How to make a 20ml solution with a pH of 4.85 given .1M Acetic Acid and .15M Sodium acetate? (no water can be added)?

CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) ….. pKa = 4.74

At equilibrium :
pH = pKa - log([CH₃COOH]/[CH₃COO⁻])
4.85 = 4.74 - log([CH₃COOH]/[CH₃COO⁻])
log([CH₃COOH]/[CH₃COO⁻]) = -0.11
[CH₃COOH]/[CH₃COO⁻] = 10⁻⁰·¹¹
[CH₃COOH]/[CH₃COO⁻] = 0.776
[CH₃COOH]ₒ/[CH₃COO⁻]ₒ ≈ 0.776

Let V ml be the volume of CH₃COO⁻ added.
Then, the volume of CH₃COOH added = (20 - V) ml

[CH₃COOH]ₒ/[CH₃COO⁻]ₒ ≈ 0.776
0.1(20 - V) / 0.15V) = 0.776
2 - 0.1V = 0.116V
0.216V = 2.0
V = 9.3

Volume of 0.15 M CH₃COONa added = 9.3 ml
Volume of 0.1 M CH₃COOH added = (20 - 9.3) ml = 10.7 ml