NH3 is a weak base (Kb = 1.8 × 10–5) so the salt NH4Cl acts as weak acid. What is the pH of a solution that is 0.043 M in NH4Cl at 25 °C?
What is the pH of a 0.043 M ammonium chloride (NH4Cl) solution at 25 °C, considering that NH3 is a weak base with a Kb value of 1.8 × 10⁻⁵, and that the salt NH4Cl behaves as a weak acid in solution?
1 Answers
Hydrolysis of NH4Cl can be written as NH4Cl + H2O ==> Cl^- + NH3 + H+ or simply NH4^+ + H2O ==> NH3 + H3O^+
Kw = KaKb
Ka = Kw/Kb = 1×10^-14/1.8×10^-5 = 5.56×10^-10
Ka = 5.56×10^-10 = [NH3][H3O+]/[NH4Cl] = (x)(x)/0.043 -x and assuming x is small relative to 0.043 M, then ignore it
x^2 = 2.39×10^-11
x = 4.9×10^-6 = [H3O+] (note: this is small relative to 0.043 so assumption was valid, and no need for quadratic)
pH = -log 4.9×10^-6
pH = 5.3
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