal volumes does not require that concentration to double or halve. The amounts and concentrations of each acid-derived pair change proportionally during mixing. Because both equilibria can remain satisfied at the original common hydrogen-ion concentration, no additional net ionization adjustment is necessary under the stated assumptions. Different acid strengths and concentrations can therefore yield the same pH and retain that pH after mixing. The result follows from simultaneous equilibrium and charge-balance consistency, not from averaging pH values. \(\textbf{Final answer:}\) \[ \boxed{pH_P=pH_Q=pH_{\mathrm{mixture}}\approx3.00} \]

Explanation: \(\textbf{Include the ionic contribution of water:}\) The HCl concentration is comparable to the hydrogen-ion concentration arising from water autoionization. Therefore, the approximation \[ [\mathrm{H^+}]=C_{\mathrm{HCl}} \] is not sufficiently accurate. \(\textbf{Construct the charge-balance equation:}\) Complete HCl dissociation gives \[ [\mathrm{Cl^-}]=C \] Let \[ h=[\mathrm{H^+}] \] Electrical neutrality requires \[ h=C+[\mathrm{OH^-}] \] Using the water ionic product, \[ [\mathrm{OH^-}]=\frac{K_w}{h} \] Thus, \[ h=C+\frac{K_w}{h} \] \(\textbf{Solve for the positive hydrogen-ion concentration:}\) \[ h^2-Ch-K_w=0 \] The positive root is \[ h=\frac{C+\sqrt{C^2+4K_w}}2 \] \(\textbf{Evaluate the solution at T₁:}\) \[ h_1=\frac{10^{-7}+\sqrt{(10^{-7})^2+4(10^{-14})}}2 \] \[ h_1=\frac{10^{-7}+\sqrt5\times10^{-7}}2 \] \[ h_1\approx1.618\times10^{-7}\,\mathrm{M} \] Therefore, \[ pH_1=-\log_{10}(1.618\times10^{-7}) \] \[ pH_1\approx6.79 \] \(\textbf{Evaluate the same HCl concentration at T₂:}\) \[ h_2=\frac{10^{-7}+\sqrt{(10^{-7})^2+4(10^{-12})}}2 \] \[ h_2\approx1.051\times10^{-6}\,\mathrm{M} \] Therefore, \[ pH_2=-\log_{10}(1.051\times10^{-6}) \] \[ pH_2\approx5.98 \] \(\textbf{Determine the neutral reference pH at each temperature:}\) Neutrality requires \[ [\mathrm{H^+}]=[\mathrm{OH^-}]=\sqrt{K_w} \] At \(T_1\), \[ pH_{\mathrm{neutral},1}=7.00 \] At \(T_2\), \[ pH_{\mathrm{neutral},2}=6.00 \] \(\textbf{Classify the two solutions using charge balance:}\) In each solution, the chloride ions supply additional negative charge that must be balanced by hydrogen ions. Therefore, \[ [\mathrm{H^+}]\gt [\mathrm{OH^-}] \] Both solutions are acidic. At \(T_2\), the pH of approximately 5.98 is only slightly below the neutral pH of 6.00, but the solution remains acidic. \(\textbf{Interpret the temperature effect:}\) Increasing Kw raises the hydrogen-ion concentration contributed by water itself. At such a low strong-acid concentration, that contribution has a large effect on the measured pH. However, a change in the neutral reference pH must not be confused with a transition from acidic to basic behaviour. The defining comparison is between hydrogen-ion and hydroxide-ion concentrations at the actual temperature. The experiment links extreme dilution, water autoionization, temperature-dependent neutrality and the logarithmic pH scale. \(\textbf{Final answer:}\) \[ \boxed{pH_{T_1}\approx6.79,\qquad pH_{T_2}\approx5.98} \]

Explanation: \(\textbf{Interpret the concentration change:}\) The analytical concentration decreases from \[ C_1=0.100\,\mathrm{M} \] to \[ C_2=0.00100\,\mathrm{M} \] The dilution factor is \[ \frac{C_1}{C_2}=100 \] A strong monoprotic acid supplies approximately one hydrogen ion per formula unit under the stated conditions. Its hydrogen-ion concentration therefore decreases by a factor of one hundred. Its pH should rise by approximately two units. \(\textbf{Interpret the behaviour of P:}\) The graph shows \[ pH_{P,1}=1.00 \] \[ pH_{P,2}=3.00 \] The hydrogen-ion concentrations are \[ [\mathrm{H^+}]_{P,1}=0.100\,\mathrm{M} \] \[ [\mathrm{H^+}]_{P,2}=0.00100\,\mathrm{M} \] These agree with essentially complete ionization. Thus, P exhibits strong-electrolyte