Question

50mL of solution in which the analytical concentration of benzoic acid is 0.124 M is titrated with 0.261 M NaOH; if V_e is the volume needed to reach the endpoint, what is the pH at 0 Ve, 1/4 Ve, 1/2 Ve, Ve, and 1.212 Ve?

Answer 1

a) pH at Ve = 0

benzoic acid <==> benzoate- + H+

let x amotunt has dissociated,

Ka = [benzoate-][H+]/[benzoic acid]

6.4 x 10^-5 = x^2/0.124

x = [H+] = 2.82 x 10^-3 M

pH = -log[H+] = 2.55

b) pH at 1/4 Ve

moles of benzoic acid = 0.124 M x 50 mL = 6.2 mmol

moles of NaOH added = 6.2/4 = 1.55 mmol

Volume of NaOH added = 1.55/0.261 = 5.94 ml

[benzoic acid] = 6.2-1.55/55.94 = 0.083 M

[benzoate] = 1.55/55.94 = 0.03 M

pH = pKa + log([base]/[acid])

     = 4.20 + log(0.03/0.083)

     = 3.76

c) pH at 1/2 Ve

This is half equivalence point,

moles of acid present = moles of salt formed

pH = pKa = 4.20

d) pH at Ve

moles of acid = moles of NaOH

moles of benzoate = 0.124 M x 50 mL = 6.2 mmol

moles of NaOH added = 6.2 mmol

Volume of NaOH added = 6.2/0.261 = 23.75 ml

[benzoate] = 6.2/73.75 = 0.084 M

benzoate- + H2O <==> benzoic acid + OH-

let x amount has hydrolyzed,

Kb = Kw/Ka = 1 x 10^-14/6.4 x 10^-5 = x^2/6.4 x 10^-5

x = [OH-] = 3.62 x 10^-6 M

pOH = 5.44

pH = 14 - pOH = 8.56

e) pH at 1.212Ve

moles of benzoic acid = 0.124 M x 50 mL = 6.2 mmol

moles of NaOH added = 6.2 x 1.212 = 7.51 mmol

excess NaOH = 1.31 moles

Volume of NaOH added = 7.51/0.261 = 28.77 ml

[NaOH] = 1.31/78.77 = 0.017 M

pOH = 1.78

pH = 14 - pOH = 12.22