Question

The solubility of a saturated solution of calcium fluoride is $2 \times 10^{-4}$ moles per litre. Its solubility product is

• A. $12 \times 10^{-2}$
• B. $14 \times 10^{-4}$
• C. $22 \times 10^{-11}$
• D. $32 \times 10^{-12}$

Answer 1

Answer: (D)

The solubility (s) of saturated solution of $CaF _{2}$ is $2 \times 10^{-4} M$.
$\underset{s}{CaF _{2}} \rightleftharpoons \underset{s}{Ca ^{2+}}+\underset{s}{2 F ^{-}}$
$\therefore$ Solubility product, $K _{ sp }=\left[ Ca ^{2+}\right]\left[ F ^{-}\right]^{2}$
$= S \cdot(2 S )^{2}=4 S ^{3}$
$=4 \times\left(2 \times 10^{-4}\right)^{3}$
$=32 \times 10^{-12}$