Question

50.00 mL 0.0416 M citric acid is titrated with 0.1288 M NaOH.

1) Calculate the equivalence point volume of the NaOH solution. (4 pts)

2) Identify the species, at the equivalence point, which determines the pH of the titration solution and calculate its concentration. (6 pts)

3) Calculate the pH of the titration solution at the equivalence point. (10 pts)

Answer 1

1)

citric acid is triprotic acid = H3A

H3A   + 3 NaOH -----------------------------> Na3A + 3H2O

C1V1 / n1 = C2 V2 / n2

0.0416 x 50 / 1 = 0.1288 x V2 / 3

V2 = 48.45 mL

volume of NaOH needed = 48.45 mL

b)

at equivalence point total volume = 50.00 + 48.45 = 98.45 mL

millimoles of A-3 = 50 x 0.0416 = 2.08

A-3 concentration = 2.08 / 98.45

                            = 0.0211 M

at equivalence point only citrate ion is present .

citrate ion concentration = 0.0211 M

3)

A-3   + H2O --------------------------> HA-2 + OH-

0.0211                                       0            0

0.0211-x                                     x           x

Kb1 = [HA-2][OH-]/ [A-3]

1.0 x 10^-14 / 4.0 × 10⁻⁷ = 2.5 x 10^-8 = x^2 / 0.0211 - x

x^2 + 2.5 x 10^-8 x - 5.28 x 10^-10 = 0

x = 2.30 x 10^-5

[OH-] = 2.30 x 10^-5 M

pOH = -log [OH-]

pOH = -log (2.30 x 10^-5)

pOH = 4.64

pH + pOH = 14

pH = 9.36