A) 3
B) 4
C) 5
D) 6
Correct Answer: B
Solution :
| [b] \[{{H}_{2}}\xrightarrow{{}}2{{H}^{+}}+2{{e}^{-}}\] |
| \[\frac{H{{g}_{2}}C{{l}_{2}}+2{{e}^{-}}\xrightarrow{{}}2Hg+2C{{l}^{-}}}{{{H}_{2}}+H{{g}_{2}}C{{l}_{2}}\xrightarrow{{}}2{{H}^{+}}+2C{{l}^{-}}+2Hg}\] |
| \[{{E}_{initial}}=E{}^\circ +\frac{0.059}{2}\log \frac{1}{{{[{{H}^{+}}]}^{2}}\,{{[C{{l}^{-}}]}^{2}}}=0.6\] |
| \[{{E}_{\text{final}}}=E{}^\circ +\frac{0.059}{2}\log \frac{1}{{{({{10}^{-7}})}^{2}}\,{{[C{{l}^{-}}]}^{2}}}=0.777\] |
| subtracting we get |
| \[0.177=0.059\log \frac{[{{H}^{+}}]}{{{10}^{-7}}}\] |
| \[\Rightarrow \] \[\frac{[{{H}^{+}}]}{{{10}^{-7}}}=1000\] |
| \[\Rightarrow \] \[[{{H}^{+}}]={{10}^{-4}}\]\[\Rightarrow \,\,\,\,pH=4.\] |
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