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NEET CHEMISTRYMedium

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

A

5.3×10−11 mol L−15.3 \times 10^{-11}\text{ mol L}^{-1}

B

5.3×10−8 mol L−15.3 \times 10^{-8}\text{ mol L}^{-1}

C

5.3×10−9 mol L−15.3 \times 10^{-9}\text{ mol L}^{-1}

D

5.3×10−10 mol L−15.3 \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+0.1≈0.1 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: 5.3×10−11=(S)(0.1)25.3 \times 10^{-11} = (S)(0.1)^2 5.3×10−11=S×10−25.3 \times 10^{-11} = S \times 10^{-2} S=5.3×10−1110−2=5.3×10−9 mol L−1S = \frac{5.3 \times 10^{-11}}{10^{-2}} = 5.3 \times 10^{-9}\text{ mol L}^{-1}.

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