behaviour in the stated range. \(\textbf{Identify the weak-acid concentration relationship:}\) For a weak monoprotic acid HA, \[ K_a=\frac{[\mathrm{H^+}]^2}{C-[\mathrm{H^+}]} \] When ionization is sufficiently small, \[ [\mathrm{H^+}]\approx\sqrt{K_aC} \] Thus, a hundredfold reduction in C produces approximately a tenfold reduction in hydrogen-ion concentration. The corresponding pH rise is approximately one unit. \(\textbf{Compare that prediction with Q:}\) Q's measured pH rises from approximately 4.00 to 5.00 during the same dilution. This agrees with the characteristic approximate square-root dependence of weak-acid ionization. Therefore, Q is the weak electrolyte. \(\textbf{Infer Ka from the more concentrated Q solution:}\) At \[ C=0.100\,\mathrm{M} \] the observed pH is approximately 4.00. Hence, \[ [\mathrm{H^+}]\approx1.00\times10^{-4}\,\mathrm{M} \] Using the weak-acid approximation, \[ K_a\approx\frac{[\mathrm{H^+}]^2}{C} \] \[ K_a\approx\frac{(10^{-4})^2}{0.100} \] \[ K_a\approx1.00\times10^{-7} \] \(\textbf{Check the second point:}\) At \[ C=0.00100\,\mathrm{M} \] the predicted hydrogen-ion concentration is \[ [\mathrm{H^+}]\approx\sqrt{(10^{-7})(10^{-3})} \] \[ [\mathrm{H^+}]\approx1.00\times10^{-5}\,\mathrm{M} \] Therefore, \[ pH\approx5.00 \] This agrees with Q's second measured point. \(\textbf{Interpret the experimental identification:}\) The distinction is not based solely on comparing the initial pH values. The key evidence is how pH changes as analytical concentration changes. The strong acid's ionization is already essentially complete, whereas the weak acid becomes fractionally more ionized during dilution. That greater fractional ionization partly offsets the decrease in analytical concentration. The graph therefore reveals a difference in ionization behaviour as well as allowing the weak acid's equilibrium constant to be estimated. \(\textbf{Final answer:}\) \[ \boxed{K_a(Q)\approx1.00\times10^{-7}} \] Therefore, P is the strong acid and Q is the weak acid.

Explanation: \(\textbf{Establish the ionic-product expression for P:}\) For the dissolution equilibrium \[ \mathrm{MX_2(s)\rightleftharpoons M^{2+}+2X^-} \] the ionic product is \[ Q_{sp,P}=[\mathrm{M^{2+}}][\mathrm{X^-}]^2 \] Using the original concentrations, \[ Q_{sp,P,1}=(0.0100)(0.0200)^2 \] \[ Q_{sp,P,1}=4.00\times10^{-6} \] This exceeds the specified Ksp, confirming initial supersaturation. \(\textbf{Apply the twofold dilution to P:}\) Each ion concentration is halved before precipitation: \[ [\mathrm{M^{2+}}]'=0.00500\,\mathrm{M} \] \[ [\mathrm{X^-}]'=0.0100\,\mathrm{M} \] Therefore, \[ Q_{sp,P,2}=(0.00500)(0.0100)^2 \] \[ Q_{sp,P,2}=5.00\times10^{-7} \] Since \[ Q_{sp,P,2}\lt 1.00\times10^{-6} \] P becomes unsaturated. \(\textbf{Examine the different dissolution stoichiometry in Q:}\) For \[ \mathrm{NY(s)\rightleftharpoons N^++Y^-} \] the ionic product is \[ Q_{sp,Q}=[\mathrm{N^+}][\mathrm{Y^-}] \] Initially, \[ Q_{sp,Q,1}=(0.0100)(0.0100) \] \[ Q_{sp,Q,1}=1.00\times10^{-4} \] This exceeds its Ksp of \(2.00\times10^{-5}\). \(\textbf{Apply the same dilution to Q:}\) After doubling the solution volume, \[ [\mathrm{N^+}]'=[\mathrm{Y^-}]'=0.00500\,\mathrm{M} \] Therefore, \[ Q_{sp,Q,2}=(0.00500)(0.00500) \] \[ Q_{sp,Q,2}=2.50\times10^{-5} \] Since \[ Q_{sp,Q,2}\gt 2.00\times10^{-5} \] Q remains supersaturated. \(\textbf{Explain the different dilution factors for ionic products:}\) For MX₂, dilution changes one first-power concentration and one squared concentration. Doubling the volume consequently reduces its ionic product by \[ 2^3=8 \] For NY, the ionic product contains two first-power concentrations. The same dilution reduces its ionic product by \[ 2^2=4 \] The two solutions also begin at different degrees of supersaturation relative to their respective Ksp values. \(\textbf{Interpret the experimental