Question

If the solubility of $Ca{F}_2$ in water at ${18}^{\circ }C$ is $2.05\times {10}^{-4}mol{L}^{-1}$. Calculate its solubility product at the given temperature.

• A. $3.45\times {10}^{-11}$
• B. $4.45\times {10}^{-12}$
• C. $5.45\times {10}^{-10}$
• D. $6.66\times {10}^{-9}$

Answer 1

The correct option is A $3.45\times {10}^{-11}$

$Ca{F}_2\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right)Initial:C00Equilibrium:C-ss2s$

$K_{sp}=\left[C{a}^{2+}\right][F^{-}{]}^2$

$K_{sp}=s\times (2s{)}^2=4s^3$

$K_{sp}=4s^3=4\times (2.05\times {10}^{-4}{)}^3$

$K_{sp}=3.45\times {10}^{-11}$