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 0.027 M in NH4Cl at 25 °C?

Answer 1

first convert the Kb in to Ka using

Ka x Kb = 1.0 x 10^-14

Ka =  1.0 x 10^-14 / Kb

put the given Kb value in above equation

Ka = 1.0 x 10^-14 / 1.8 x 10^-5

Ka = 5.55 x 10^-10

NH4Cl is week acid

since it is week acid there will be a equilibrium it will not dissociate 100%

now construct ICE table

NH₄⁺ <---------> H⁺ + NH₃

I 0.027 0 0

C -x +x +x

E 0.027-x +x +x

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

put all the given values

5.55 x 10^-10 = [x][x] / [0.027-x]

x² + x 5.55 * 10^-10 - 1.4985 * 10^-11 = 0

solve the quadratic equation

x = 3.87 x 10^-7 M

now x means concentration of H+

[H+] = 3.87 x 10^-6 M

pH = -log[H+]

pH = -log[3.87 x 10^-6 ]

pH = 5.41