Question

The molar solubility of $CaF_2$ (K$_{sp} = 5.3 \times 10^{-11}$) in $0.1\, M$ solution of $NaF$ will be

• A. $ 5.3 \times 10^{-11} mol L ^{-1}$
• B. $5.3 \times 10^{-8} mol L ^{-1}$
• C. $ 5.3 \times 10^{-9} mol L ^{-1}$
• D. $5.3 \times 10^{-10} mol L ^{-1}$

Answer 1

Answer: (C)

$\underset{(a-s')}{CaF_2(s)} \rightleftharpoons \underset{s'}{Ca^{+2}(aq)} + \underset{2s'}{2F^- (aq)}$
$\underset{\overset{C}{0}}{NaF(aq)} \rightarrow \underset{\overset{0}{C}}{Na^+ (aq)} + \underset{\overset{0}{C}}{F^- (aq)}$
In solution- $[F^-] = (2s' + C)$
$[F^-] \approx C$ (due to common ion effect)
$K_{sp(caF_2)} = [Ca^{+2}] . [F^-]^2$
$K_{sp(CaF_2)} = s' . C^2$
$s' = \frac{5.3 \times 10^{-11}}{(10^{-1})^2}$
$ s' = 5.3 \times 10^{-9} \,mol\,L^{-1}$