Why use Kw to find Ka for NH4+ in a 1 M NH4Br solution

[a.a]

Homework Statement

The approximate pH of a 1 M solution of NH4Br [Kb(NH3) = 1.8 x 10-5] is___

Homework Equations

The Attempt at a Solution

Ka(NH4) = 10^-14 / 1.8 x 10^-5 =
=5.5555 x 10^-10

therefore, 5.5555 x 10^-10 = [NH3][H+]/[NH4+]

Since 1 H+ forms for every NH3 that forms and that [NH3] + [NH4+] = 1,
5.5555 x 10^-10 = [H+][H+]/(1-[H+])
Since Ka is small, 1-[H+] approximately = 1
[H+] = 2.357 x 10^-5
ph = 4.63

where did the 10^-14 come from and why are we saying this:
5.5555 x 10^-10 = [NH3][H+]/[NH4+]

 

[Borek]

10^-14 is Kw.

The other part is about Bronsted-Lowry theory. NH₄⁺ is a Bronsted-Lowry acid. Expression that you have listed is its dissociation constant.

 

[a.a]

what exactlly is pKa and pKb?

 

[Borek]

pKa = -log (Ka)