Question

^{Sulphurous acid (H₂SO₃) has K_a1 = 1.7 × 10–2 and K_a2 = 6.4 × 10–8. The pH of 0.588 M H₂SO₃ is _________. (Round off to the nearest integer).}

Answer 1

Solution:-

Step 1:- To find the value of α, by using formula $\mathrm{Ka}=\frac{\mathrm{Products}}{\mathrm{Reactants}}$

Given,

$$\begin{gathered} {\mathrm{Ka}}_1=1.7*{10}^{-2} \\ {\mathrm{Ka}}_2=6.4*{10}^{-8} \end{gathered}$$

pH of solution is due to first dissociation only since ${\mathrm{K}}_{\mathrm{a}1}>>{\mathrm{K}}_{\mathrm{a}2}$

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

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_3\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\mathrm{aq})+{\mathrm{HSO}}^{3-}(\mathrm{aq}) \\ \mathrm{t}=0\mathrm{C}00 \\ \mathrm{t}=\mathrm{t}0.588-\mathrm{\alpha }\mathrm{\alpha }\mathrm{\alpha } \end{gathered}$$

$$\begin{gathered} \mathrm{Ka}=\frac{\mathrm{Products}}{\mathrm{Reactants}} \\ \mathrm{Ka}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HSO}}^{3-}\right]}{\left[{\mathrm{H}}_2{\mathrm{SO}}_3\right]} \\ \frac{1.7}{100}=\frac{\left[\mathrm{\alpha }\right]\left[\mathrm{\alpha }\right]}{\left[0.588-\mathrm{\alpha }\right]} \\ 1.7*0.588-1.7\mathrm{\alpha }=100{\mathrm{\alpha }}^2 \\ 100{\mathrm{\alpha }}^2+1.7\mathrm{\alpha }-1=0 \\ \mathrm{To}\mathrm{solve}\mathrm{for}\mathrm{\alpha },\mathrm{we}\mathrm{use}\mathrm{the}\mathrm{quadratic}\mathrm{formula} \\ \mathrm{\alpha }=\frac{-b\pm \sqrt{{\mathit{b}}^{\mathit{2}}\mathit{-}\mathit{4}\mathit{a}\mathit{c}}}{2\mathrm{a}} \\ \mathrm{\alpha }=\frac{-1.7+\sqrt{{(1.7)}^2-4*100*1}}{2*100} \\ \mathrm{\alpha }=\left[{\mathrm{H}}^{+}\right]=0.0916 \end{gathered}$$

Step 2: Using the concentration to find pH

$$\begin{gathered} \mathrm{pH}=-\log [{\mathrm{H}}^{+}] \\ \mathrm{pH}=-\log (0.0916) \\ \mathrm{pH}=1.036\cong 1 \end{gathered}$$

Final answer :- The pH of 0.588 M H₂SO₃ is 1.