chem help please?

____________ NH₄⁺(aq) __ ⇌ __ NH₃ (aq) __ + __ H⁺(aq) ___ Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)
Initial: _____ 0.0100 M __________ 0 M _________ 0 M
Change: ______ -y M ___________ +y M ________ +y M
At eqm: __ (0.0100 - y) M ________ y M _________ y M

As Ka is very small, the dissociation of NH₄⁺ is to a very small extent.
Hence, assume that 0.0100 ≫ y
At eqm [NH₄⁺] = (0.0100 - y) M ≈ 0.0100 M

Ka = [NH₃] [H⁺] / [NH₄⁺]
(1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = y² / 0.0100
y = √[0.0100 × (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)]
y = 2.36 × 10⁻⁶

pH = -log[H⁺] = -log(2.36 × 10⁻⁶) = 5.63