A) \[1.7\times {{10}^{-6}}\]
B) \[1.7\times {{10}^{-16}}\]
C) \[1.7\times {{10}^{-18}}\]
D) \[1.7\times {{10}^{-12}}\]
Correct Answer: B
Solution :
Solubility of \[{{M}_{2}}S\]salt is \[3.5\times {{10}^{-6}}M\] \[\underset{\text{Solubility}\,3.5\times {{10}^{-6}}M}{\mathop{{{M}_{2}}S}}\,\rightleftharpoons \underset{2\times 3.5\times {{10}^{-6}}M}{\mathop{2{{M}^{+}}}}\,+\underset{3.5\times {{10}^{-6}}M}{\mathop{{{S}^{2-}}}}\,\](On 100% ionisation) \[\therefore \]\[{{K}_{sp}}\](Solubility product of\[{{M}_{2}}S\])\[={{[{{M}^{+}}]}^{2}}[{{S}^{2-}}]\] \[=(7.0\times {{10}^{-6}})2(3.5\times {{10}^{-6}})=171.5\times {{10}^{-16}}\]
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