Question

#2

What is the purpose of this experiment? The following data were collected in the titration of 10.0 mL of 0.10 M weak acid, HA, with 0.10 M NaOH solution. Tabulate and plot three graphs: titration curve, first derivative and second derivative, and find the equivalence point. An example of the procedure ids given below V' is the average of two consecutive volume: 21.00 + 20.50/2 = 20.75 and f' is calculated by taking the difference in pH and dividing by the difference in volume of NaOH

Answer 1

Titration curves are pH vs Volume of titrant.

mL 0.10 M NaOH	pH	v'(mL)	f'($\Delta$pH/v')	v''(mL)	f''($\Delta$($\Delta$pH/v'')
0.00	3.00	0.5	1.05	1.125	-0.5893333333
1.00	4.05	1.75	0.3133333333	2.75	-0.0606666667
2.50	4.52	3.75	0.192	5	0
5.00	5.00	6.25	0.192	7.1875	0.0554666667
7.50	5.48	8.125	0.296	8.625	0.2773333333
8.75	5.85	9.125	0.5733333333	9.375	1.3333333333
9.50	6.28	9.625	1.24	9.725	7.4666666667
9.75	6.59	9.825	2.7333333333	9.8875	126.1333333333
9.90	7.00	9.95	18.5	10	0
10.00	8.85	10.05	18.5	10.1125	-127.2
10.10	10.70	10.175	2.6	10.275	-7
10.25	11.09	10.375	1.2	11	-0.7146666667
10.50	11.39	11.625	0.3066666667	12.75	-0.0928395062
12.75	12.08	13.875	0.0977777778
15.00	12.30

The end point usually occurs at the point of maximum deflection, where the absolute value of the first derivative reaches a maximum and the second derivative changes sign

Titration Curve 14.00 12.00 10.00 8.00 6.00 4.00 2.00 0.00- 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 Volume of NaOH (mL)

Second Derivative Plot 150 100 50 0 0 2 4 6 8 12 14 100 150 V" (mL)

First Derivative Plot 20 18 16 14 12 エ10 4 2 0 2 4 6 10 12 14 16 v(mL)

So the equivalance point is 10.0 mL