Calculate pH of 0.1M NH4NO3?

It is found that Kb of NH₃ = 1.8 × 10⁻⁵
NH₄⁺ is the conjugate acid of NH₃.
Then, Ka of NH₄⁺ = Kw/Ka(NH₃) = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵)

_________ NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) .... Ka = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵)
Initial ____ 0.1 M ____________ 0 M ____ 0 M
Change ___ -y M ____________ +y M ___ +y M
At eqm __ (0.1 + y) M _________ y M ____ y M

Assume that 0.1 ≫ y,
then at eqm, [NH₄⁺ ] = (0.1 + y) M ≈ 0.1 M

Ka = [NH₃] [H₃O⁺] / [NH₄⁺]
y² / 0.1 = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵)
y = 7.45 × 10⁻⁶

pH = -log[H₃O⁺] = -log(7.45 × 10⁻⁶) = 5.1

Here's the answer I gave for the other time you asked the same question:
So, the solution has NH4+ ions at a concentration of 0.10 M

NH4+ acts as an acid and undergoes the ionization:
NH4+ <--> NH3 + H+
Ka = [NH3][H+]/[NH4+]

You are given a Kb for NH3, so you need to calculate Ka for NH4+. Nice thing is that Ka X Kb = Kw = 1.0X10^-14. So:
Ka = 1.0X10^-14 / 1.8X10^-5 = 5.6X10^-10

Now, in the solution, let [H+] = [NH3] = x. Because Ka is small, you can assume that [NH4+] = 0.10. So:
Ka = 5.6X10^-10 = x^2 / 0.1
x = [H+] = 7.4X10^-6 M
pH = 5.13

Idk