A) \[\frac{{{E}_{1}}+{{E}_{2}}}{0.118}\]
B) \[\frac{{{E}_{2}}-{{E}_{1}}}{0.118}\]
C) \[-\frac{{{E}_{1}}+{{E}_{2}}}{0.118}\]
D) \[\frac{{{E}_{1}}-{{E}_{2}}}{0.118}\]
Correct Answer: A
Solution :
| [a] \[{{E}_{1}}={{E}^{O}}-\frac{0.059}{1}\log {{\left[ {{H}^{+}} \right]}_{1}}\] |
| \[{{E}_{2}}={{E}^{O}}-\frac{0.059}{1}\log {{\left[ {{H}^{+}} \right]}_{2}}\] |
| on adding (also \[E_{H}^{o}=0\]) |
| \[{{E}_{1}}+{{E}_{2}}=-\frac{0.059}{1}\left[ \log {{\left[ {{H}^{+}} \right]}_{1}}+\log {{\left[ {{H}^{+}} \right]}_{2}} \right]\] |
| Now for \[C{{H}_{3}}COOHC{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}\] |
| \[\left[ {{H}^{+}} \right]=\frac{{{K}_{a}}\left[ C{{H}_{3}}COOH \right]}{\left[ C{{H}_{3}}CO{{O}^{-}} \right]}\] |
| \[\therefore \] \[{{\left[ {{H}^{+}} \right]}_{1}}={{K}_{a}}\frac{y}{x}\] |
| \[\,{{\left[ {{H}^{+}} \right]}_{2}}={{K}_{a}}\frac{x}{y}\] |
| \[\therefore \]\[{{E}_{1}}+{{E}_{2}}=-\frac{0.059}{1}\left[ \log \frac{{{K}_{a}}y}{x}+\log \frac{{{k}_{a}}x}{y} \right]\] |
| \[=-0.059\left[ 2\log {{K}_{a}} \right]\] |
| \[\log {{K}_{a}}=\frac{{{E}_{1}}+{{E}_{2}}}{2\times \left( -0.059 \right)}\] |
| \[\log {{K}_{a}}=-\frac{{{E}_{1}}+{{E}_{2}}}{0.118}\] |
| or \[p{{K}_{a}}=\frac{{{E}_{1}}+{{E}_{2}}}{0.118}\] |
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