Question

3. The following information is provided for oxalic acid: H1ャNatCA H2C2Oa + H2O <-> H30' +HC204" Kal= 5.4 x 10" d) Sketch the shape of the pH titration curve of the 0.200 M Na,C20, with 0.200 M HCI. i) Label the pH of the solution at the beginning of the pH titration ii) Label the pH of the solution at the first and second equivalence. il) Label the pKbı and pKaz on the curve.

Answer 1

3. Titration

i) initial pH

C2O4^2- + H2O <==> HC2O4- + OH-

Kb1 = 1 x 10^-14/5.4 x 10^-5 = x^2/0.2

x = [OH-] = 1.36 x 10^-5 M

pOH = -log[OH-] = 4.87

pH = 14 - pOH = 9.13

ii) pH at first equivalence point

= 1/2(pKa1 + pKa2) = 1/2(1.27 + 4.27) = 2.77

pH at second equivalence point

with 50 mlNa2C2O4 solution

[H2C2O4] formed at second eq. point = 0.2 M x 50 ml/100 ml = 0.1 M

H2C2O4 + H2O <==> H3O+ + HC2O4-

Ka1 = 5.4 x 10^-2 = x^2/0.1

x = [H3O+] = 0.0735 M

pH = -log[H3O+] = 1.134

iii) pKab1 is first half-equivalence point

pKb2 is second half-equivalence point

Points

HCl(ml)	pH
0	8-Jan
10	4.87
25	4.27
50	2.77
75	1.27
100	1.134