Sulphurous acid (H₂SO₃) has Ka₁ = 1.7 x 10^-2 and Ka₂ = 6.4 x 10^-8. The pH of 0.588M H₂SO₃
is ____. (Round off to the Nearest Integer).

H₂SO₃ is dibasic acid, But according to data Ka₂<<<Ka₁ So pH is only due to Ka₁
${\mathrm{H}}_2{\mathrm{S}}_3\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{HSO}}_3^{-}\left(\mathrm{aq}\right){\mathrm{Ka}}_1=1.7\times {10}^{-2}$
t = 0   C   0   0
t=t  Dissociated Produced 
           C$\mathrm{\alpha }$         C$\mathrm{\alpha }$         C$\mathrm{\alpha }$ 
t = t_eq Left    Present
         C- C$\mathrm{\alpha }$       C$\mathrm{\alpha }$               C$\mathrm{\alpha }$ 
        C(1-$\mathrm{\alpha }$)        C$\mathrm{\alpha }$              C$\mathrm{\alpha }$
${\mathrm{ka}}_1=\frac{{\mathrm{C}}^2{\mathrm{\alpha }}^2}{\mathrm{C}(1-\mathrm{\alpha })}\text{ as }\mathrm{\alpha }<<1\text{ so }1-\mathrm{\alpha }$ is ignored
$\begin{array}{l}{\mathrm{Ka}}_1={\mathrm{C\alpha }}^2\mathrm{\alpha }=\sqrt{\frac{{\mathrm{Ka}}_1}{\mathrm{C}}}\\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\mathrm{C\alpha }=\mathrm{C}\times \sqrt{\frac{{\mathrm{Ka}}_1}{\mathrm{C}}}=\sqrt{{\mathrm{Ka}}_1\mathrm{C}}={\left(1.7\times {10}^{-2}\times 0.588\right)}^{\frac{1}{2}}\\ ={\left(0.999\times {10}^{-2}\right)}^{\frac{1}{2}}={\left(1\times {10}^{-2}\right)}^{\frac{1}{2}}\\ \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=-\log {\left({10}^{-2}\right)}^{\frac{1}{2}}=\frac{1}{2}\times (-1)\times (-2\log 10)=1\end{array}$