For carbonate hydrolysis, $CO X_{3} X^{2 -} + H X_{2} O HCO X_{3} X^{-} + OH X^{-}$ , $K_{h} = K_{w} / K_{a 2}$ with $K_{a 2} (H X_{2} CO X_{3}) = 4.7 \times 1 0^{- 11}$ . Neglect the second step. Find the percentage degree of hydrolysis of $0.10 M$ $Na X_{2} CO X_{3}$ by solving the quadratic exactly (do not use $h = K_{h} / C$ ). Give the answer to two decimals.

Numerical answer type, answer in %, tolerance 0.02.

Answer: 4.5075 %

K_{h} = 2.13 \times 1 0^{- 4}

. With

K_{h} = C h^{2} / (1 - h)

C h^{2} + K_{h} h - K_{h} = 0

, so

h = [- K_{h} + K_{h}^{2} + 4 C K_{h}] / (2 C) = 0.04507

, i.e.

4.51%

. (The approximate formula would give

4.61%