500.0 mL of 0.160 M NaOH is added to 535 mL of 0.200 M weak acid (Ka = 2.51 × 10-5). What is the pH of the resulting buffer?

Denote HA as the chemical formula of the weak acid.

Initial number of moles of NaOH = (0.160 mol/L) × (500.0/1000 L) = 0.0800 mol
Initial number of moles of HA = (0.200 mol/L) × (535/1000 L) = 0.107 mol

Consider the reaction between NaOH and HA.
NaOH(aq) + HA(aq) → NaA(aq) + H₂O(l) …. Kₐ

After reaction :
Number of moles of HA left = (0.107 - 0.0800) = 0.027 M
Number of moles of A⁻ formed = Number of moles of NaA formed = 0.0800 M
Then, [HA]/[A⁻] = 0.027/0.0800

pH = pKₐ - log([HA]/[A⁻])
pH = -log(2.51 × 10⁻⁵) - log(0.027/0.0800)
pH = 5.07 ≈ 5.1