Calculate the pH of a solution prepared by mixing 15.0 mL of 0.10 M NaOH and 30.0 mL of a 0.10 M benzoic acid solution. (Benzoic acid is monprotic; its dissociation constant is 6.5 x 10^-5.).

Moles = molarity x mass

Moles of NaOH = (0.10 mol/L) (0.015 L) = 0.0015 mol

Moles of benzoic acid = (0.100 mol/L) (0.0300 L) = 0.003 mol

NaOH and benzoic acid react in a 1:1 molar ratio. This means they are a buffer.

After the reaction

moles of sodium benzoate = 0.00150 mol

moles of benzoic acid = 0.00150 mol

pH = pKa + log [base / acid]

pH = - log (6.5 x 10^-5)

pH = 4.187

pH = 4.19