NH3 is a weak base ( 𝐾b=1.8×10−5 ) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.051 M in NH4Cl at 25 °C?

Answer 1

To find the pH of a salt solution, look at the hydrolysis of that salt.

NH₄Cl is the salt of a weak base (NH4OH) and a strong acid (HCl). The pH of the salt solution should therefore be acidic.

NH₄Cl + H₂O ==> NH₄OH + HCl .. hydrolysis reaction

NH₄⁺ + H₂O ==> NH₄OH + H⁺ .. net ionic hydrolysis equation

The Ka for NH₄⁺ can be calculated from the Kb of the conjugate base

Ka = Kw/Kb = 1x10^-14 / 1.8x10^-5

Ka = 5.56x10^-10

Ka = [ NH₄OH][H⁺] / [NH₄⁺] = (x)(x) / 0.051 - x

5.56x10^-10 = x² / 0.051 - x (assume x is small relative to 0.051 and ignore it)

5.56x10^-10 = x² / 0.051

x² = 2.83x10^-11

x = [H⁺] = 5.3x10^-6 (this is insignificant relative to 0.051 so above assumption was valid)

pH = -log [H⁺] = - log 5.3x10^-6

pH = 5.27 (acidic as predicited)