Question

A hydrogen electrode placed in a buffer solution of $C{H}_{3}COONa$ and acetic acid in the ratios $x:y$ and $y:x$ has oxidation potential values $E_{1}V$ and $E_{2}V$ respectively at ${25}^{\circ }C.$ The $p{K}_a$ value of acetic acid would be:
(Pressure of $H_2$ is $1atm$)

• 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. $\frac{E_1-E_2}{0.118}$

Answer 1

The correct option is A $\frac{E_1+E_2}{0.118}$

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$
$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$
$\therefore \left[H^{+}\right]=\frac{K_a\left[C{H}_{3}COOH\right]}{\left[C{H}_{3}CO{O}^{-}\right]}$

For the first cell, $\left[H^{+}\right]=\frac{K_a\times y}{x}$
$H_2\to 2H^{+}+2e^{-}$
According to Nernst equation at ${25}^{\circ }C$,
$E_1=E_{H_2/H^{+}}=-\frac{0.0591}{2}\log {\left(\frac{y{K}_a}{x}\right)}^2$
$\therefore E_1=-0.0591\log \left(\frac{y{K}_a}{x}\right)$

For the second cell, $\left[H^{+}\right]=\frac{K_a\times x}{y}$
$H_2\to 2H^{+}+2e^{-}$
According to Nernst equation at ${25}^{\circ }C$,
$E_2=E_{H_2|H^{+}}=-\frac{0.0591}{2}\log {\left(\frac{x{K}_a}{y}\right)}^2$
$\therefore E_2=-0.0591\log \left(\frac{x{K}_a}{y}\right)$

$E_1+E_2=-0.0591\log K_a^2$
$E_1+E_2=-0.118\log K_a$
$\therefore p{K}_a=\frac{E_1+E_2}{0.118}$