Calculate the pH of 310 mL of a 0.80 M aqueous solution of NH3 at 25 °C given that the Ka of NH4Cl is 5.6×10-10.?

Kb(NH₃) = Kw/Ka(NH₄Cl) = (1.00 × 10⁻¹⁴) / (5.6 × 10⁻¹⁰)

_______________ NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + H₂O(l) …… Kb(NH₃)
Initial: __________ 0.80 M __________ 0 M _____ 0 M
Change: _________ -y M ___________ +y M ____ +y M
Equilibrium: ___ (0.80 - y) M ________ y M _____ y M

As Ka(NH₃) is very small, the dissociation of NH₃ is to a very small extent.
Hence, 0.80 ≫ y
[NH₃] at equilibrium = (0.80 - y) M ≈ 0.80 M

At equilibrium: Kb(NH₃) = [NH₄⁺] [OH⁻] / [NH₃]
(1.00 × 10⁻¹⁴) / (5.6 × 10⁻¹⁰) = y² / 0.80
y = √{0.80 × (1.00 × 10⁻¹⁴) / (5.6 × 10⁻¹⁰)}
y = 0.0038

pOH = -log[OH⁻] = -log(0.0038) = 2.4
pH = pKw - pOH = 14.0 - 2.4 = 11.6