What is the pH of 0.256 M ammonium chloride NH4Cl?

What is the pH of a 0.04 mol dm-3 solution of ammonium chloride (NH4Cl)?

• Answer:

  Ammonium ions act as an acid and transfer protons to water: NH4+ + H2O <--> NH3 + H3O+ Ka for this reaction is: Ka = [NH3][H3O+]/[NH4+] Depending on the text book, you may have Ka given, or you may have to calculate it from Kb of ammonia. For any conjugate acid-base pair, Ka X Kb = Kw = 1X10^-14. In my text, Kb for NH3 is 1.8X10^-5. So, Ka = 1X10^-14 / 1.8X10^-5 = 5.6X10^-10 Then, in this solution, [H3O+] = [NH3] = x, and [NH4+] = 0.04 - x. Let's assume x is small compared to 0.04, so Ka, then becomes: Ka = x^2 / 0.04 = 5.6X10^-10 x = 4.7X10^-6 M = [H3O+] pH = 5.33