What is the pH of a 0.0125M acetic acid solution?

Refer to: https://depts.washington.edu/eooptic/links/acidstrength.html
Ka for CH₃COOH = 1.8 × 10⁻⁵

_____________ CH₃COOH(aq) + H₂O⁺(aq) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) ____ Ka = 1.8 × 10⁻⁵
Initial: _________ 0.0125 M ________________ 0 M ________ 0 M
Chanege: _________ -y M _________________ +y M _______ +y M
Equilibirum: __ (0.0125 - y) M

As Ka is very small, the dissociation of CH₃COOH is to a very small extent.
We can assume that 0.0125 ≫ y
[CH₃COOH] at equilibrium = (0.0125 - y) M ≈ 0.0125 M

At equilibrium: Ka = [CH₃COO⁻] [H₃O⁺] / [CH₃COOH]
1.8 × 10⁻⁵ = y² / 0.0125
y = √[(1.8 × 10⁻⁵) × 0.0125] M = 4.74 × 10⁻⁴

pH = -log[H₃O⁺] = -log(4.74 × 10⁻⁴) = 3.32