result:}\) Dilution lowers every ionic concentration but does not guarantee that every initially supersaturated solution becomes unsaturated. The final saturation classification depends on the ionic-product expression and the relevant Ksp. The stoichiometric powers in that expression determine how strongly dilution changes Qsp. The assumption that no precipitation occurs before the instantaneous comparison is important because precipitation would otherwise alter the ionic inventories. The problem distinguishes a concentration disturbance from the subsequent formation of solid and demonstrates why the same dilution can produce different saturation outcomes. \(\textbf{Final answer:}\) \[ \boxed{Q_{sp,P,2}=5.00\times10^{-7}\lt K_{sp,P},\qquad Q_{sp,Q,2}=2.50\times10^{-5}\gt K_{sp,Q}} \]

Explanation: \(\textbf{Determine the original weak-acid ionization:}\) HA ionizes according to \[ \mathrm{HA\rightleftharpoons H^++A^-} \] For an initially pure weak-acid solution, \[ [\mathrm{H^+}]\approx\sqrt{K_aC} \] Therefore, \[ [\mathrm{H^+}]_1\approx\sqrt{(10^{-5})(0.100)} \] \[ [\mathrm{H^+}]_1\approx1.00\times10^{-3}\,\mathrm{M} \] The original degree of ionization is approximately \[ \alpha_1\approx\frac{1.00\times10^{-3}}{0.100} \] \[ \alpha_1\approx0.0100 \] Thus, approximately one percent of the original HA is ionized. \(\textbf{Identify the effect of adding NaA:}\) NaA is treated as a strong electrolyte: \[ \mathrm{NaA\rightarrow Na^++A^-} \] It introduces a substantial concentration of A⁻, which is already a product of HA ionization. The common ion suppresses additional HA ionization. The equilibrium constant remains unchanged because the temperature is fixed. \(\textbf{Estimate the new hydrogen-ion concentration:}\) The added A⁻ concentration is approximately \(0.0500\,\mathrm{M}\). The weak-acid equilibrium gives \[ K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \] Rearranging, \[ [\mathrm{H^+}]_2\approx K_a\frac{[\mathrm{HA}]}{[\mathrm{A^-}]} \] Substituting, \[ [\mathrm{H^+}]_2\approx(10^{-5})\frac{0.100}{0.0500} \] \[ [\mathrm{H^+}]_2\approx2.00\times10^{-5}\,\mathrm{M} \] \(\textbf{Compare the weak-acid ionization fractions:}\) The new approximate ionization fraction is \[ \alpha_2\approx\frac{2.00\times10^{-5}}{0.100} \] \[ \alpha_2\approx2.00\times10^{-4} \] Therefore, \[ \alpha_2\lt \alpha_1 \] The weak acid is substantially less ionized after adding its conjugate-base salt. \(\textbf{Count the principal ions in the original solution:}\) Before salt addition, the main mobile ions generated by HA are H⁺ and A⁻. Their combined concentrations are approximately \[ [\mathrm{H^+}]+[\mathrm{A^-}] \approx2(1.00\times10^{-3}) \] \[ \approx0.00200\,\mathrm{M} \] \(\textbf{Count the added ions after dissolving NaA:}\) The added strong electrolyte supplies approximately \[ [\mathrm{Na^+}]=0.0500\,\mathrm{M} \] and at least approximately \[ [\mathrm{A^-}]=0.0500\,\mathrm{M} \] from its dissociation. Thus, the combined concentration of these two mobile-ion species alone is approximately \(0.100\,\mathrm{M}\), much greater than the original weak-acid ionic concentration. \(\textbf{Interpret the apparent contradiction:}\) The common-ion effect decreases ionization of the weak acid. However, the common-ion salt itself introduces mobile ions into solution. The total ionic population can therefore increase while the fraction of HA molecules that are ionized decreases. These are different quantities and must not be treated as interchangeable. One concerns the reversible ionization equilibrium of HA, while the other includes the ions supplied directly by NaA. The experiment integrates strong-electrolyte dissociation, weak-electrolyte ionization and common-ion suppression. It also shows why a reduction in weak-acid ionization does not necessarily imply a reduction in the solution's overall ionic content.

