What is the pH of a 0.0300 M solution of HONH₃Cl (Kb of HONH₂ is 1.1 × 10⁻⁸)?

Answer 1

HONH3Cl is an acid. Ignoring the Cl ion, we have HONH₃⁺. The Ka for this acid can be found from the Kb of the conjugate base as follows:

KaKb = Kw

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

Ka = 9.09x10^-7

HONH3 ==> H+ + HONH2

Ka = [H+][HONH2] / [HONH3]

9.09x10^-7 = (x))x) / 0.03 - x and assume x is small relative to 0.03 and ignore

9.09x10^-7 = (x))x) / 0.03

x² = 2.7x10^-8

x = 1.65x10^-4 M = [H+]

pH = -log [H+]

pH = 3.78