For $CH X_{3} NH X_{2}$ , $K_{b} = 4.4 \times 1 0^{- 4}$ at 298 K. Solving the quadratic exactly (no approximation for the degree of ionisation), the pH of a 0.0010 M aqueous solution is closest to ( $K_{w} = 1.0 \times 1 0^{- 14}$ ; $lo g 4.79 = 0.680$ )

Answer: A

With

x = [OH X^{-}]

x^{2} / (0.0010 - x) = 4.4 \times 1 0^{- 4}

, so

x^{2} + 4.4 \times 1 0^{- 4} x - 4.4 \times 1 0^{- 7} = 0

and

x = 4.789 e - 04

M. Then pOH = 3.32 and pH = 10.68. The approximation

x = K_{b} c

gives 10.82 and is not valid because ionisation is nearly 48 per cent.