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  4. A saturated solution of CaF2 is 2× 10-4mol/L Its solubility product constant is:

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Q. A saturated solution of $ Ca{{F}_{2}} $ is $ 2\times {{10}^{-4}}mol/L $ Its solubility product constant is:

KEAMKEAM 2003

Solution:

Solubility of $ Ca{{F}_{2}}=2\times {{10}^{-4}}mol\,{{L}^{-1}} $ Each mole of $ Ca{{F}_{2}} $ dissolving in $ {{H}_{2}}O $ gives one mole of $ C{{a}^{2+}} $ and two moles of $ {{F}^{-}} $ ions. $ Ca{{F}_{2}}C{{a}^{2+}}+2{{F}^{-}} $ $ 2\times {{10}^{-4}}M.2\times 2\times {{10}^{-4}}M $ $ {{K}_{sp}}=[C{{a}^{2+}}]{{[{{F}^{-}}]}^{2}} $ $ =[2\times {{10}^{-4}}]{{[2\times 2\times {{10}^{-4}}]}^{2}} $ $ {{K}_{sp}}=3.2\times {{10}^{-11}} $