Question

A hydrogen electrode is placed in a buffer solution of acetic acid and sodium acetate in the ratio $y:x$ and $x:y$ has electrode potential values $E_1$ and $E_2$ volts respectively. $p{K}_a$ value for 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)

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

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{⊝}+H^{\oplus }$

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

The reaction occurring at the hydrogen electrode.

$H^{+}+e^{-}⟶1/2H_2$

The equation in the first case: $E_1=-\frac{0.059}{1}\log \frac{1}{\left[H^{+}\right]}$

$=0.059\log \left[H^{+}\right]=0.059\log (K_a\times \frac{y}{x})$

Similarly, $E_2=0.059\log \left[K_a\left(\frac{x}{y}\right)\right]$

Now, $E_1+E_2=0.059[\log K_a\times \frac{y}{x}+\log K_a\frac{x}{y}]$

$=0.059\log {K_a}^2=0.018\log K_a$

$\therefore \log K_a=-\frac{E_1+E_2}{0.118}$

$\Rightarrow p{K}_a=\frac{E_1+E_2}{0.118}$