What is the pH of a 2.62x10^-1 M NH3 solution? pKa for NH4+=9.241?

Method 1 :

[NH₃]ₒ = 2.62 × 10⁻¹ M = 0.262 M
pKb of NH₃ = pKw - (pKa of NH₄⁺) = 14.000 - 9.241 = 4.759
Kb of NH₃ = 10⁻⁴˙⁷⁵⁹

______________ NH₃(aq) _ + _ H₂O(l) _ ⇌ _ NH₄⁺(aq) _ + _ OH⁻(aq) ____ Kb of NH₃ = 10⁻⁴˙⁷⁵⁹
Initial: ________ 0.262 M _________________ 0 M ________ 0 M
Change: _______ -y M ___________________ +y M _______ +y M
Equilibrium: _ (0.262 - y) M _______________ y M ________ y M

As Kb is very small, the dissociation of NH₃ is negligible.
It can be assumed that [NH₃] at equilibrium = (0.262 - y) M ≈ 0.262 M

At equilibrium:
Kb = [NH₃] [OH⁻] / [NH₃]
10⁻⁴˙⁷⁵⁹ = y² / 0.262
y = √(10⁻⁴˙⁷⁵⁹ × 0.262) M = 2.14 × 10⁻³

pOH = -log[OH⁻] = -log(2.14 × 10⁻³) = 2.67
pH = pKw - pOH = 14.0 - 2.67 = 11.33


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Method 2 :

pKb of NH₃ = pKw - (pKa of NH₄⁺) = 14.000 - 9.241 = 4.759

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) …… pKb = 4.759

pOH = (1/2)(pKb - log[NH₃]) = (1/2){4.759 - log(0.262)} = 2.67
pH = pKw - pOH = 14.0 - 2.67 = 11.33