Explanation: \(\textbf{Distinguish dissolved molecular B from its ionized form:}\) The specified gas-dissolution relationship fixes the concentration of un-ionized B(aq). The aqueous base-ionization equilibrium then determines how much BH⁺ and OH⁻ form. These equilibria are connected, but they describe different processes. \(\textbf{Determine the initial dissolved molecular concentration:}\) \[ [\mathrm{B(aq)}]_1=kp_{B,1} \] Substituting, \[ [\mathrm{B(aq)}]_1=(0.100)(1.00) \] \[ [\mathrm{B(aq)}]_1=0.100\,\mathrm{M} \] \(\textbf{Apply the weak-base equilibrium:}\) \[ K_b=\frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B(aq)}]} \] With negligible water autoionization, \[ [\mathrm{BH^+}]\approx[\mathrm{OH^-}] \] Therefore, \[ [\mathrm{OH^-}]^2\approx K_b[\mathrm{B(aq)}] \] \(\textbf{Calculate the initial hydroxide concentration:}\) \[ [\mathrm{OH^-}]_1\approx\sqrt{(10^{-7})(0.100)} \] \[ [\mathrm{OH^-}]_1\approx1.00\times10^{-4}\,\mathrm{M} \] Thus, \[ [\mathrm{BH^+}]_1\approx1.00\times10^{-4}\,\mathrm{M} \] \(\textbf{Calculate the initial ionized fraction:}\) The total dissolved B-derived concentration is \[ C_{B,\mathrm{total}}=[\mathrm{B(aq)}]+[\mathrm{BH^+}] \] Hence, \[ f_1=\frac{[\mathrm{BH^+}]_1}{[\mathrm{B(aq)}]_1+[\mathrm{BH^+}]_1} \] \[ f_1\approx\frac{0.000100}{0.100+0.000100} \] \[ f_1\approx9.99\times10^{-4} \] \(\textbf{Determine the molecular concentration at higher pressure:}\) The gas pressure increases fourfold. Therefore, \[ [\mathrm{B(aq)}]_2=(0.100)(4.00) \] \[ [\mathrm{B(aq)}]_2=0.400\,\mathrm{M} \] \(\textbf{Calculate the new hydroxide concentration:}\) \[ [\mathrm{OH^-}]_2\approx\sqrt{(10^{-7})(0.400)} \] \[ [\mathrm{OH^-}]_2\approx2.00\times10^{-4}\,\mathrm{M} \] Consequently, \[ \frac{[\mathrm{OH^-}]_2}{[\mathrm{OH^-}]_1}=2.00 \] The hydroxide concentration doubles. \(\textbf{Calculate the new ionized fraction:}\) \[ f_2\approx\frac{0.000200}{0.400+0.000200} \] \[ f_2\approx5.00\times10^{-4} \] Therefore, \[ \frac{f_2}{f_1}\approx0.500 \] The fraction of dissolved B-derived material present as BH⁺ becomes approximately half its original value. \(\textbf{Interpret the competing concentration effects:}\) The increased gas pressure raises the concentration of un-ionized molecular B in water. The base-ionization equilibrium responds by producing more BH⁺ and OH⁻. However, the molecular concentration increases fourfold while the ionized concentration increases only twofold. The fraction ionized therefore decreases even though the absolute hydroxide-ion concentration increases. This distinction parallels the difference between absolute ionic concentration and percentage ionization in weak electrolytes. The experiment integrates gas–liquid dissolution, weak-base ionization and the interpretation of fractional versus absolute equilibrium amounts. \(\textbf{Final answer:}\) \[ \boxed{\frac{[\mathrm{OH^-}]_2}{[\mathrm{OH^-}]_1}=2.00,\qquad\frac{f_2}{f_1}\approx0.500} \]

