Question

What would be the solution pH when you add 10 mmol of ammonium acetate (NH4Ac, NH4 + = ammonium, Ac— = acetate) to 1 liter of water? What is acid-base reaction that takes place? And to what extent does that reaction occur at equilibrium (i.e., what % of the initial NH4 + or Ac— has reacted)?

Answer 1

mol of NH4+ = 10

mol of Ac- = 10

find pH solution:

NH4+ + H2O <-> NH3 + H3O+

CH3COO- + H2O <--> CH3COOH + OH-

add all

NH4+ + H2O + CH3COO- + H2O <-> NH3 + H3O+ + CH3COOH + OH-

note that

NH4+ + CH3COO- + 2H2O <-> NH3 + 2H2O + CH3COOH

forms

NH4+ + CH3COO- <-> NH3 + CH3COOH

given

Ka of CH3COOH = [CH3COO-][H+]/CH3COOH] = 1.8*10^-5

Kb of NH3 = [NH4+][OH-]/[NH3] = 1.8*10^-5

since

Ka = Kb

and equimolar,, i.e. 10 mmol of NH4+ and 10 mmol of CH3COO-

this must have a pH = 7

extent of reaction will be

%M = [H+]/M* 100%

[H+] = 10^-7

M = 10*10^-3

% ion = (10^-7)/(10*10^-3) * 100 = 0.001 %