Question

Bromine in excess is dropped to a $0.01 \, \text{M}$ $SO_{2}$ . All of $SO_{2}$ is oxidized to $H_{2}SO_{4}$ and the excess $Br_{2}$ is removed by flushing with gaseous $N_{2}$ . Determine the $\text{pH}$ of the resulting solution assuming $K_{a 1}$ of $H_{2}SO_{4}$ very large & $\text{K}_{\text{a2}} = 10^{- 2} .$ Take the value of $log \, \left(\right.3.24\left.\right) \, = \, 0.51$ .

Answer 1

$\underset{(0.01-\mathrm{x})}{\mathrm{HSO}_4^{-}}(\mathrm{aq}) \rightleftharpoons \underset{(0.03+\mathrm{x})}{\mathrm{H}^{+}}(\mathrm{aq})+\underset{\mathrm{x}}{\mathrm{SO}_4^{2+}} ; \mathrm{Ka}_2=10^{-2}$

$\mathrm{Ka}_2=10^{-2}=\frac{\mathrm{x}(0.03+\mathrm{x})}{(0.01-\mathrm{x})}$

$\text{x} = 2.36 \times 10^{- 3} = 0.00236$

$\left[\text{H}^{\text{+}}\right]_{\text{Total}} = 0.03 + 0.00236$

$= 0.03236 = 3.24 + 0.00236$

$\text{pH} = - \log 3.240 \times 10^{- 2}$

$= 2 - \log 3.24 = 1.49$