Chemistry · 2021
JEE Main · 16 March 2021, Shift 2 · Q56
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 M H_2 SO_3 is \ \ \ \. (Round off to the Nearest Integer).
Sulphurous acid $\displaystyle \left(\mathrm{H}_2 \mathrm{SO}_3\right)$ has $\displaystyle \mathrm{Ka}_1=1.7 \times 10^{-2}$ and $\displaystyle \mathrm{Ka}_2=6.4 \times 10^{-8}$. The pH of $\displaystyle 0.588$ M $\displaystyle \mathrm{H}_2 \mathrm{SO}_3$ is \_\_\_\_. (Round off to the Nearest Integer).
Official answer
1
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