Sulphurous acid H 2 SO 3 has Ka 1 = 1 . 7 × 10 - 2 and Ka 2 = 6 . 4 × 10 - 8 . The pH of 0 . 588…

${\mathrm{H}}_2{\mathrm{SO}}_3$ $[$Dibasic acid$]$

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

$\Rightarrow$ $\mathrm{pH}$ of solution is due to First dissociation only since ${{\mathrm{K}}_{\mathrm{a}}}_1>>{\mathrm{Ka}}_2$

$\Rightarrow$ First dissociation of ${\mathrm{H}}_2{\mathrm{SO}}_3$

${\mathrm{H}}_2{\mathrm{SO}}_3\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{\oplus }\left(\mathrm{aq}\right)+{\mathrm{HSO}}_3^{-}\left(\mathrm{aq}\right) : {\mathrm{ka}}_1=1.7\times {10}^{-2}$

$\mathrm{t}= 0$

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

$\Rightarrow {\mathrm{Ka}}_1=\frac{1.7}{100}=\frac{\left[{\mathrm{H}}^{\oplus }\right]\left[{\mathrm{HSO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{SO}}_3\right]}$

$\Rightarrow \frac{1.7}{100}=\frac{{\mathrm{x}}^2}{\left(0.588-\mathrm{x}\right)}$

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

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

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

Therefore $\mathrm{pH}$ of sol. is :

$\mathrm{pH}=-\log \left[{\mathrm{H}}^{\oplus }\right]$

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