A) \[4.0\times {{10}^{-10}}M\]
B) \[1.6\times {{10}^{-4}}M\]
C) \[1.0\times {{10}^{-4}}M\]
D) \[2.0\times {{10}^{-6}}M\]
Correct Answer: C
Solution :
\[M{{X}_{2}}{{\underset{s}{\mathop{M}}\,}^{2+}}+\underset{2s}{\mathop{2{{X}^{-}}}}\,\] \[{{K}_{sp}}=[{{M}^{2+}}]{{[{{X}^{-}}]}^{2}}\] If solubility be s then \[{{K}_{sp}}=(s){{(2s)}^{2}}=4{{s}^{3}}\] \[4{{s}^{2}}=4\times {{10}^{-12}}\] \[\therefore \] \[s=1\times {{10}^{-4}}M\] \[\therefore \] \[{{M}^{2+}}=s=1\times {{10}^{-4}}M\]
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