Explanation: \(\textbf{Establish the adjacent-species relationships:}\) The two ionization equilibria are \[ \mathrm{H_2A\rightleftharpoons H^++HA^-} \] \[ \mathrm{HA^-\rightleftharpoons H^++A^{2-}} \] Their equilibrium expressions give \[ \frac{[\mathrm{HA^-}]}{[\mathrm{H_2A}]}=\frac{K_{a1}}{[\mathrm{H^+}]} \] and \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]}=\frac{K_{a2}}{[\mathrm{H^+}]} \] Both ratios change with the maintained pH. \(\textbf{Evaluate the distribution at pH 4.00:}\) Here, \[ [\mathrm{H^+}]=10^{-4}\,\mathrm{M} \] Therefore, \[ \frac{[\mathrm{HA^-}]}{[\mathrm{H_2A}]}=\frac{10^{-3}}{10^{-4}}=10 \] and \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]}=\frac{10^{-7}}{10^{-4}}=0.001 \] Thus, \[ [\mathrm{H_2A}]:[\mathrm{HA^-}]:[\mathrm{A^{2-}}] =1:10:0.01 \] The HA⁻ fraction is \[ f_P=\frac{10}{11.01} \] \[ f_P\approx0.9083 \] \(\textbf{Evaluate the distribution at pH 5.00:}\) Now, \[ [\mathrm{H^+}]=10^{-5}\,\mathrm{M} \] The adjacent ratios are \[ \frac{[\mathrm{HA^-}]}{[\mathrm{H_2A}]}=100 \] and \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]}=0.0100 \] Therefore, \[ [\mathrm{H_2A}]:[\mathrm{HA^-}]:[\mathrm{A^{2-}}] =1:100:1 \] Hence, \[ f_Q=\frac{100}{102} \] \[ f_Q\approx0.9804 \] H₂A and A²⁻ have equal equilibrium concentrations at this pH. \(\textbf{Evaluate the distribution at pH 6.00:}\) Here, \[ [\mathrm{H^+}]=10^{-6}\,\mathrm{M} \] The ratios become \[ \frac{[\mathrm{HA^-}]}{[\mathrm{H_2A}]}=1000 \] \[ \frac{[\mathrm{A^{2-}}]}{[\mathrm{HA^-}]}=0.100 \] Dividing the species ratio by 1000 gives \[ [\mathrm{H_2A}]:[\mathrm{HA^-}]:[\mathrm{A^{2-}}] =0.001:1:0.100 \] Therefore, \[ f_R=\frac1{1.101} \] \[ f_R\approx0.9083 \] \(\textbf{Assess Statement I:}\) The calculations give \[ f_P=f_R\approx0.9083 \] Thus, Statement I is correct. The minor-species distributions differ, but the intermediate HA⁻ has the same overall fraction. \(\textbf{Assess Statement II:}\) The HA⁻ fraction at pH 5.00 is approximately 0.9804. This exceeds its fraction at pH 4.00 or 6.00. Statement II is correct. \(\textbf{Assess Statement III:}\) At pH 5.00, the calculated distribution is \(1:100:1\). The concentrations of H₂A and A²⁻ are therefore equal. Statement III is correct. \(\textbf{Assess Statement IV:}\) At pH 6.00, the fraction of H₂A is substantially smaller than at pH 4.00. Because the total analytical concentrations are specified to be equal, its absolute concentration is also smaller. Statement IV is incorrect. \(\textbf{Interpret the distribution symmetry:}\) The intermediate species HA⁻ is favoured over a broad pH region between the two pKa values. At lower pH, a greater proportion remains as H₂A. At higher pH, a greater proportion becomes A²⁻. The equal intermediate fractions at pH 4.00 and 6.00 arise from the symmetric placement of those pH values around the midpoint of the two pKa values. This equality does not mean that the three individual species concentrations are identical in P and R. The question connects successive acid-ionization constants with species distributions rather than treating the intermediate ion as a fixed fraction. Statements I, II and III are correct.

