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 pka 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)
V Vakul

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

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