g12 chem HBr and NH3 pH at equivalence?

HBr is a strong acid which completely ionized to form H₃O⁺ ions.

The balanced equation for the reaction between HBr and NH₃.
H₃O⁺(aq) + NH₃(aq) → NH₄⁺(aq) + H₂O(ℓ)
Mole ratio H₃O⁺ : NH₃ : NH₄⁺ = 1 : 1 : 1

Moles of H₃O⁺ reacted = Moles of HBr = (0.10 mol/L) × (50.0/1000 L) = 0.005 mol
Moles of NH₄⁺ formed = 0.005 mol
Moles of NH₃ reacted = 0.005 mol
Volume of NH₃ solution needed = (0.005 mol) / (0.10 mol/L) = 0.05 L
Volume of the final solution = [0.05 + (50.0/1000)] L = 0.1 L
[NH₄⁺] in the final solution = (0.005 mol) / (0.1 L) = 0.05 mol/L

NH₄⁺ ions slightly dissociate in water. Thus construct an ICE table.
         NH₄⁺(aq) + H₂O(ℓ) ⇌ NH₃(aq) + H₃O⁺(aq)     Ka = Kw / Kb(NH₃)
Initial:   0.0.5 M            0 M       0 M
Change:   -y M            +y M      +y M
Eqm:   (0.05 - y) M         y M       y M
       ≈ 0.05 M due to very small Ka

At equilibrium:
Ka = [NH₃] [H₃O⁺] / [NH₄⁺]
Kw / Kb(NH₃) = [NH₃] [H₃O⁺] / [NH₄⁺]
(1 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = y² / 0.05
y = √[0.05 × (1 × 10⁻¹⁴) / (1.8 × 10⁻⁵)] = 5.27 × 10⁻⁶
pH = -log[H₃O⁺] = -log(5.27 × 10⁻⁶) = 5.28

This problem is worked exactly as the previous question you asked: 
https://answers.yahoo.com/question/index?qid=20200819142456AAVMRGf

Use that one as an example to solve this one.