Sodium sulfite is 0.15 M in water at 298 K. $K_{a 2}$ of $H X_{2} SO X_{3}$ is $6.2 \times 1 0^{- 8}$ ( $K_{w} = 1.0 \times 1 0^{- 14}$ ). Considering only the first hydrolysis step of $SO X_{3} X^{2 -}$ and using $[OH X^{-}] = K_{h} c$ , calculate the pH to two decimal places.

Numerical answer type, tolerance 0.02.

Answer: 10.19

SO X_{3} X^{2 -} + H X_{2} O HSO X_{3} X^{-} + OH X^{-}

has

K_{h} = K_{w} / K_{a 2} = 1.613 e - 07

[OH X^{-}] = 1.613 e - 07 \times 0.15 = 1.555 e - 04

M. pOH

= 3.808

, pH

= 10.19