Question

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.010 M in NH4Cl at 25 °C?

Answer 1

NH4+   <----------->    NH3   + H+

I        0.01 0         0

C         -a +a       +a

E      0.01-a a          a

Ka of NH4+ = [NH3][H+]/[NH4+]

Kw/Kb of NH3 = a²/(0.01 - a)

10^-14/1.8 x 10^-5 = a²/(0.01 - a)

5.556 x 10^-10 =  a²/(0.01 - a)

a² + 5.556 x 10^-10a - 5.556 x 10^-12 = 0

The positive root of the quadratic equation is:

a = 2.357 x 10^-6

[H+] = a = 2.357 x 10^-6 M

pH = -log[H+] = 5.63