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 \[3.5\times {{10}^{-6}}\,M\] \[\underset{3.5\times {{10}^{-6}}\,M}{\mathop{{{M}_{2}}S}}\,\rightleftharpoons \,\underset{2\times 3.5\times {{10}^{-6}}\,M}{\mathop{2{{M}^{+}}}}\,+\underset{3.5\times {{10}^{-6}}\,M}{\mathop{{{S}^{2}}}}\,\] on 100% ionisation \[{{K}_{sp}}\] (Solubility product of \[{{M}_{2}}S\]) \[{{[{{M}^{+}}]}^{2}}[S]\] \[={{(7.0\times {{10}^{-6}})}^{2}}(3.5\times {{10}^{-6}})\] \[=171.5\times {{10}^{-18}}\] \[=1.71\times {{10}^{-16}}\]
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