A) \[1\times {{10}^{-9.8}}\]
B) \[1\times {{10}^{-19.6}}\]
C) \[2\times {{10}^{-10}}\]
D) \[2.64\times {{10}^{-14}}\]
Correct Answer: A
Solution :
[a] \[2A{{g}^{+}}+{{H}_{2}}\xrightarrow{{}}2{{H}^{+}}+2Ag\] \[E=E{}^\circ -\frac{0.0591}{2}\log \frac{{{[{{H}^{+}}]}^{2}}}{{{P}_{{{H}_{2}}}}\times {{[A{{g}^{+}}]}^{2}}}\] \[0.222=0.7995-\frac{0.0591}{2}log\frac{1}{{{[A{{g}^{+}}]}^{2}}}\] \[[A{{g}^{+}}]={{10}^{-9.8}}\] \[{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]=({{10}^{-9.8}})\times (1)={{10}^{-9.8}}\]You need to login to perform this action.
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