Question

Trial 1

mmol of NaOH	pH
0.0	2.03
4.77	2.24
6.05	2.62
6.91	2.88
7.28	3.28
7.38	3.44
7.47	6.55
7.66	6.84
7.76	7.06
7.85	7.22
8.04	7.40
8.56	7.78
8.89	8.00
9.18	8.16
9.60	8.41
9.74	8.61
10.17	8.95
10.45	9.24
10.74	9.52
11.16	9.75
11.59	9.99
11.92	10.19
12.29	10.42
12.53	10.67
12.91	10.86
13.19	11.08
13.76	11.32
14.66	11.52
16.18	11.73
17.50	11.90
18.21	12.10

Trial 2

mmol of NaOH	pH
0.0	1.94
5.58	2.18
6.19	2.38
6.67	2.66
6.95	2.84
7.19	3.15
7.33	3.62
7.42	6.08
7.57	6.85
7.71	7.03
7.94	7.24
8.14	7.40
8.56	7.81
8.99	8.04
9.37	8.30
9.65	8.53
9.93	8.77
10.07	8.95
10.26	9.14
10.45	9.33
10.88	9.60
11.30	9.84
11.59	10.06
12.30	11.50
16.08	11.70
17.50	11.90
18.49	12.00

Plot a titration cure of mmol of NaOH versus pH with each trial on the same graph. Determine the start point of the titration on the graph of each trial as well as pKa1, pKa2 and pKa3 (if necessary) and determine the isoelectric point of the amino acid titrated

Answer 1

The titration curve is plotted and the isoelectric point is identified i.e the pH at which zwitter ionic form is present. It is the pH at which amino acid is neutral.

pK_a1  is observed from the loss of H+ from the acidic carboxyl group at low pH and pK_a2 is observed from the loss of H⁺ from the more basic amino group at high pH.

In the buffer region i.e below isoelectric point, pH= pKa according to the Henderson–Hasselbalch equation.

By extrapolating the midpoint of each buffering region i.e of the plateau, pKa can be found out.

From graph,

pKa1= 2.8 pKa2= 9.2 pI= (pKa1 + pKa2)/2 = (2.8 + 9.2)/2 = 6.0