Question

A hydrogen electrode placed in a buffer solution of $C{H}_{3}COONa$ and $C{H}_{3}COOH$ in the ratios $x:y$ and $y:x$ has electrode potential values $E_1$ volts and $E_2$ volts respectively at 25${}^0$C . The $p{K}_a$ value of acetic acid is :

[$E_1$and $E_2$ are oxidation potentials]

• 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

Answer: (A)

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

The expression for the oxidation potential for hydrogen electrode is :
$E_1=E_1^0-\frac{0.0592}{n}log{\left[H^{+}\right]}_1=0.0592log\frac{1}{{\left[H^{+}\right]}_1}$
$E_2=E_2^0-\frac{0.0592}{n}log{\left[H^{+}\right]}_2=0.0592log\frac{1}{{\left[H^{+}\right]}_2}$
But, ${pK}_a=-log{K}_a=log\frac{y}{x{\left[H^{+}\right]}_1}=log\frac{y}{x}+log\frac{1}{{\left[H^{+}\right]}_1}$
Hence, $log\frac{1}{{\left[H^{+}\right]}_1}={pK}_a+log\frac{x}{y}$
Similarly, $log\frac{1}{{\left[H^{+}\right]}_2}={pK}_a+log\frac{y}{x}$
Hence, $E_1+E_2=0.0592\times 2\times {pK}_a$
${pK}_a=\frac{E_1+E_2}{0.118}$

Hence, option A is correct.