Question

The solubility of $Ca{F}_2(K_{sp}=3.4\times {10}^{-11})$ in $0.1M$ solution of $NaF$ would be

• A. $3.4\times {10}^{-12}M$
• B. $3.4\times {10}^{-10}M$
• C. $3.4\times {10}^{-9}M$
• D. $3.4\times {10}^{-13}M$

Answer 1

The correct option is C $3.4\times {10}^{-9}M$

$Ca{F}_2\to C{a}^{+2}+2F^{-}$

$S$ $0.1$

$Ca{F}_2:K_{sp}=3.4\times {10}^{-11}=\left[C{a}^{+2}\right][F^{-}{]}^2$

$S=\frac{3.4\times {10}^{-11}}{(0.1{)}^2}=3.4\times {10}^{-9}$