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A hydrogen electrode placed in a buffer solution of $CH_3COONa$ and acetic acid in the ration's x:y and y:x has electrode potential values $E_1$ volt and $E_2$ volt at $25^{\circ}$ C. The pk_a value of acetic acid is ( $E_1 \:and \:E_2$ are oxidation potential.)

Option 1)
$\frac{E_1 +E_2}{0.118}$

Option 2)
$\frac{E_2+E_1}{0.118}$

Option 3)
$-\frac{E_1+E_2}{0.118}$

Option 4)
$\frac{E_1 -E_2}{0.118}$

Answers (1)

As we learnt in

M(s) is Solid -

[M] = 1

- wherein

$E_{M^{n+}/M}=E_{M^{n+}/M}^{0}-\frac{RT}{nf}ln\frac{1}{[M^{n+}]}$

By Nernst equation, we have

$E_{M^{n+}/M}=E^{0}_{M^{n+}/M}-\frac{RT}{nF}\, \, ln \frac{[M]}{[M^{n+}]}$

The reaction undergoes as follows

$\underset{y}{CH_{3}COOH}\, \, \rightleftharpoons\underset{x}{CH_{3}COOH^{-}} \, \,+\, \, H^{+}$

$E_{1}=E_{1}^{0}-\frac{0.0592}{n}\, \, log\left [ H^{+} \right ]= 0.0592\, \, log\, \, \frac{1}{\left [ H^{+} \right ]}_{1}\ \: \: \: ............(1)$

Similarly $E_{2}=0.0592\, \, log\, \, \frac{1}{\left [ H^{+} \right ]}_{2}\ \: \: \: \: ..........(2)$

Also for the same reaction we have

$Ka=\frac{x\left [ H^{+} \right ]_{1}}{y}$

$pKa=-log\, \, \frac{x\left [ H^{+} \right ]_{1}}{y}=log\, \, \left ( \frac{y}{x\left [ H^{+} \right ]_{1}} \right )$

$=log_{y}-log_{x}-log\left [ H^{+} \right ]_{1}$

$\Rightarrow log\frac{1}{\left [ H^{+} \right ]1}=pKa+log\frac{x}{y}\ \: \: \: \: ...........(3)$

Similarly $log\frac{1}{\left [ H^{+} \right ]_{2}}=pKa+log\frac{y}{x}\ \: \: \: \: \: \: .........(4)$

Adding (1) & (2) and (3) & (4)

we get $E_{1}+E_{2}=0.0592\times 2\times pKa$

$pKa=\frac{E_{1}+E_{2}}{0.118}$

Option 1)

$\frac{E_1 +E_2}{0.118}$

Correct

Option 2)

$\frac{E_2+E_1}{0.118}$

Incorrect

Option 3)

$-\frac{E_1+E_2}{0.118}$

Incorrect

Option 4)

$\frac{E_1 -E_2}{0.118}$

Incorrect

Posted by

Vakul

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