equilibrium

50 cm^3 of 1M NH3 is titrated with 1M HCl. What is the pH value before any HCl added ?

NH3(aq) + H2O(l) = NH4^+(aq) + OH^-(aq)
Kb = Kw/Ka = (1 x 10^-14)/(5.6 x 10^-10) = 1.79 x 10^-5 M

At equilibrium :
Let [OH^-] = a M
Then [NH4+] = a M
and [NH3] = (1 - a) M

Kb = [NH4^+][OH^-]/[NH3]
a^2/(1 - a) = 1.79 x 10^-5
a^2 + (1.79 x 10^-5)a - (1.79 x 10^-5) = 0
a = 0.00422
pOH = -log(0.00422) = 2.37
pH = 14 - 2.37 = 11.63


What is the pH after 25cm^3 of HCl ADDDED?

H^+(aq) + NH3(g) → NH4^+(aq)
Neglected any dissociation:
Total volume = 25 + 50 = 75 cm^3
[NH4^+]o = 1 x (25/75) = 0.333 M
[NH3]o = 1 x (50/75) - 1 x (25/75) = 0.333 M

Consider the dissociation of NH4^+(aq) :
NH4^+(aq) = NH3(aq) + H^+(aq)
pH ≈ pKa - log([NH4^+]o/[NH3]o)
pH ≈ -log(5.6 x 10^-10) - log(0.333/0.333) = 9.25


What is the pH at equivalence pt , when 50cm^3 of HCl has added?

H^+(aq) + NH3(g) → NH4^+(aq)
Neglected any dissociation:
Total volume = 50 + 50 = 100 cm^3
[NH4^+]o = 1 x (50/100) = 0.5 M
[NH3]o = 1 x (50/100) - 1 x (50/100) = 0 M

Consider the dissociation of NH4^+(aq) :
NH4^+(aq) = NH3(aq) + H^+(aq)
Let [H^+] = b M
Then [NH3] = b M
and [NH4^+] = (0.5 - b) M
Ka = [NH3][H^+]/[NH4^+]
b^2/(0.5 - b) = 5.6 x 10^-10
b^2 + (5.6 x 10^-10)b - (2.8 x 10^-10) = 0
b = 1.67 x 10^-5
pH = -log(1.67 x 10^-5) = 4.78