JEE PYQ: Ionic Equilibrium Question 2
Question 2 - 2021 (16 Mar Shift 2)
Type: Numerical
Sulphurous acid $(\text{H}2\text{SO}3)$ has $K{a_1} = 1.7 \times 10^{-2}$ and $K{a_2} = 6.4 \times 10^{-8}$. The pH of $0.588$ M $\text{H}_2\text{SO}_3$ is ____. (Round off to the Nearest Integer)
Show Answer
Answer: 1
Solution
$\text{H}2\text{SO}3$ is a dibasic acid. pH is due to first dissociation only since $K{a_1} \gg K{a_2}$.
First dissociation: $\text{H}_2\text{SO}_3 \rightleftharpoons \text{H}^+ + \text{HSO}3^-$; $K{a_1} = 1.7 \times 10^{-2}$
$K_{a_1} = \frac{x^2}{0.588 - x}$
$1.7 \times 0.588 - 1.7x = 100x^2$
Solving: $x = 0.09186$
$\text{pH} = -\log(0.09186) = 1.036 \approx 1$
BETA
