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 0.040 M in NH4Cl at 25 °C?

Answer 1

use:
Ka = Kw/Kb
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Ka = (1.0*10^-14)/Kb
Ka = (1.0*10^-14)/1.8*10^-5
Ka = 5.556*10^-10

NH3      + H2O ----->     NH4+   +   H+
4*10^-2                    0         0
4*10^-2-x                  x         x


Ka = [H+][NH4+]/[NH3]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.556*10^-10)*4*10^-2) = 4.714*10^-6

since c is much greater than x, our assumption is correct
so, x = 4.714*10^-6 M



So, [H+] = x = 4.714*10^-6 M


use:
pH = -log [H+]
= -log (4.714*10^-6)
= 5.3266
Answer: 5.33