Question

The solubility of $ CaF_2 $ in pure water is $ 2.3 × 10^{−4}\, mol\, dm^{−3} $ . Its solubility product will be :

• A. $ 4.6 × 10^{−4} $
• B. $ 4.6 × 10^{−8} $
• C. $ 6.9 × 10^{−12} $
• D. $ 4.9 × 10^{−11} $

Answer 1

Answer: (D)

First write dissociation reaction of $CaF _{2}$ then find relation between solubility and solubility product to find correct answer.

Given, solubility of

$CaF _{2}=2.3 \times 10^{-5} \,mol \,dm ^{-3}$

$\underset{\text{moles after dissociation}}{{CaF_{2}}} \ce{<=>} \underset{x}{Ca^{2+}}+\underset{2x}{2F^{-}}$

$ K_{s p} =\left[ Ca ^{2+}\right]\left[ F ^{-}\right]^{2} $

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

$=4 x^{3} 4$

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

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