Question

Bromine in excess is dropped to a $0.01\text{M}$ $S{O}_2$ . All of $S{O}_2$ is oxidized to $H_{2}S{O}_4$ and the excess $B{r}_2$ is removed by flushing with gaseous $N_2$ . Determine the $\text{pH}$ of the resulting solution assuming $K_{a1}$ of $H_{2}S{O}_4$ very large & ${\text{K}}_{\text{a2}}=10^{-2}.$ Take the value of $log(3.24)=0.51$ .

Answer 1

Answer: 1.49

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

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

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

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

$=0.03236=3.24+0.00236$

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

$=2-\log 3.24=1.49$