Question

a)You obtain a 0.817 M solution of NH4Cl. Knowing that the Kb of NH3 is 1.8 * 10-5, what is the Ka of NH4+?

b) What is the [H+] in the solution in part a?

c) What is the pH of the solution in part a?

Answer 1

Ka x Kb = Kw (ionic product of water = 10^-14)

Ka x 1.8 x 10^-5 = 10^-14

Ka = 10^-14 / 1.8 x 10^-5

Ka =5.56 x 10^-10

now NH4 can dissociate in the folowing way

            NH4+ + H2O ------> NH3 + H3O+

I            0.817           0    0    

C -x    +x    +x

E 0.817-x +x +x

Ka = [NH3][H3O+] / [NH4+]

5.56 x 10^-10 = [x][x] / [0.817-x]

x² + x5.56 x 10^-10 -4.543 x 10^-10

Part A answer x = 0.0000213 M = 2.13x 10^-5 = [H3O+]

Part B

pH = -log[H3O+]

pH = -log[0.0000213]

pH = 4.672