Calculating pH of buffers?

Refer to: https://ch301.cm.utexas.edu/data/section2.php?target=ka-kb-constants.php
Ka for CH₃COOH (acetic acid) = 1.8 × 10⁻⁵

[CH₃COOH] = (5.0 M) × (15/2000) = 0.0375 M
[CH₃COO⁻] = (2.5 M) × (20/2000) = 0.025 M

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

Henderson-Hasselbalch equation:
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
pH = -log(1.8 × 10⁻⁵) + log(0.025/0.0375)
pH = 4.57

I am assuming that you means "20 mL of 2.5 M sodium acetate".

Equilibrium (letting HAc = acetic acid and Ac- = acetate ion)

HAc <--> Ac- + H+

Ka = 1.8X10^-5 = [H+][Ac-]/[HAc]

Concentration HAc = 15 mL X 5 M / 2000 mL = 0.0375 M HAc
Concentration Ac- = 20 mL X 2.5 M / 2000 mL = 0.025 M

Ka = 1.8X10^-5 = [H+](0.025)/0.0375
[H+] = 2.7X10^-5
pH = 4.57

You could also use the Henderson-Hasselbalch equation:
pH = pKa + log [Ac-]/[HAc]
pH = 4.74 + log (0.025/0.0375) = 4.56