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 :
\[B{{i}_{2}}{{S}_{3}}{{\underset{2s}{\mathop{2Bi}}\,}^{3+}}+\underset{3s}{\mathop{3{{s}^{2-}}}}\,\] \[{{K}_{sp}}={{(2S)}^{2}}{{(3S)}^{3}}=108\,{{S}^{5}}\] \[108\,{{S}^{5}}=1\times {{10}^{-17}}\] \[S={{\left( \frac{1\times {{10}^{-17}}}{108} \right)}^{{\scriptstyle{}^{1}/{}_{2}}}}=1.56\times {{10}^{-4}}\] \[MnS\underset{s}{\mathop{M{{n}^{2+}}}}\,+\underset{s}{\mathop{{{S}^{2-}}}}\,\] \[{{K}_{SP}}={{S}^{2}}\] \[S=\sqrt{{{K}_{SP}}}=\sqrt{8\times {{10}^{-37}}}=2.64\times {{10}^{-8}},\] \[CuS\underset{s}{\mathop{C{{u}^{2+}}}}\,+\underset{s}{\mathop{{{S}^{2-}}}}\,\] \[{{K}_{SP}}={{S}^{2}}\] \[S=\sqrt{{{K}_{SP}}}=\sqrt{8\times {{10}^{-37}}}=0.87\times {{10}^{-18}}\] \[A{{g}_{2}}S\underset{2s}{\mathop{2A{{g}^{+}}}}\,+\underset{s}{\mathop{{{S}^{2-}}}}\,\] \[{{K}_{SP}}=4{{S}^{3}}\] \[4{{S}^{3}}=6\times {{10}^{-51}}\] \[S=\sqrt[3]{\frac{6\times {{10}^{-51}}}{4}}=1.14\times {{10}^{-17}}\] The solubility of \[B{{i}_{2}}{{S}_{3}}\] is maximum. Hence, it is the most soluble.
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