Explanation: \(\textbf{Distinguish free X⁻ from total dissolved salt:}\) Each dissolved formula unit of MX supplies one M⁺ and one X-derived unit. However, the X⁻ ion may accept a proton and form HX. Therefore, the total amount of MX dissolved need not equal the free X⁻ concentration. Let S denote molar solubility. Then, \[ [\mathrm{M^+}]=S \] and \[ S=[\mathrm{X^-}]+[\mathrm{HX}] \] \(\textbf{Relate protonated and free X species:}\) The acid-ionization constant is \[ K_a=\frac{[\mathrm{H^+}][\mathrm{X^-}]}{[\mathrm{HX}]} \] Rearranging, \[ \frac{[\mathrm{HX}]}{[\mathrm{X^-}]}=\frac{[\mathrm{H^+}]}{K_a} \] Therefore, \[ S=[\mathrm{X^-}]\left(1+\frac{[\mathrm{H^+}]}{K_a}\right) \] \(\textbf{Apply the dissolution saturation condition:}\) For MX, \[ K_{sp}=[\mathrm{M^+}][\mathrm{X^-}] \] Since \[ [\mathrm{M^+}]=S \] we have \[ K_{sp}=S[\mathrm{X^-}] \] Substituting the expression for total solubility, \[ K_{sp}=\frac{S^2}{1+[\mathrm{H^+}]/K_a} \] Therefore, \[ S=\sqrt{K_{sp}\left(1+\frac{[\mathrm{H^+}]}{K_a}\right)} \] \(\textbf{Determine the total solubility at pH 4.00:}\) In P, \[ [\mathrm{H^+}]_P=1.00\times10^{-4}\,\mathrm{M} \] Thus, \[ 1+\frac{[\mathrm{H^+}]_P}{K_a} =1+\frac{10^{-4}}{10^{-5}} \] \[ =11.0 \] Therefore, \[ S_P=\sqrt{(1.10\times10^{-8})(11.0)} \] \[ S_P=\sqrt{1.21\times10^{-7}} \] \[ S_P\approx3.48\times10^{-4}\,\mathrm{M} \] \(\textbf{Determine the total solubility at pH 6.00:}\) In Q, \[ [\mathrm{H^+}]_Q=1.00\times10^{-6}\,\mathrm{M} \] Therefore, \[ 1+\frac{[\mathrm{H^+}]_Q}{K_a} =1+\frac{10^{-6}}{10^{-5}} \] \[ =1.10 \] Thus, \[ S_Q=\sqrt{(1.10\times10^{-8})(1.10)} \] \[ S_Q=\sqrt{1.21\times10^{-8}} \] \[ S_Q=1.10\times10^{-4}\,\mathrm{M} \] \(\textbf{Calculate the solubility ratio:}\) \[ \frac{S_P}{S_Q} =\sqrt{\frac{11.0}{1.10}} \] \[ \frac{S_P}{S_Q}=\sqrt{10} \] \[ \frac{S_P}{S_Q}\approx3.16 \] \(\textbf{Interpret the pH dependence:}\) The intrinsic Ksp is the same in both experiments because temperature is unchanged. The difference arises because lower pH favours protonation of X⁻ into HX. That protonation reduces the free X⁻ concentration available for the Ksp expression. Additional MX can consequently dissolve while maintaining the same saturation ionic product. The total molar solubility therefore increases in the more acidic solution. This illustrates why a salt containing the conjugate base of a weak acid can have pH-dependent solubility. The calculation combines acid-base speciation, a material balance and the solubility-product condition. \(\textbf{Final answer:}\) \[ \boxed{\frac{S_P}{S_Q}=\sqrt{10}\approx3.16} \]

Explanation: \(\textbf{Establish the initial ionic product:}\) The dissolution equilibrium is \[ \mathrm{MX(s)\rightleftharpoons M^++X^-} \] Initially, the ion concentrations are \[ [\mathrm{M^+}]_0=[\mathrm{X^-}]_0=0.0100\,\mathrm{M} \] Therefore, \[ Q_{sp,0}=(0.0100)(0.0100) \] \[ Q_{sp,0}=1.00\times10^{-4} \] Since \[ Q_{sp,0}\gt K_{sp} \] the original solution is supersaturated and can precipitate MX. \(\textbf{Follow precipitation in pathway P:}\) The original amounts of M⁺ and X⁻ are equal. Formation of one mole of