Calculating pH?

Question

Can anyone explain how to calculate the pH of 0.300 M HONH3Cl? and how to calculate a mixture containing 0.300 M HONH2 (Kb =1.1*10^-8) and 0.300 M HONH3Cl?

Answer 1

HONH2 + H2O <==> HONH3+ + OH-, Kb = 1.1*10^-8
Let me first show the solution to the second part.
Salt HONH3Cl in the aquous solution completely dissociate into HONH3+ and Cl-. Cl- is a spectator and can be ignored.
Kb = 1.1*10^-8 = [HONH3+] *[OH-] /[HONH2] = [OH-]
pH = 14 + log([OH-])
= 14 + log(1.1*10^-8)
= 6.04
Now come to the first part.
Let [HONH2] = X. Since we did not add HONH2, all HONH2 must from reaction of HONH3+ with water, and from the following reaction we know that []H3O+] = [HONH2] = X.
Re-write the reaction as:
HONH2 + H3O+ <==> HONH3+ + H2O
where H3O+ can be written as H+.
Hence Kb/Kw = 1.1*10^6 = [HONH3+] *[OH-] /{[HONH2]*[H3O+]*[OH-]}
= [HONH3+] /{[HONH2]*[H3O+]}
= (0.3 - X)/X^2
Or: (1.1*10^6)X^2 + X - 0.3 = 0
A positive solution to this quadratic is: X = 0.0005218
Hence pH = -log([H+]) = -log(0.0005218) = 3.28