A) \[s={{({{K}_{sp}})}^{1/3}}\]
B) \[s={{({{K}_{sp}}/2)}^{1/3}}\]
C) \[s={{({{K}_{sp}}/4)}^{1/3}}\]
D) \[s={{({{K}_{sp}}/2)}^{1/2}}\]
Correct Answer: C
Solution :
\[Ca{{F}_{2}}\underset{s}{\mathop{C{{a}^{2+}}}}\,+\underset{2}{\mathop{2}}\,{{\underset{s}{\mathop{F}}\,}^{-}}\] Let \[{{K}_{sp}}=[C{{a}^{2+}}]{{[{{F}^{-}}]}^{2}}=s\times {{(2s)}^{2}}=4{{s}^{3}}\] \[s=\sqrt[3]{\frac{{{K}_{sp}}}{4}}\] or \[s={{\left( \frac{{{K}_{sp}}}{4} \right)}^{1/3}}\]
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