Question

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} $

Answer 1

Answer: (B)

$CaF_2 \qquad$ $\underset{\text{s}}{\ce{Ca^{2+}}}$ $+\underset{\text{2s}}{\ce{2F^- }}$
$K_{SP} = s(2s)^2 = 4s^3 $
$K_{SP} = 4 (2.3 \times 10^{-6})^3$
$ 48.668 \times 10^{-18} (mol\,dm^{-3})^3$