A) \[2.0\times {{10}^{-4}}mol/L\]
B) \[2\times {{10}^{-3}}mol/L\]
C) \[4.0\times {{10}^{-4}}mol/L\]
D) \[8\times {{10}^{-3}}mol/L\]
Correct Answer: A
Solution :
Sol. \[Ca{{F}_{2}}\xrightarrow{{}}\underset{s}{\mathop{C{{a}^{2+}}}}\,+\underset{s}{\mathop{2{{F}^{-}}}}\,\] \[{{K}_{sp}}=(s){{(2s)}^{2}}-4{{s}^{3}}\] \[S={{\left( \frac{{{K}_{sp}}}{4} \right)}^{1/3}}={{\left( \frac{3.2\times {{10}^{-11}}}{4} \right)}^{1/3}}\] \[=2\times {{10}^{-4}}mol/L\]
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