Solubility product of a sparingly soluble salt MF2 is 9×10−18M3.At the same temperature, solublility of the salt MF2 in 0.3 M KF solution will be

MF2⇌M+2s+2F−2sKF→K+0.3 M+F−0.3 MKspMF2=M+2F−2 =s2s+0.322s+0.3≈0.3∵s is too smallKsp=s0.32s=Ksp0.3×0.3=9×10−180.3×0.3=10−16 M

Solubility product of a sparingly soluble salt MF2 is 9×10−18M3.At the same temperature, solublility of the salt MF2 in 0.3 M KF solution will be

MF2⇌M+2s+2F−2sKF→K+0.3 M+F−0.3 MKspMF2=M+2F−2 =s2s+0.322s+0.3≈0.3∵s is too smallKsp=s0.32s=Ksp0.3×0.3=9×10−180.3×0.3=10−16 M

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Solubility product of a sparingly soluble salt $({MF}_{2})$ is $9 \times 10^{- 18} M^{3}$ .At the same temperature, solublility of the salt $({MF}_{2})$ in 0.3 M KF solution will be

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$10^{- 6} M$

$3 \times 10^{- 15} M$

$10^{-16} M$

$3 \times 10^{- 12} M$

answer is C.

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Detailed Solution

$\begin{array}{l} {MF}_{2} ⇌ \underset{s}{M^{+ 2}} + \underset{2 s}{2 F^{-}} \\ KF \to \underset{0 .3 M}{K^{+}} + \underset{0 .3 M}{F^{-}} \\ K_{sp} ({MF}_{2}) = [M^{+ 2}] {[F^{-}]}^{2} \\ = s {(2 s + 0 .3)}^{2} \\ (2 s + 0 .3) \approx 0 .3 (∵ s is too small) \\ K_{sp} = s {(0 .3)}^{2} \\ s = \frac{K_{sp}}{0 .3 \times 0 .3} = \frac{9 \times 10^{- 18}}{0 .3 \times 0 .3} = 10^{- 16} M \end{array}$