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)

Key Idea: First write dissociation reaction of $Ca{F}_2$ then find relation between solubility and solubility product to find correct answer. Given, solubility of $Ca{F}_2=2.3\times {10}^{-4}mold{m}^{-3}$ $\underset{\begin{array}{c}molesafter\\ dissociation\end{array}}{Ca{F}_2}\underset{x}{C{a}^{2+}}+\underset{2x}{2F^{-}}$ $K_{sp}=[C{a}^{2+}]{[F^{-}]}^2$ $=x\times {(2x)}^2$ $=4x^3$ $\therefore$ $K_{sp}=4\times {(2.3\times {10}^{-4})}^3$ $=4.9\times {10}^{-11}$