# Sulphurous acid H2SO3

Question:

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 $\mathrm{pH}$ of $0.588 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{3}$ is_____________ . (Round off to the Nearest Integer)

Solution:

$\mathrm{H}_{2} \mathrm{SO}_{3}$ [Dibasic acid]

$\mathrm{c}=0.588 \mathrm{M}$

$\Rightarrow \quad \mathrm{pH}$ of solution $\mathrm{p}$ due to First dissociation only since $\mathrm{K}_{\mathrm{a}},>\mathrm{Ka}_{2}$

$\Rightarrow$ First dissociation of $\mathrm{H}_{2} \mathrm{SO}_{3}$

$\mathrm{H}_{2} \mathrm{SO}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{\oplus}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}): \mathrm{ka}_{1}=1.7 \times 10^{-2}$

$\mathrm{t}=0 \quad \mathrm{C}$

$\begin{array}{llll}\mathrm{t} & \mathrm{C}-\mathrm{x} & \mathrm{x} & \mathrm{x}\end{array}$

$\Rightarrow \quad \mathrm{Ka}_{1}=\frac{1.7}{100}=\frac{\left[\mathrm{H}^{\oplus}\right]\left[\mathrm{HSO}_{3}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{SO}_{3}\right]}$

$\Rightarrow \frac{1.7}{100}=\frac{x^{2}}{(0.58-x)}$

$\Rightarrow \quad 1.7 \times 0.588-1.7 x=100 x^{2}$

$\Rightarrow \quad 100 x^{2}+1.7 x-1=0$

$\Rightarrow \quad\left[\mathrm{H}^{\oplus}\right]=\mathrm{x}=\frac{-1.7+\sqrt{(1.7)^{2}+4 \times 100 \times 1}}{2 \times 100}=0.09186$

Therefore $\mathrm{pH}$ of sol. is : $\mathrm{pH}=-\log \left[\mathrm{H}^{\oplus}\right]$

$\Rightarrow \quad \mathrm{pH}=-\log (0.09186)=1.036 \simeq 1$