Identify the pH of 0.5888 M Sulphurous acid ${\mathrm{H}}_2{\mathrm{SO}}_3$

Given data: It is given to identify the pH of 0.5888 M $H_{2}S{O}_3$

Explanation

$$\begin{gathered} {\mathrm{H}}_2\mathrm{S}{{\mathrm{O}}_3}_{\left(\mathrm{aq}\right)} \rightleftharpoons \mathrm{HS}{{\mathrm{O}}^{3-}}_{\left(\mathrm{aq}\right)} + {{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)} \\ 0.5888= \mathrm{C} {\mathrm{C\alpha }}_1 ({\mathrm{C\alpha }}_1+{\mathrm{C\alpha }}_1{\mathrm{\alpha }}_2) \\ \mathrm{HS}{{{\mathrm{O}}_3}^{-}}_{\left(\mathrm{aq}\right)} \rightleftharpoons \mathrm{S}{{\mathrm{O}}_3}^{2-} + {\mathrm{H}}^{+} \\ {\mathrm{C\alpha }}_1(1-{\mathrm{\alpha }}_2) {\mathrm{C\alpha }}_1{\mathrm{\alpha }}_2 {\mathrm{C\alpha }}_1 + {\mathrm{C\alpha }}_1{\mathrm{\alpha }}_2 \end{gathered}$$

$$\begin{gathered} {\mathrm{\alpha }}_1 =\sqrt{\frac{17\times {10}^{-12}}{0.588}}=\sqrt{\frac{17}{289\times 2}} \\ {\mathrm{H}}^{+} ={\mathrm{C\alpha }}_1 \\ \sqrt{{\mathrm{K\alpha }}_1\times \mathrm{C}} =\sqrt{17\times {10}^{-3}\times 0.588} \\ =0.099 \\ \mathrm{pH} =-{\log }_{10}0.099=1 \end{gathered}$$