Help with Chemistry problems.?

Denote the monoprotic acid as HA.

Consider the dissociation of HA:
___________ HA(aq) ___ ⇌ ___ H⁺(aq) ___+ ___ A⁻(aq) ______ Ka = 2.0 × 10⁻⁶
Initial: ______ 0.48 ___________ 0 M _________ 0 M
Change: _____ -y M __________ +y M ________ +y M
At eqm: __ (0.48 - y) M ________ y M _________ y M

As Ka is very small, and thus y is very small.
We can assume that 0.48 ≫ y
and thus [HA] at eqm = (0.48 - y) M ≈ 0.48 M

At eqm:
Ka = [H⁺] [A⁻] / [HA]
y² / 0.48 = 2.0 × 10⁻⁶
y = √(0.48 × 2.0 × 10⁻⁶)
y = 9.80 × 10⁻⁴
[H⁺] at eqm = 9.80 × 10⁻⁴ M

pH = -log[H⁺] = -log(9.80 × 10⁻⁴) = 3.0