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?

1 Answer

  • 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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