MX removes one mole of each ion. Therefore, their dissolved concentrations remain equal during precipitation. At equilibrium with solid present, \[ K_{sp}=[\mathrm{M^+}][\mathrm{X^-}] \] Let the remaining free-ion concentration be s. Then, \[ s^2=1.00\times10^{-6} \] \[ s=1.00\times10^{-3}\,\mathrm{M} \] \(\textbf{Determine the dissolved inventory before filtration:}\) The solution volume is \(1.00\,\mathrm{L}\). Therefore, \[ n_{\mathrm{M^+,dissolved}}=(10^{-3})(1.00) \] \[ n_{\mathrm{M^+,dissolved}}=0.00100\,\mathrm{mol} \] The same amount of X⁻ remains dissolved. The amount of solid precipitated is \[ n_{\mathrm{MX(s)}}=0.0100-0.00100 \] \[ n_{\mathrm{MX(s)}}=0.00900\,\mathrm{mol} \] \(\textbf{Account for filtration:}\) All solid MX is removed from P. The filtrate therefore retains only \[ 0.00100\,\mathrm{mol} \] of each dissolved ion. The removed solid is no longer available to dissolve during the later dilution. \(\textbf{Apply the tenfold dilution of P:}\) The filtrate volume increases to \(10.0\,\mathrm{L}\). Thus, \[ [\mathrm{M^+}]_{P,f}=\frac{0.00100}{10.0} \] \[ [\mathrm{M^+}]_{P,f}=1.00\times10^{-4}\,\mathrm{M} \] The final X⁻ concentration is the same. Its ionic product is \[ Q_{sp,P,f}=(10^{-4})^2 \] \[ Q_{sp,P,f}=1.00\times10^{-8}\lt K_{sp} \] The filtrate is unsaturated, but no MX solid remains to dissolve. \(\textbf{Follow the alternative order in pathway Q:}\) Q undergoes dilution before any precipitation occurs. The original amount of each ion remains \[ 0.0100\,\mathrm{mol} \] in the new volume of \(10.0\,\mathrm{L}\). Therefore, \[ [\mathrm{M^+}]_{Q,f}=\frac{0.0100}{10.0} \] \[ [\mathrm{M^+}]_{Q,f}=1.00\times10^{-3}\,\mathrm{M} \] Similarly, \[ [\mathrm{X^-}]_{Q,f}=1.00\times10^{-3}\,\mathrm{M} \] \(\textbf{Check Q's saturation condition:}\) \[ Q_{sp,Q,f}=(10^{-3})(10^{-3}) \] \[ Q_{sp,Q,f}=1.00\times10^{-6} \] This equals Ksp. The diluted solution is exactly saturated under the stipulated model, so no net precipitation is required. \(\textbf{Interpret the different final outcomes:}\) The two pathways begin with identical solutions and end at the same volume and temperature. However, P loses a substantial amount of its dissolved material through precipitation followed by filtration. Q retains the entire original ionic inventory because dilution takes place before precipitation. The difference is not caused by a change in Ksp or by a failure of equilibrium thermodynamics. It arises because filtration removes material from one system and changes its subsequent conservation constraints. The order of physical separation and dilution can therefore determine different final concentrations when the operations do not preserve the same total material inventory. The experiment combines precipitation equilibrium, solution dilution and the practical significance of removing an equilibrium phase. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{M^+}]_{P,f}=1.00\times10^{-4}\,\mathrm{M},\qquad [\mathrm{M^+}]_{Q,f}=1.00\times10^{-3}\,\mathrm{M}} \]

