Sulphurous acid $\left(\mathrm{H}_{2} \mathrm{SO}_{3}\right)$ has $\mathrm{Ka}_{1}=1.7 \times 10^{-2}$ and $\mathrm{Ka}_{2}=6.4 \times 10^{-8}$. The pH of $0.588 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{3}$ is ______. (Round off to the nearest integer).

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Equilibrium

Sulphurous acid

(H_{2} SO_{3})

has

Ka_{1} = 1.7 \times 1 0^{- 2}

and

Ka_{2} = 6.4 \times 1 0^{- 8}

. The pH of

0.588 M H_{2} SO_{3}

is ______. (Round off to the nearest integer).

Answer: 1

Solution

The second dissociation constant is six orders of magnitude smaller than the first, so the hydrogen ion concentration is set by the first dissociation alone and the acid may be treated as monoprotic:

[H^{+}] = K_{a 1} C = 1.7 \times 1 0^{- 2} \times 0.588 = 1.0 \times 1 0^{- 1} M

Taking the negative logarithm,

pH = - lo g (1.0 \times 1 0^{- 1})

, so the pH of the solution is 1.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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