Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\underset{\left(\text{0}\text{.01-x}\right)}{{\text{HSO}}_{\text{4}}^{-}}\left(\text{aq}\right)\rightleftharpoons \underset{\left(\text{0}\text{.03+x}\right)}{{\text{H}}^{+}}\left(\text{aq}\right)+\underset{\text{x}}{{\text{SO}}_{\text{4}}^{\text{2+}}};{\text{Ka}}_{\text{2}}={10}^{-2}$

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

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

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

$=0.03236=3.24+0.00236$

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

$=2-\log 3.24=1.49$