What mass of sodium benzoate should be added to 160.0 mL of a 0.15 M benzoic acid solution in order to obtain a buffer with a pH of 4.25?

Molar mass of C₆H₅COONa (sodium benzoate) = (12.0×7 + 1.0×5 + 16.0×2 + 23.0) g/mol = 144.0 g/mol
From Wikipedia, pKₐ of C₆H₅COOH = 4.20 (to 3 sig. fig.)

C₆H₅COOH(aq) + H₂O(l) ⇌ C₆H₅COO⁻(aq) + H₃O⁺(aq)

pH = pKa + log([C₆H₅COO⁻]/[ C₆H₅COOH])
4.25 = 4.20 + log([C₆H₅COO⁻]/0.15)
log([C₆H₅COO⁻]/0.15) = 0.05
[C₆H₅COO⁻]/0.15 = 10^0.05
[C₆H₅COO⁻] = 0.15 × (10^0.05) mol/L = 0.1683 mol/L

[C₆H₅COONa]ₒ ≈ [C₆H₅COO⁻] = 0.1683 mol/L

No. of moles of C₆H₅COONa added = (0.1683 mol/L) × (160.0/1000 L) = 0.2693 mol
Mass of C₆H₅COONa added = (0.2693 mol) × (144.0 g/mol) = 3.88 g