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 $3.4\times {10}^{-9}M$

$Ca{F}_2\rightleftharpoons C{a}^{2+}+2F^{-}x2xNaF\to \underset{0.1}{N{a}^{+}}+\underset{0.1}{F^{-}}\left[C{a}^{2+}\right]=x\left[F^{-}\right]=[2x+0.1]\approx 0.1M{K}_{sp}=\left[C{a}^{2+}\right][F^{-}{]}^2\Rightarrow 3.4\times {10}^{-11}=x(0.1{)}^2\Rightarrow x=3.4\times {10}^{-9}M$