A) \[5\times {{10}^{-12}}\]
B) \[25\times {{10}^{-10}}\]
C) \[1\times {{10}^{-13}}\]
D) \[5\times {{10}^{-13}}\]
Correct Answer: D
Solution :
\[\text{Solubility}\,\underset{0.5\,\times \,{{10}^{-4}}M}{\mathop{\text{M}{{\text{X}}_{2}}}}\,\xrightarrow{{}}\underset{0.5\,\times \,{{10}^{-4}}M}{\mathop{{{\text{M}}^{2+}}}}\,+\underset{2\,\times \,0.5\,\times \,{{10}^{-4}}M}{\mathop{\text{2}{{\text{X}}^{-}}}}\,\](On 100% ionisation) \[\therefore \]\[{{K}_{sp}}\]of \[M{{X}_{2}}=[{{M}^{2+}}]{{[{{X}^{-}}]}^{2}}\] \[=(0.5\times {{10}^{-4}}){{(1.0\times {{10}^{-4}})}^{2}}\] \[=0.5\times {{10}^{-12}}=5\times {{10}^{-13}}\]
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