Q. The solubility of $CaF_{2}$ in pure water is $2.3 \times 10^{- 6}\ mol\ dm^{- 3}$. Its solubility product will be
A) $4.8 \times 10^{- 18}$
B) $48.66 \times 10^{- 18}$
C) $4.9 \times 10^{- 11}$
D) $48.66 \times 10^{- 15}$
Explanation
(b)
$CaF_{2} \rightleftharpoons Ca^{2 +} + 2F^{-}$
$s$ $2s$
$K_{sp} = s(2s)^{2} = 4s^{3}$
$K_{sp} = 4\left( 2.3 \times 10^{- 6} \right)^{3}$
$= 48.668 \times 10^{- 18}\ \left( mol\ dm^{- 3} \right)^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.