Question

The pH of an aqueous solution of 0.184 M ammonium bromide, NH4Br (aq), is

Answer 1

Kb of NH3 = 1.8*10^-5

we have below equation to be used:
Ka of NH4+ = 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+
0.184                    0         0
0.184-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)*0.184) = 1.011*10^-5

since c is much greater than x, our assumption is correct
so, x = 1.011*10^-5 M



So, [H+] = x = 1.011*10^-5 M


we have below equation to be used:
pH = -log [H+]
= -log (1.011*10^-5)
= 5.00
Answer: 5.00