Question

Find solubility product of a saturated solution of $ A{{g}_{2}}Cr{{O}_{4}} $ in water at 298 K, if the emf of the cell. $ Ag|A{{g}^{+}} $ (saturated $ A{{g}_{2}}Cr{{O}_{4}} $ solution) || $ A{{g}^{+}}(0.1M)|Ag $ is $ 0.164\,V $ at $ 298\,K. $

• 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}} $

Answer 1

Answer: (A)

$ {{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}} $