A) \[B{{i}_{2}}{{S}_{3}}({{K}_{sp}}=1\times {{10}^{-17}})\]
B) \[MnS({{K}_{sp}}=7\times {{10}^{-16}})\]
C) \[CuS({{K}_{sp}}=8\times {{10}^{-37}})\]
D) \[A{{g}_{2}}S({{K}_{sp}}=6\times {{10}^{-51}})\]
Correct Answer: A
Solution :
\[B{{i}_{2}}{{S}_{3}}\underset{2s}{\mathop{2B{{i}^{3+}}}}\,+\underset{3s}{\mathop{3{{s}^{2-}}}}\,\] \[{{K}_{sp}}={{(2s)}^{2}}(3{{s}^{3}})=108{{s}^{5}}\] \[108{{s}^{5}}=1\times {{10}^{-17}}\] \[s=\left( \frac{1\times {{10}^{-17}}}{108} \right)=1.56\times {{10}^{-4}}\] \[MnS\underset{s}{\mathop{M{{n}^{2+}}}}\,+\underset{s}{\mathop{{{S}^{2-}}}}\,\] \[{{K}_{sp}}={{s}^{2}}\] \[s=\sqrt{{{K}_{sp}}}=\sqrt{7\times {{10}^{-16}}}\] \[=2.64\times {{10}^{-8}}\] \[CuS\underset{s}{\mathop{C{{u}^{2+}}}}\,+\underset{s}{\mathop{{{S}^{2-}}}}\,\] \[{{K}_{sp}}={{s}^{2}}\] \[8\times {{10}^{-37}}={{s}^{2}}\] \[s=\sqrt{8\times {{10}^{-37}}}\] \[=0.89\times {{10}^{-18}}\] \[A{{g}_{2}}S\underset{2s}{\mathop{2A{{g}^{+}}}}\,+\underset{s}{\mathop{{{S}^{2-}}}}\,\] \[{{K}_{sp}}=4{{s}^{3}}\] \[4{{s}^{3}}=6\times {{10}^{-51}}\] \[s=\sqrt[3]{\frac{6\times {{10}^{-51}}}{4}}\] \[=\sqrt[3]{1.5\times {{10}^{-51}}}=1.14\times {{10}^{-17}}\] The solubility of\[B{{i}_{2}}{{S}_{3}}\]is maximum. Hence, it is the most stable.
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