Question

The molar solubility of $CaF _{2}\left(K_{ sp }=5.3 \times 10^{-11}\right)$ 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)

Let the solubility of $CaF _{2}$ in $0.1 \,M \,NaF$ is ' $S$' mol $L ^{-1}$
$CaF _{2}(s) \rightleftharpoons \underset{S}{Ca ^{2+}}(a q)+\underset{2 S}{2 F} ^{-}(a q)$
$NaF ( aq ) \rightleftharpoons \underset{0.1\, M}{Na ^{+}}+ \underset{0.1\,M}{F ^{-}(a q)}$
$ {\left[ F ^{-}\right] } =2 S+0.1 $
$ K_{ sp } $ of $CaF _{2} =\left[ Ca ^{2+}\right]\left[ F ^{-}\right]^{2} $
$=[S][2 S+0.1]^{2}$
$=5.3 \times 10^{-11}=[S][2 S+0.1]^{2}$
$\Rightarrow 5.3 \times 10^{-11}=[S][0.1]^{2} [\because 2 S \ll 0.1]$
$[S]=\frac{5.3 \times 10^{-11}}{(0.1)^{2}}=5.3 \times 10^{-9} \,mol\, L ^{-1}$