A) \[A{{g}_{2}}S>HgS>CuS\]
B) \[HgS>CuS>A{{g}_{2}}S\]
C) \[HgS>A{{g}_{2}}S>CuS\]
D) \[A{{g}_{2}}S>CuS>HgS\]
Correct Answer: D
Solution :
Solubility product\[({{K}_{sp}})\]of\[CuS={{10}^{-31}}\] liability of \[CuS=\sqrt{{{K}_{sp}}}=\sqrt{{{10}^{-31}}}\] \[=3.16\times {{10}^{-16}}mol/lit\] Similarly, solubility of\[A{{g}_{2}}S\] \[=\sqrt[3]{\frac{{{K}_{sp}}}{4}}\] \[(\because for\,A{{g}_{2}}S,4{{s}^{3}}={{K}_{sp}})\] \[=\sqrt{\frac{{{10}^{-42}}}{4}}\] \[=6.3\times {{10}^{-15}}mol/litre\] Solubility of\[HgS=\sqrt{{{K}_{sp}}}\] \[=\sqrt{{{10}^{-54}}}\] \[={{10}^{-27}}mol/lit\] Hence, the correct order of solubility is \[A{{g}_{2}}S>CuS>HgS\]
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