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 0.041 M in NH4Cl at 25 °C?

Answer 1

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

0.041                                        0            0

0.041 - x                                   x             x

Ka = x^2 / 0.041 - x

Kw / Kb = x^2 / 0.041 - x

10^-14 / 1.8 × 10–5 = x^2 / 0.041 - x

5.56 x 10^-10 = x^2 / 0.041 - x

x = 4.77 x 10^-6

[H3O+] = 4.77 x 10^-6 M

pH = -log [H3O+] = -log (4.77 x 10^-6)

pH = 5.32