Equilibrium - Result Question 65
####68. The molar solubility of $CaF_2(K _{s p}=5.3 \times 10^{-11})$ in $0.1 M$ solution of $NaF$ will be
[NEET Odisha 2019]
(a) $5.3 \times 10^{-10} mol L^{-1}$
(b) $5.3 \times 10^{-11} mol L^{-1}$
(c) $5.3 \times 10^{-8} mol L^{-1}$
(d) $5.3 \times 10^{-9} mol L^{-1}$
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Solution:
(d) $CaF_2 \rightarrow Ca^{2+}+2 F^{-}$ $t=0 \quad 0 \quad 0$
At eqm. $s \quad 2 s$
$NaF \longrightarrow Na^{+}+F^{-}$
$\begin{matrix} 0.1 & 0.1 & 0.1\end{matrix} $
Due to common ion effect of $NaF$, solubility of $CaF_2$ is further supressed. Therefore, the concentration of $F^{-}$will be mainly due to $NaF$.
$K _{s p}=[Ca^{2+}][F^{-}]_2$
$K _{s p}^{s p}=(s)(0.1+2 s)^{2} \quad 0.1+2 s \approx 0.1$
$s=\frac{K _{sp}}{(0.1)^{2}}=\frac{5.3 \times 10^{-11}}{(0.1)^{2}}=5.3 \times 10^{-9} mol L^{-1}$
