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MCQEasyJEE 2024Solubility Product

JEE Chemistry 2024 Question with Solution

For a sparingly soluble salt AB2AB_2, the equilibrium concentrations of A2+A^{2+} ions and B−B^- ions are 1.2×10−4 M1.2 \times 10^{-4} \, \text{M} and 0.24×10−3 M0.24 \times 10^{-3} \, \text{M}, respectively. The solubility product of AB2AB_2 is:

  • A

    0.069×10−120.069 \times 10^{-12}

  • B

    6.91×10−126.91 \times 10^{-12}

  • C

    0.276×10−120.276 \times 10^{-12}

  • D

    27.65×10−1227.65 \times 10^{-12}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The equilibrium concentrations are [A2+]=1.2×10−4 M[A^{2+}] = 1.2 \times 10^{-4} \, \text{M} and [B−]=0.24×10−3 M[B^-] = 0.24 \times 10^{-3} \, \text{M} for the sparingly soluble salt AB2AB_2.

Find: The solubility product KspK_{sp} of AB2AB_2.

The dissociation equilibrium is

AB2(s)⇌A2+(aq)+2B−(aq)AB_2 (s) \rightleftharpoons A^{2+} (aq) + 2B^- (aq)

So, the solubility product expression is

Ksp=[A2+][B−]2K_{sp} = [A^{2+}][B^-]^2

Substituting the given values,

Ksp=(1.2×10−4)(0.24×10−3)2K_{sp} = (1.2 \times 10^{-4})(0.24 \times 10^{-3})^2

First, calculate the square term:

(0.24×10−3)2=0.0576×10−6(0.24 \times 10^{-3})^2 = 0.0576 \times 10^{-6}

Now multiply:

Ksp=(1.2×10−4)(0.0576×10−6)=6.912×10−12K_{sp} = (1.2 \times 10^{-4})(0.0576 \times 10^{-6}) = 6.912 \times 10^{-12}

Rounding to the given options,

Ksp=6.91×10−12K_{sp} = 6.91 \times 10^{-12}

Therefore, the correct option is B.

Using Direct Equilibrium Concentrations

Given: [A2+]=1.2×10−4 M[A^{2+}] = 1.2 \times 10^{-4} \, \text{M} and [B−]=0.24×10−3 M[B^-] = 0.24 \times 10^{-3} \, \text{M}.

Find: The value of KspK_{sp}.

Use the equilibrium relation for AB2AB_2:

Ksp=[A2+][B−]2K_{sp} = [A^{2+}][B^-]^2

Insert the concentrations directly:

Ksp=(1.2×10−4)(0.24×10−3)2K_{sp} = (1.2 \times 10^{-4})(0.24 \times 10^{-3})^2 =(1.2×10−4)(0.0576×10−6)= (1.2 \times 10^{-4})(0.0576 \times 10^{-6}) =6.912×10−12= 6.912 \times 10^{-12}

Hence, the nearest listed value is 6.91×10−126.91 \times 10^{-12}, so the correct option is B.

Common mistakes

  • Using the wrong solubility product expression such as Ksp=[A2+][B−]K_{sp} = [A^{2+}][B^-]. This is incorrect because the stoichiometric coefficient of B−B^- is 22, so its concentration must be squared. Use Ksp=[A2+][B−]2K_{sp} = [A^{2+}][B^-]^2 instead.

  • Treating 0.24×10−30.24 \times 10^{-3} as 2.4×10−32.4 \times 10^{-3} or squaring only 10−310^{-3} and not 0.240.24. This gives an incorrect magnitude. Square the entire quantity 0.24×10−30.24 \times 10^{-3} together.

  • Assuming the solubility ss must first be calculated. Here the equilibrium ionic concentrations are already given, so KspK_{sp} can be obtained directly from the formula without introducing an unnecessary variable.

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