A) \[5.0\times {{10}^{-9}}M\]
B) \[2.0\times {{10}^{-5}}M\]
C) \[2.2\times {{10}^{-4}}M\]
D) \[1.0\times {{10}^{-4}}M\]
Correct Answer: A
Solution :
In\[CaC{{l}_{2}}\],\[[C{{l}^{-}}]=2\times 0.004=0.008mol\,\,{{L}^{-1}}\]\[[C{{l}^{-}}]\] from \[AgCl\] is too small and is neglected \[{{k}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]\] \[4\times {{10}^{-10}}=[A{{g}^{+}}]\times 0.08\] \[[A{{g}^{+}}]=\frac{4\times {{10}^{-10}}}{0.08}\] \[=\frac{4}{8}\times {{10}^{-8}}\] \[=5.0\times {{10}^{-9}}M\]
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