Question

A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH after the addition of 13.0 mL of NaOH? The K_a of benzoic acid is 6.3x10^-5.

Answer 1

Volume of solution = V = 25 m/ Molarity of benzoic acid = 0.15 M therefore Moles of benzoic acid = Vol. In Lip) times Molarity = 25/100 times 0.15 = 3.75 times 10^-3 moles Similarly, Moles of NaOH added = 0.15 times 13/1000 = 1.95 times 10^-3 moles HA rightarrow benzoic acid therefore HA + OH^- rightwardsharpoonoverleftwardsharpoon H_2O + A^circledminus Initially (3.75 times 10^-3) (3.75 times 10^-3) At equilibrium (3.75 - 1.95) times 10^-3 - 1.95 times 10^-3 1.95 times 10^-3 therefore Moles of HA remaining = (3.75 - 1.95) times 10^-3 at equilibrium = 1.8 times 10^-3 at equilibrium = 1.8 times 10^-3 therefore P^H = P^ka + log ([A^-]/[MA]) Given Pk_a = -log_10 (k_a) = -log_10 (6.3 times 10^-5) = 4.2 therefore P^H = 4.2 + log (1.95 times 10^-3/1.8 times 10^-3) therefore P^H = 4.235