Chemistry mcq? Calculate pH?

Let y M be [OH⁻] M at equilibrium.

For a solution containing 0.1 M NH₄Cl and 0.01 M NH₄OH :
NH₄OH(aq) ⇌ NH₄⁺(aq) + OH⁻(aq) ...... Kb = 1.8 × 10⁻⁵

Initial concentration :
[NH₄OH]ₒ = 0.01 M
[NH₄⁺]ₒ = 0.1 M
[OH⁻]ₒ = 0 M

At equilibrium :
[NH₄OH] = (0.01 - y) M ≈ 0.01 M, assuming 0.01 ≫ y due to very small Kb and common ion effect
[NH₄⁺] = (0.1 + y) M ≈ 0.1 M, assuming 0.01 ≫ y due to very small Kb and common ion effect
[OH⁻] = y M

pOH = pKb - log([NH₄OH]/[NH₄⁺]) = -log(1.8 × 10⁻⁵) - log(0.01/01) = 5.7
pH = 14 - pOH = 14 - 5.7 = 8.3