Question

Will someone please answer the THIRD question underneaththefirst two, thank you

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

Titration

3) Equivalence point

Citric acid is a triprotic acid

Ist equivalence point

Volume NaOH added = 0.0416 M x 50 ml/0.1288 M = 16.15 ml

[H2citr-] formed = 0.0416 M x 50 ml/66.15 ml = 0.03 M

Ka2 = 1.7 x 10^-5 = x^2/0.03

x = [H3O+] = 7.14 x 10^-4 M

pH = -log[H3O+] = 3.15

IInd equivalence point

Volume NaOH added = 2 x 0.0416 M x 50 ml/0.1288 M = 32.30 ml

[Hcitr^2-] formed = 0.0416 M x 50 ml/82.30 ml = 0.025 M

Ka3 = 6.4 x 10^-6 = x^2/0.025

x = [H3O+] = 4 x 10^-4 M

pH = -log[H3O+] = 3.40

IIIrd equivalence point

Volume NaOH added = 3 x 0.0416 M x 50 ml/0.1288 M = 48.45 ml

[citr^3-] formed = 0.0416 M x 50 ml/98.45 ml = 0.021 M

citr3- + H2O <==> Hcitr^2- + OH-

Kb1 = 1 x 10^-14/6.4 x 10^-6 = x^2/0.021

x = [OH-] = 5.74 x 10^-6 M

pOH = -log[OH-] = 5.24

pH = 14 - pOH = 8.76