A) \[4.6\times {{10}^{-4}}\]
B) \[4.6\times {{10}^{-8}}\]
C) \[6.9\times {{10}^{-12}}\]
D) \[4.9\times {{10}^{-11}}\]
Correct Answer: D
Solution :
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 \[Ca{{F}_{2}}=2.3\times {{10}^{-4}}mol\,d{{m}^{-3}}\] \[\underset{\begin{smallmatrix} moles\text{ }after \\ dissociation \end{smallmatrix}}{\mathop{Ca{{F}_{2}}}}\,\underset{x}{\mathop{C{{a}^{2+}}}}\,+\underset{2x}{\mathop{2{{F}^{-}}}}\,\] \[{{K}_{sp}}=[C{{a}^{2+}}]{{[{{F}^{-}}]}^{2}}\] \[=x\times {{(2x)}^{2}}\] \[=4{{x}^{3}}\] \[\therefore \] \[{{K}_{sp}}=4\times {{(2.3\times {{10}^{-4}})}^{3}}\] \[=4.9\times {{10}^{-11}}\]
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