Question

A hydrogen electrode placed in a solution containing sodium acetate and acetic acid in the concentration ratios of $x:y$ and $y:x$ has electrode potential values $E_1$ and $E_2$ volts, respectively, at ${25}^{\circ }\text{C}$. The $p{K}_a$ value of acetic acid is:

• A. $\frac{-(E_1+E_2)}{2\times 0.059}$
• B. $\frac{(E_1+E_2)}{2\times 0.059}$
• C. $\frac{(E_2-E_1)}{2\times 0.059}$
• D. $\frac{-(E_1+E_2)}{0.059}$

Answer 1

The correct option is A $\frac{-(E_1+E_2)}{2\times 0.059}$

In case of a concentration cell,
$Pt|H_2|C{H}_{3}COONa|C{H}_{3}COOH||C{H}_{3}COOH|C{H}_{3}COONa|H_2|Pt$
For anode,
$E_1=-0.059log\left[H_{ox}^{+}\right]$
$E_1=0.059\left[p{H}_{ox}\right]$ ---1

For cathode,
$E_2=-0.059log\left[H_{red}^{+}\right]$
$E_2=0.059\left[p{H}_{red}\right]$ ---2
Adding 1 and 2,
$E1+E2=0.059\times [p{H}_1+p{H}_2]$
$E1+E2=0.059\times \left[\right(pKa+log\frac{sal{t}_{ox}}{aci{d}_{ox}})+(pKa+log\frac{sal{t}_{red}}{aci{d}_{red}})$
$E1+E2=0.059\times \left[\right(pKa+log\frac{x}{y})+(pKa+log\frac{y}{x})$
$E1+E2=0.059\times [2pKa+log1]$
$E1+E2=0.059\times 2\times pKa$
$pKa=\frac{E_1+E_2}{2\times 0.059}$