Question

The ionization constant of an acid, $k_a$, is the measure of strength of an acid. The $k_a$ values of acetic acid, hypochlorous acid and formic acid are $1.74\times {10}^{-5},3.0\times {10}^{-8}$ and $1.8\times {10}^{-4}$ respectively. Which of the following orders of $pH$ of $0.1mold{m}^{-3}$ solutions of these acids is correct?

• A. acetic acid > hypochlorous acid > formic acid
• B. hypochlorous acid > acetic acid > formic acid
• C. formic acid > acetic acid > hypochlorous acid
• D. formic acid > hypochlorous acid > acetic acid

Answer 1

The correct option is C formic acid > acetic acid > hypochlorous acid

We know,

$\left[H^{+}\right]=\sqrt{k_a+C}$

Where $k_a$ is acid dissociation constant and $C$ is the concentration of $H^{+}$ ion of the solution.

$\therefore$ For give same value of $C$

$\left[H^{+}\right]\propto \sqrt{k_a}$

Also $pH=-lo{g}_{10}\left[H^{+}\right]$

$pH\propto \frac{1}{log\sqrt{k_a}}$

$\therefore$ Formic acid with $k_a=1.8\times {10}^{-4}$ is the most acidic and hypochlorous acid with $k_a=3\times {10}^{-8}$ is the least acidic.

Acidic Strength:
formic acid > acetic acid > hypochlorous acid

Hence, option $\left(D\right)$ is correct.