What is the pH of a 0.0460 M solution of HONH₃Cl (Kb of HONH₂ is 1.1 × 10⁻⁸)?

Answer 1

HONH₃⁺ + H₂O ==> HONH₂ + H₃O⁺ ... hydrolysis of the weak conjugate acid, HONH₃⁺

Ka*Kb = 1x10^-14

Ka = 1x10^-14 / 1.1x10^-8

Ka = 9.1x10^-7

Ka = 9.1x10^-7 = [HONH₂][ H₃O⁺] / [HONH₃⁺]

9.1x10^-7 = (x)(x) / 0.0460 - x (assume x is small relative to 0.0460 and neglect it)

9.1x10^-7 = (x)(x) / 0.0460

x² = 4.19x10^-8

x = 2.0x10^-4 M = [H₃O⁺] and this is small relative to 0.046 (less than 0.5%, so ignoring it was valid)

pH = -log [H₃O⁺]

pH = -log 2.0x10^-4

pH = 3.7