What is the pH of a 0.0380 M solution of HONH₃Cl (Kb of HONH₂ is 1.1 × 10⁻⁸)?

Answer 1

HONH₃Cl is the conjugate acid of HONH₂. It will react with water as follows:

HONH₃Cl + H₂O ==> HONH₂ + H₃O⁺ + Cl^-

Since it is acting as an acid, we will use the Ka for HONH₃Cl and determine the pH of the solution. Looking at the net ionic equation for simplicity, we have ...

HONH₃⁺ ==> HONH₂ + H⁺

Ka = [HONH₂] [H⁺] / [HONH₃⁺]

Ka = Kw/Kb = 1x10^-14 / 1.1x10^-8 = 9.09x10^-7

9.09x10^-7 = (x)(x) / 0.0380-x (assume x is small relative to 0.0380 and ignore in denominator)

9.09x10^-7 = (x)(x) / 0.0380

x² = 3.45x10^-8

x = [H⁺] = 1.86x10^-4 (note: this is less than 5% of 0.038 so above assumption is valid)

pH = -log [H⁺] = -log 1.86x10^-4

pH = 3.730