QJEEM21Ch56b0: JEE Main 2021 Chemistry - Class 11 JEEM21Ch56b0 Solution: Sulphurous acid (H 2 SO 3 ) has Ka 1 = 1.7 10 2 and Ka 2 = 6.4…

Step-by-Step Verified Solution

$K{a_1}$ of {H_2}S{O_3} > > K{a_2} of ${H_2}S{O_3}$

$\therefore$ The contribution of H⁺ from 2^nd dissociation of H₂SO₃ can be neglected.



$\Rightarrow$ ${{c{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$

$\Rightarrow {{0.588{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$

$\Rightarrow 58.8{\alpha ^2} = 1.7 - 1.7\alpha$

$\Rightarrow 58.8{\alpha ^2} + 1.7\alpha - 1.7 = 0$

$\alpha = {{ - 1.7 + \sqrt {{{1.7}^2} + 4 \times 1.7 \times 58.8} } \over {2 \times 58.8}} = 0.156$

$[{H^ + }] = c\alpha = 0.092$

$pH = - \log [{H^ + }]$

$= 1.036$

$\approx 1$