Sulphurous acid ( H _ 2 SO _ 3 ) has Ka _ 1 = 1 . 7 × 10 ^ - 2 and...

${H}_{2}{SO}_{3}$ $[$Dibasic acid$]$

$c = 0 . 588M$

$⇒$ $pH$ of solution is due to First dissociation only since ${{K}_{a}}_{1} > > {Ka}_{2}$

$⇒$ First dissociation of ${H}_{2}{SO}_{3}$

${H}_{2}{SO}_{3} ( aq ) ⇌ {H}^{⊕} ( aq ) + {HSO}_{3}^{-} ( aq ) : {ka}_{1} = 1 . 7 \times {10}^{- 2}$

$t = 0$

$tC - xxx$

$⇒ {Ka}_{1} = \dfrac{1 . 7}{100} = \dfrac{\left[{H}^{⊕}\right]\left[{HSO}_{3}^{-}\right]}{\left[{H}_{2}{SO}_{3}\right]}$

$⇒ \dfrac{1 . 7}{100} = \dfrac{{x}^{2}}{\left(0 . 588 - x\right)}$

$⇒ 1 . 7 \times 0 . 588 - 1 . 7x = 100{x}^{2}$

$⇒ 100{x}^{2} + 1 . 7x - 1 = 0$

$⇒ \left[{H}^{⊕}\right] = x = \dfrac{- 1 . 7 + \sqrt{( 1 . 7{)}^{2} + 4 \times 100 \times 1}}{2 \times 100} = 0 . 09186$

Therefore $pH$ of sol. is :

$pH = - \log \left[{H}^{⊕}\right]$

$⇒ pH = - \log \left(0 . 09186\right) = 1 . 036 ≃ 1$