# Solve the following

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 $0.588 \mathrm{MH}_{2} \mathrm{SO}_{3}$ is_______________.  (Round off to the Nearest Integer)

Solution:

(1)

$\mathrm{H}_{2} \mathrm{SO}_{3}[$ Dibasic acid $] \mathrm{c}=0.588 \mathrm{M}$

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

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

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

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

$\mathrm{t} \quad \mathrm{C}-\mathrm{x} \quad \mathrm{x} \quad \mathrm{x}$

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

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

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

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

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

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

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