Kb for methylamine (CH3NH2) is 4.4 x 10-4. What is the pH of a 0.210M solutions of methylamine?

CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq) .... Kb = 4.4 × 10⁻⁴

At equilibrium :
Let [CH₃NH₃⁺] = [OH⁻] = y M
Then [CH₃NH₂] = (0.210 - y) M

Kb = [CH₃NH₃⁺][ OH⁻]/[ CH₃NH₂]
y²/(0.210 - y) = 4.4 × 10⁻⁴
y² + (4.4 × 10⁻⁴)y - (9.24 × 10⁻⁵) = 0
y = {-(4.4 × 10⁻⁴) ± √[(4.4 × 10⁻⁴) - 4(-9.24 × 10⁻⁵)]} / 2
y = 9.40 × 10⁻³ or y = -9.83 × 10⁻³ (rejected)

At equilibrium :
[OH⁻] = 9.40 × 10⁻³ M
pOH = -log[OH⁻] = -log(9.40 × 10⁻³) = 2.02
pH = 14 - 2.02 = 11.98