Question

The solubility of $Ca{F}_2$ in pure water is $2.3\times 10^{-4}mold{m}^{-3}$ . Its solubility product will be :

• A. $4.6\times 10^{-4}$
• B. $4.6\times 10^{-8}$
• C. $6.9\times 10^{-12}$
• D. $4.9\times 10^{-11}$

Answer 1

Answer: (D)

First write dissociation reaction of $Ca{F}_2$ then find relation between solubility and solubility product to find correct answer.

Given, solubility of

$Ca{F}_2=2.3\times 10^{-5}mold{m}^{-3}$

$\underset{\text{moles after dissociation}}{Ca{F}_2}\rightleftharpoons \underset{x}{C{a}^{2+}}+\underset{2x}{2F^{-}}$

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

$=x\times (2x{)}^2$

$=4x^{3}4$

$\therefore K_{sp}-4\times {\left(2.3\times 10^{-4}\right)}^3$

$=4.9\times 10^{-11}$