Chemistry help! acid and bases?

According to: http://www4.ncsu.edu/~franzen/public_html/CH201/data/Acid_Base_Table.pdf
Ka for HCN = 6.2 × 10⁻¹⁰
(Difference sources may give slightly different values of Ka. This may lead to slightly different answers.)

Consider the dissociation of 0.040 M HCN:
___________ HCN(aq) _ + _ H₂O(l) ⇌ _ H₃O⁺(aq) _ + _ CN⁻(aq) ____ Ka = 6.2 × 10⁻¹⁰
Initial: _____0.040 M _______________ 0 M ________ 0 M
Change: _____ -y M ________________ +y M _______ +y M
At eqm: _ (0.040 - y) M _____________ y M _________ y M

As Ka is very small, the dissociation of HCN is negligible, i.e.
[HCN] at eqm = (0.040 - y) M ≈ 0.040 M

At eqm:
Ka = [H₃O⁺] [CN⁻] / [HCN]
6.2 × 10⁻¹⁰ = y² / 0.040
y = √{0.040 × (6.2 × 10⁻¹⁰)} = 5.0 × 10⁻⁶

pH = -log[H₃O⁺] = -log(5.0 × 10⁻⁶) = 5.3
[HCN] = (0.040 - y) M = {0.040 - (5.0 × 10⁻⁶)} M ≈ 0.040 M
[H₃O⁺] = y M = 5.0 × 10⁻⁶ M
[CN⁻] = y M = 5.0 × 10⁻⁶ M