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 0.019 M in NH4Cl at 25 °C?

Answer 1

first find the

Ka of the NH3

Ka x Kb = 1 x 10^-14

Ka x 1.8 × 10^–5 = 1 x 10^-14

Ka = 1 x 10^-14 / 1.8 × 10^–5

Ka = 5.56 x 10^-10

       NH4Cl (aq) <====> HCl (aq) + NH3 (aq)

I       0.019                         0                    0

C      -x                             +x                   +x

E   0.019 -x                       +x                   +x

Ka = [NH3] [HCl] / [NH4Cl]

5.56 x 10^-10 = [x] [x] / [0.019 - x]

x² + x 5.56 * 10^-10 - 1.0564 x 10^-11 = 0

solve the quadratic equation

x = 3.25 x 10^-6 = [HCl]

since HCl is strong acid concentration of HCl = concentration of H+

pH = -log[3.25 x 10^-6 ] = 5.5