Treat H2SO4 as fully ionised in its first step and HSO4- as a weak acid with K_a2 = 1.2x10^-2. For a 0.10 M H2SO4 solution, solve the second equilibrium exactly (quadratic) and give the pH to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3ia035

Treat $H X_{2} SO X_{4}$ as fully ionised in its first step and $HSO X_{4} X^{-}$ as a weak acid with $K_{a 2} = 1.2 \times 1 0^{- 2}$ . For a 0.10 M $H X_{2} SO X_{4}$ solution, solve the second equilibrium exactly (quadratic) and give the pH to two decimal places.

Numerical answer type, tolerance 0.01.

After step 1,

[H X^{+}] = [HSO X_{4} X^{-}] = 0.10

M. For step 2:

K_{a 2} = (0.10 + x) x / (0.10 - x)

, giving

x^{2} + 0.112 x - 1.2 \times 1 0^{- 3} = 0

x = 0.00985

[H X^{+}] = 0.10985

M and pH

= 0.96