Sulfuric acid is taken as completely ionised in the first step, with K_a2 = 1.2x10^-2 for HSO4-. The pH of 0.10 M H2SO4 (second step solved exactly) is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia199

Sulfuric acid is taken as completely ionised in the first step, with $K_{a 2} = 1.2 \times 1 0^{- 2}$ for $HSO X_{4} X^{-}$ . The pH of $0.10 M$ $H X_{2} SO X_{4}$ (second step solved exactly) is

First step gives

0.10 M

H X^{+}

and

0.10 M

HSO X_{4} X^{-}

. For the second step

(0.10 + x) x / (0.10 - x) = 0.012

, i.e.

x^{2} + 0.112 x - 0.0012 = 0

, so

x = 0.00985 M

. Then

[H X^{+}] = 0.1098 M

and

pH = 0.96