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NEET CHEMISTRYEquilibriumMedium

Question

The molar solubility of CaF2\text{CaF}_2 (Ksp=5.3×1011K_{sp} = 5.3 \times 10^{-11}) in 0.1 M0.1\text{ M} solution of NaF will be:

A

5.3×1011 mol L15.3 \times 10^{-11}\text{ mol L}^{-1}

B

5.3×108 mol L15.3 \times 10^{-8}\text{ mol L}^{-1}

C

5.3×109 mol L15.3 \times 10^{-9}\text{ mol L}^{-1}

D

5.3×1010 mol L15.3 \times 10^{-10}\text{ mol L}^{-1}

Step-by-Step Solution

The dissociation of CaF2\text{CaF}_2 is given by the equilibrium: CaF2(s)Ca2+(aq)+2F(aq)\text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) Let the molar solubility of CaF2\text{CaF}_2 in 0.1 M0.1\text{ M} NaF be SS. The concentration of Ca2+\text{Ca}^{2+} will be SS. The concentration of F\text{F}^- will be the sum of F\text{F}^- from CaF2\text{CaF}_2 (2S2S) and F\text{F}^- from NaF (0.1 M0.1\text{ M}). Due to the common ion effect, the dissociation of CaF2\text{CaF}_2 is suppressed, making 2S2S negligible compared to 0.1 M0.1\text{ M}. Thus, [F]=2S+0.10.1 M[\text{F}^-] = 2S + 0.1 \approx 0.1\text{ M}. The solubility product expression is: Ksp=[Ca2+][F]2K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 Substitute the values: 5.3×1011=(S)(0.1)25.3 \times 10^{-11} = (S)(0.1)^2 5.3×1011=S×1025.3 \times 10^{-11} = S \times 10^{-2} S=5.3×1011102=5.3×109 mol L1S = \frac{5.3 \times 10^{-11}}{10^{-2}} = 5.3 \times 10^{-9}\text{ mol L}^{-1}.

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.

CHEMISTRYEquilibriumsolubilitytextcafsolution

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