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)
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