Question

The molar solubility of CaF$_2$ (K$_{sp}$ = 5.3 × 10$^{-11}$) in 0.1 M solution of NaF will be

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

Answer 1

The Correct Option is C

${\underset{(a-s')}{CaF_2(s)} <=> \underset{s'}{Ca^{+2}(aq)} + \underset{2s'}{2F^- (aq)}}$
${\underset{\overset{C}{0}}{NaF(aq)} -> \underset{\overset{0}{C}}{Na^+ (aq)} + \underset{\overset{0}{C}}{F^- (aq)}}$
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}$