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Ionic Equilibrium question

2011 · Shift 2 · Q6
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Ionic Equilibrium question

2011 · Shift 2 · Q6

JEE AdvancedChemistryIonic EquilibriumNumerical+2 / −1
In 1 L saturated solution of AgClAgClAgCl [Ksp(AgClAgClAgCl) = 1.6 ×\times× 10-10], 0.1 mol of CuClCuClCuCl [Ksp(CuClCuClCuCl) = 1.0 ×\times× 10-6] is added. The resultant concentration of Ag+Ag^+Ag+ in the solution is 1.6 ×\times× 10-x. The value of "x" is
Numerical answer
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Correct answer: 7

Step-by-step Solution:

  1. Identify the relevant equilibria: We have two sparingly soluble salts, AgClAgClAgCl and CuClCuClCuCl, in the same solution. Both will establish a solubility equilibrium, and since they share a common ion (Cl−Cl^-Cl−), these equilibria are linked. The dissolution reactions are: AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)AgCl(s)⇌Ag+(aq)+Cl−(aq) CuCl(s)⇌Cu+(aq)+Cl−(aq)CuCl(s) \rightleftharpoons Cu^+(aq) + Cl^-(aq)CuCl(s)⇌Cu+(aq)+Cl−(aq)

  2. Write the solubility product expressions (Ksp): The equilibrium conditions are described by their respective Ksp values: Ksp(AgCl)=[Ag+][Cl−]=1.6×10−10...(i)K_{sp}(AgCl) = [Ag^+][Cl^-] = 1.6 \times 10^{-10} \quad ...(i)Ksp​(AgCl)=[Ag+][Cl−]=1.6×10−10...(i) Ksp(CuCl)=[Cu+][Cl−]=1.0×10−6...(ii)K_{sp}(CuCl) = [Cu^+][Cl^-] = 1.0 \times 10^{-6} \quad ...(ii)Ksp​(CuCl)=[Cu+][Cl−]=1.0×10−6...(ii) Both these equations must be satisfied simultaneously in the final solution.

  3. Analyze the source of the common ion, Cl−Cl^-Cl−: The chloride ions in the solution come from the dissolution of both AgClAgClAgCl and CuClCuClCuCl. From the stoichiometry of the dissolution reactions, the total concentration of chloride ions is the sum of the concentrations of the cations: [Cl−]total=[Ag+]+[Cu+]...(iii)[Cl^-]_{total} = [Ag^+] + [Cu^+] \quad ...(iii)[Cl−]total​=[Ag+]+[Cu+]...(iii)

  4. Compare the solubilities and make an approximation: We are given that Ksp(CuCl)=1.0×10−6K_{sp}(CuCl) = 1.0 \times 10^{-6}Ksp​(CuCl)=1.0×10−6 and Ksp(AgCl)=1.6×10−10K_{sp}(AgCl) = 1.6 \times 10^{-10}Ksp​(AgCl)=1.6×10−10. Since Ksp(CuCl)≫Ksp(AgCl)K_{sp}(CuCl) \gg K_{sp}(AgCl)Ksp​(CuCl)≫Ksp​(AgCl), the solubility of CuClCuClCuCl is much greater than that of AgClAgClAgCl. This means that the contribution of CuClCuClCuCl to the total chloride ion concentration will be much larger than that of AgClAgClAgCl. Therefore, we can say that [Cu+]≫[Ag+][Cu^+] \gg [Ag^+][Cu+]≫[Ag+]. Based on this, we can approximate the total chloride concentration in equation (iii): [Cl−]total=[Ag+]+[Cu+]≈[Cu+][Cl^-]_{total} = [Ag^+] + [Cu^+] \approx [Cu^+][Cl−]total​=[Ag+]+[Cu+]≈[Cu+]

  5. Calculate the concentration of Cu+Cu^+Cu+ and Cl−Cl^-Cl−: We substitute the approximation [Cl−]total≈[Cu+][Cl^-]_{total} \approx [Cu^+][Cl−]total​≈[Cu+] into the Ksp expression for CuClCuClCuCl (equation ii): [Cu+][Cl−]=[Cu+][Cu+]=[Cu+]2≈1.0×10−6[Cu^+][Cl^-] = [Cu^+][Cu^+] = [Cu^+]^2 \approx 1.0 \times 10^{-6}[Cu+][Cl−]=[Cu+][Cu+]=[Cu+]2≈1.0×10−6 Solving for [Cu+][Cu^+][Cu+]: [Cu+]≈1.0×10−6=1.0×10−3 M[Cu^+] \approx \sqrt{1.0 \times 10^{-6}} = 1.0 \times 10^{-3} \text{ M}[Cu+]≈1.0×10−6​=1.0×10−3 M Since [Cl−]total≈[Cu+][Cl^-]_{total} \approx [Cu^+][Cl−]total​≈[Cu+], we have: [Cl−]total≈1.0×10−3 M[Cl^-]_{total} \approx 1.0 \times 10^{-3} \text{ M}[Cl−]total​≈1.0×10−3 M We must verify that the solution is saturated with CuClCuClCuCl. The amount of CuClCuClCuCl that needs to dissolve to achieve this concentration in 1 L is 1.0×10−31.0 \times 10^{-3}1.0×10−3 mol. Since 0.1 mol of CuClCuClCuCl was added, which is much greater than 1.0×10−31.0 \times 10^{-3}1.0×10−3 mol, the solution will indeed be saturated with CuClCuClCuCl, and solid CuClCuClCuCl will be present at equilibrium. Our assumption is valid.

  6. Calculate the resultant concentration of Ag+Ag^+Ag+: Now we can use the calculated total chloride concentration in the Ksp expression for AgClAgClAgCl (equation i) to find the final concentration of Ag+Ag^+Ag+: [Ag+][Cl−]total=1.6×10−10[Ag^+][Cl^-]_{total} = 1.6 \times 10^{-10}[Ag+][Cl−]total​=1.6×10−10 [Ag+](1.0×10−3)=1.6×10−10[Ag^+](1.0 \times 10^{-3}) = 1.6 \times 10^{-10}[Ag+](1.0×10−3)=1.6×10−10 [Ag+]=1.6×10−101.0×10−3[Ag^+] = \frac{1.6 \times 10^{-10}}{1.0 \times 10^{-3}}[Ag+]=1.0×10−31.6×10−10​ [Ag+]=1.6×10−7 M[Ag^+] = 1.6 \times 10^{-7} \text{ M}[Ag+]=1.6×10−7 M

  7. Determine the value of 'x': The problem states that the resultant concentration of Ag+Ag^+Ag+ in the solution is 1.6×10−x1.6 \times 10^{-x}1.6×10−x. Comparing this with our calculated value: 1.6×10−x=1.6×10−71.6 \times 10^{-x} = 1.6 \times 10^{-7}1.6×10−x=1.6×10−7 Therefore, the value of 'x' is 7.

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