Question

Below is the data for three titrations. In this experiment we have 25 mL (approx. 0.03M) of various monoprotic acids and we slowly add 50 mL of NaOH while monitoring pH. We add NaOH to change pH about 0.15 after each addition. We add unti lthe total 50 mL of base solution (NaOH) is added. Knowing this, for each trial, what is the equivalence point(s) and the pK_a

Run 1
mL	pH
0	1.89
3	1.89
6	2.3
7	2.29
9	2.48
12	2.83
14	3.59
15.5	10.23
16.5	11.69
18	11.97
20	12.12
22	12.27
27	12.4
31	12.46
34	12.54
38	12.58
43	12.66
50	12.68
Run 2
mL	pH
0	3.86
1	3.8
3	4.1
4	4.29
4.3	4.48
6	4.61
7.8	4.76
9.4	4.88
11.3	5.01
12.8	5.13
14.8	5.27
16.8	5.44
18.4	5.64
19.8	5.8
21.6	6.02
21.8	6.18
22.1	6.33
22.9	6.78
23.4	7.62
24.1	10.25
24.5	11.24
25.8	11.64
29.5	12.02
33	12.18
36.1	12.3
46	12.46
50	12.52
Run 3
mL	pH
0	4.06
5	4.6
8	4.77
10	4.91
13	5.12
15.4	5.26
16.5	5.42
17.7	5.52
18.8	5.62
20	5.8
21	5.96
22	6.35
23	6.74
23.8	7.8

Answer 1

Plot the data for the three curves graphically as shown below:

Plot 1: Run 1

The pH at the equivalence point is roughly equal to the mid-point of the steep portion of the curve and can be easily found as

pH = (3.59 + 10.23)/2 = 6.91 (ans).

The pK_a of the acid can be easily found out from the initial pH of the acid (when no NaOH is added). Given the initial pH as 1.89, we have, pH = -log [H⁺] = 1.89

===> [H⁺] = 10^-1.89 = 0.01288

Write down the dissociation equation as

HA (aq) <====> H⁺ (aq) + A^- (aq)

At equilibrium, [H⁺] = [A^-] = 0.01288 M and [HA] = (0.03 – 0.01288) M = 0.01712 M

K_a = [H⁺][A^-]/[HA] = (0.01288)²/(0.01712) = 9.69*10^-3 ≈ 9.70*10^-3 (ans).

Plot 2: Run 2

Again, pH at the equivalence point = (7.62 + 10.25)/2 = 8.935 (ans).

Find the K_a from the initial pH as before.

Initial pH = 3.86; =====> [H⁺] = [A^-] = 10^-3.86 = 1.380*10^-4 M and [HA] ≈ 0.03 M (since only a small portion of HA is ionized, we can assume the equilibrium concentration of HA to be virtually equal to the initial concentration).

Therefore,

K_a = (1.380*10^-4)²/(0.03) = 6.348*10^-7 ≈ 6.35*10^-7 (ans).