Question

The solubility of $ Ca{{F}_{2}} $ in pure water is $ 2.3\times {{10}^{-4}}mol\text{ }d{{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}}mol\,d{{m}^{-3}} $ $ \underset{\begin{smallmatrix} moles\text{ }after \\ dissociation \end{smallmatrix}}{\mathop{Ca{{F}_{2}}}}\,\underset{x}{\mathop{C{{a}^{2+}}}}\,+\underset{2x}{\mathop{2{{F}^{-}}}}\, $ $ {{K}_{sp}}=[C{{a}^{2+}}]{{[{{F}^{-}}]}^{2}} $ $ =x\times {{(2x)}^{2}} $ $ =4{{x}^{3}} $ $ \therefore $ $ {{K}_{sp}}=4\times {{(2.3\times {{10}^{-4}})}^{3}} $ $ =4.9\times {{10}^{-11}} $