A) \[[A{{g}^{+}}]=1.8\times {{10}^{-7}}M;[P{{b}^{2+}}]=1.7\times {{10}^{-6}}M\]
B) \[[A{{g}^{+}}]=1.8\times {{10}^{-11}}M;[P{{b}^{2+}}]=8.5\times {{10}^{-5}}M\]
C) \[[A{{g}^{+}}]=1.8\times {{10}^{-9}}M;[P{{b}^{2+}}]=1.7\times {{10}^{-3}}M\]
D) \[[A{{g}^{+}}]=1.8\times {{10}^{-11}}M;[P{{b}^{2+}}]=8.5\times {{10}^{-4}}M\]
Correct Answer: C
Solution :
\[{{K}_{sp}}\] for \[AgCl=[A{{g}^{+}}][C{{l}^{-}}]\] \[\therefore \] \[[A{{g}^{+}}]=\frac{1.8\times {{10}^{-10}}}{{{10}^{-1}}}\] \[=1.8\times {{10}^{-9}}M.\] \[{{K}_{sp}}\]for\[PbC{{l}_{2}}=[P{{b}^{2}}^{+}]{{[C{{l}^{-}}]}^{2}}\] \[\therefore \] \[[P{{b}^{2+}}]=\frac{1.7\times {{10}^{-5}}}{{{10}^{-1}}\times {{10}^{-1}}}\] \[=1.7\times {{10}^{-3}}M\]You need to login to perform this action.
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