Chm, pH question?

Kb(CN⁻) = Kw/Ka(HCN) = (1.0 × 10⁻¹⁴) / (4.9 × 10⁻¹⁰) = 2.04 × 10⁻⁵

_________ CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq) __ Kb = 2.04 × 10⁻⁵
Initial:         0.750 M                      0 M           0 M
Change:       -y M                         +y M         +y M
Eqm:       (0.750 - y) M                 y M            y M
_              ≈ 0.750 M

At equilibrium :
Kb = [HCN] [OH⁻] / [CN⁻]
2.04 × 10⁻⁵ = y² / 0.750
y = √(0.750 × 2.04 × 10⁻⁵) = 3.91 × 10⁻³

p0H = -log[OH⁻] = log(3.91 × 10⁻³) = 2.41
pH = pKw - pOH = 14.00 - 2.41 = 11.59