if we make a buffer by mixing 0.030 mol hcl with 0.050 mol ch3coo-(aq) in 2.00L, what is the pH?

HCl is a strong acid. 0.03 mol HCl gives 0.03 mol H₃O⁺ in water.

CH₃COO⁻(aq) + H₃O⁺(aq) → CH₃COOH(aq) + H₂O(l)
After reaction :
[CH₃COOH] = 0.030/2 M = 0.015 M
[CH₃COO⁻] = (0.050 - 0.030)/2 M = 0.010 M

Consider the dissociation of CH₃COOH :
CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) .... Kₐ = 1.8 × 10⁻⁵

At equilibrium :
pH = pKₐ - log([CH₃COOH]/[ CH₃COO⁻])
pH = -log(1.8 × 10⁻⁵) - log(0.015/0.010)
pH = 4.6