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NEET CHEMISTRYEquilibriumMedium

Question

The solubility product of a sparingly soluble salt AX2\text{AX}_2 is 3.2×10113.2 \times 10^{-11}. Its solubility (in moles/litre) is:

A

3.1×1043.1 \times 10^{-4}

B

2×1042 \times 10^{-4}

C

4×1044 \times 10^{-4}

D

5.6×1065.6 \times 10^{-6}

Step-by-Step Solution

For a sparingly soluble salt of type AX2\text{AX}_2, its dissociation equilibrium is: AX2(s)A2+(aq)+2X(aq)\text{AX}_2(s) \rightleftharpoons \text{A}^{2+}(aq) + 2\text{X}^-(aq)

If SS is the molar solubility, then the equilibrium concentrations of the ions are: [A2+]=S[\text{A}^{2+}] = S [X]=2S[\text{X}^-] = 2S

The solubility product constant (KspK_{sp}) is given by: Ksp=[A2+][X]2K_{sp} = [\text{A}^{2+}][\text{X}^-]^2 Ksp=(S)(2S)2=4S3K_{sp} = (S)(2S)^2 = 4S^3

Given that Ksp=3.2×1011K_{sp} = 3.2 \times 10^{-11}, we can rewrite it as 32×101232 \times 10^{-12} for easier calculation: 4S3=32×10124S^3 = 32 \times 10^{-12} S3=32×10124S^3 = \frac{32 \times 10^{-12}}{4} S3=8×1012S^3 = 8 \times 10^{-12}

Taking the cube root of both sides: S=(8×1012)13=2×104 mol/LS = (8 \times 10^{-12})^{\frac{1}{3}} = 2 \times 10^{-4} \text{ mol/L}

Thus, the solubility of the salt is 2×104 mol/L2 \times 10^{-4} \text{ mol/L}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYEquilibriumsolubilityproductsparinglysolubletextax

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