Question

The solubility of $Ca{F}_2$ in water at $298K$ is $1.7\times 10^{-3}$ $gm$ per $100c{m}^3$. The solubility product of $Ca{F}_2$ at 298 $K$ is

• A. $4.14\times 10^{-11}$
• B. $4.14\times 10^{11}$
• C. $4.14\times 10^{-6}$
• D. $4.14\times 10^6$

Answer 1

Answer: (A)

Solubility of $Ca{F}_2=1.7\times 10^{-3}gm$ per $100,c{m}^3$
$=1.7\times 10^{-3}\times \frac{1000}{100}g{L}^{-1}$
$=\frac{1.7\times 10^{-2}}{78}$
$=2.18\times 10^{-4}mol{L}^{-1}$
$Ca{F}_2\rightleftharpoons C{a}^{2+}+2F^{-}$
$K_{sp}\cdot \left[C{a}^{2+}\right]{\left[F^{-}\right]}^2$
$=\left(2.18\times 10^{-4}\right){\left(2\times 2.18\times 10^{-4}\right)}^2$
$=4.14\times 10^{-11}$