A) \[2.287\times {{10}^{-12}}mo{{l}^{3}}{{L}^{-3}}\]
B) \[2.287\times {{10}^{-13}}mo{{l}^{3}}{{L}^{-3}}\]
C) \[2.287\times {{10}^{-14}}mo{{l}^{3}}{{L}^{-3}}\]
D) \[2.287\times {{10}^{-15}}mo{{l}^{3}}{{L}^{-3}}\]
Correct Answer: A
Solution :
\[{{E}_{cell}}=\frac{0.059}{1}\log \frac{[A{{g}^{+}}]RHS}{[A{{g}^{+}}]\,LHS}\] \[=0.164=\frac{0.059}{1}\log \frac{0.1}{[\text{A}{{\text{g}}^{+}}]\,\text{LHS}}\] \[\text{A}{{\text{g}}^{+}}(LHS)=1.66\times {{10}^{-4}}M\] so, \[[CrO_{4}^{2-}]=\frac{1.66\times {{10}^{-4}}}{2}\] \[{{K}_{sp}}(\text{A}{{\text{g}}_{\text{2}}}\text{Cr}{{\text{O}}_{\text{4}}})={{[A{{g}^{+}}]}^{2}}\,[\text{CrO}_{4}^{2-}]\] \[={{(1.66\times {{10}^{-4}})}^{2}}\left( \frac{1.66\times {{10}^{-4}}}{2} \right)\] \[=2.287\times {{10}^{-12}}mo{{l}^{3}}{{L}^{-3}}\]
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