Question

NH3 is a weak base (Kb = 1.8 x 10-5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.053 M in NH4Cl at 25 ℃? Number pH =

Answer 1

we have below equation to be used:

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

NH4+ + H2O -----> NH3 + H+

5.3*10^-2 0 0

5.3*10^-2-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that 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)*5.3*10^-2) = 5.426*10^-6

since c is much greater than x, our assumption is correct

so, x = 5.426*10^-6 M

so,

[H+] = x = 5.426*10^-6 M

we have below equation to be used:

pH = -log [H+]

= -log (5.426*10^-6)

= 5.27

Answer: 5.27

Feel free to comment below if you have any doubts or if this answer do not work