Sodium phenoxide, $C X_{6} H X_{5} ONa$ , is 1.0 $\times 1 0^{- 3}$ M in water at 298 K. For phenol $K_{a} = 1.0 \times 1 0^{- 10}$ . Because hydrolysis is not small, solve $K_{h} = x^{2} / (c - x)$ exactly as a quadratic for $x = [OH X^{-}]$ (do not use $x ≪ c$ ). Find the pH to two decimal places.

Numerical answer type, tolerance 0.02.

Answer: 10.43

K_{h} = K_{w} / K_{a} = 1.0 \times 1 0^{- 4}

x^{2} + K_{h} x - K_{h} c = 0

gives

x = [- K_{h} + K_{h}^{2} + 4 K_{h} c] /2 = 2.702 e - 04

M. pOH

= 3.568

, so pH

= 10.43

. (The approximation

K_{h} c

would overestimate

[OH X^{-}]

at 3.16e-04 M.)