Explanation: \(\textbf{Identify the two simultaneous dissolution equilibria:}\) Because both solids remain present, each saturated dissolution equilibrium must be satisfied: \[ \mathrm{AgCl(s)\rightleftharpoons Ag^++Cl^-} \] \[ \mathrm{AgBr(s)\rightleftharpoons Ag^++Br^-} \] The two equilibria share the same dissolved silver-ion concentration. Consequently, they cannot be treated as two independent saturated solutions. \(\textbf{Construct the individual solubility-product expressions:}\) For silver chloride, \[ [\mathrm{Ag^+}][\mathrm{Cl^-}]=9.00\times10^{-10} \] For silver bromide, \[ [\mathrm{Ag^+}][\mathrm{Br^-}]=1.60\times10^{-9} \] \(\textbf{Relate the two halide concentrations:}\) Dividing the expressions eliminates the common silver-ion concentration: \[ \frac{[\mathrm{Cl^-}]}{[\mathrm{Br^-}]} =\frac{9.00\times10^{-10}}{1.60\times10^{-9}} \] \[ \frac{[\mathrm{Cl^-}]}{[\mathrm{Br^-}]}=\frac9{16} \] Thus, bromide ions are more abundant than chloride ions in the saturated solution. \(\textbf{Apply electrical neutrality:}\) Under the stated assumptions, Ag⁺ is the only appreciable cation. Chloride and bromide are the only appreciable anions. Therefore, \[ [\mathrm{Ag^+}]=[\mathrm{Cl^-}]+[\mathrm{Br^-}] \] \(\textbf{Express both halide concentrations through silver:}\) \[ [\mathrm{Cl^-}]=\frac{9.00\times10^{-10}}{[\mathrm{Ag^+}]} \] \[ [\mathrm{Br^-}]=\frac{1.60\times10^{-9}}{[\mathrm{Ag^+}]} \] Substituting into charge balance, \[ [\mathrm{Ag^+}] =\frac{9.00\times10^{-10}+1.60\times10^{-9}}{[\mathrm{Ag^+}]} \] Therefore, \[ [\mathrm{Ag^+}]^2=2.50\times10^{-9} \] \[ [\mathrm{Ag^+}]=5.00\times10^{-5}\,\mathrm{M} \] \(\textbf{Recover the individual halide concentrations:}\) \[ [\mathrm{Cl^-}] =\frac{9.00\times10^{-10}}{5.00\times10^{-5}} \] \[ [\mathrm{Cl^-}]=1.80\times10^{-5}\,\mathrm{M} \] Similarly, \[ [\mathrm{Br^-}] =\frac{1.60\times10^{-9}}{5.00\times10^{-5}} \] \[ [\mathrm{Br^-}]=3.20\times10^{-5}\,\mathrm{M} \] \(\textbf{Calculate the chloride fraction:}\) \[ f_{\mathrm{Cl^-}} =\frac{[\mathrm{Cl^-}]}{[\mathrm{Cl^-}]+[\mathrm{Br^-}]} \] \[ f_{\mathrm{Cl^-}} =\frac{1.80\times10^{-5}}{5.00\times10^{-5}} \] \[ f_{\mathrm{Cl^-}}=0.360 \] \(\textbf{Verify simultaneous saturation:}\) For AgCl, \[ (5.00\times10^{-5})(1.80\times10^{-5}) =9.00\times10^{-10} \] For AgBr, \[ (5.00\times10^{-5})(3.20\times10^{-5}) =1.60\times10^{-9} \] Both calculated ionic products equal the respective Ksp values. \(\textbf{Interpret the shared-ion constraint:}\) Dissolving either salt supplies Ag⁺, which affects the dissolution equilibrium of the other salt. The common silver-ion concentration must therefore satisfy both solubility products and the solution's electrical neutrality simultaneously. Simply calculating the pure-water solubility of each salt separately and adding the answers would ignore their mutual common-ion effect. The final halide proportions reflect the relative solubility products, whereas the total dissolved silver concentration reflects their combined contribution. This calculation integrates two heterogeneous equilibria with ionic charge balance. \(\textbf{Final answer:}\) \[ \boxed{[\mathrm{Ag^+}]=5.00\times10^{-5}\,\mathrm{M},\qquad f_{\mathrm{Cl^-}}=0.360} \]