A) \[{{K}_{sp}}={{S}^{2}}\]
B) \[{{K}_{sp}}=4{{S}^{3}}\]
C) \[{{K}_{sp}}=108{{S}^{5}}\]
D) \[{{K}_{sp}}=S\]
Correct Answer: B
Solution :
\[PbC{{l}_{2}}\rightleftharpoons P{{b}^{2+}}+2C{{l}^{-}}\] If solubility of \[PbC{{l}_{2}}\] is S. Then, solubility of \[P{{b}^{2+}}=S\] and solubility of \[C{{l}^{-}}=2S\] On applying law of mass action \[K=\frac{[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}}{[PbC{{l}_{2}}]}\] \[K[PbC{{l}_{2}}]=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}\] \[\Rightarrow \] \[{{K}_{sp}}=S\times {{(2S)}^{2}}\] \[\therefore \]\[{{K}_{sp}}=4{{S}^{3}}\]
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