I performed an experiment to find the equivalent mass and dissociation constant
ID: 839183 • Letter: I
Question
I performed an experiment to find the equivalent mass and dissociation constant of an unknown weak acid by titrimetry.
My data is the following:
A.) Determining a satisfactory sample size for the unknown weak acid:
Mass of unknown acid, 0.12 grams
Final buret reading: 5.6 mL
Initial buret reading: 0.0mL
Volume of NaOH solution required: 5.6mL
Mass of unknown acid requiring 25mL of NaOH solution for titration: 0.54grams
B.) Titrating a sample of an unknown weak acid
Mass of weighing paper plus unknown acid: 0.91grams
Mass of weighing paper: 0.37 grams
Mass of unknown acid: 0.54grams
C.) Buret readings of first and second determinations:
First determination
Buret readings, mL | pH
0.0 / 1.51
1.0 / 1.51
1.5 / 2.47
2.0 / 2.78
2.5 / 2.94
3.0 / 3.09
3.5 / 3.14
4.0 / 3.21
4.5 /3.25
5.0 / 3.32
5.5 / 3.36
6.0 / 3.41
6.5 / 3.46
7.0 / 3.51
7.5 / 3.55
8.0 / 3.58
8.5 / 3,63
9.0 / 3.66
11.5 / 3.71
12.0 / 3.75
12.5 / 3.78
13.5 / 3.82
14.0 / 3.90
14.5 / 3.96
15.5 / 3.99
16.0 / 4.04
16.5 / 4.06
17.0 / 4.11
17.5 / 4.15
18.0 / 4.18
18.5 / 4,33
19.0 / 4.45
19.5 / 4,54
20.0 / 4.69
21.5 / 4.87
23.5 / 5.06
25.0 / 5.45
27 / 11.95
28 / 12.45
29 / 12.65
30 / 12.79
Second determination: Buret reading / pH
0.0 /2.08
0.5 / 2.53
1.0 / 2,76
1.5 / 2.85
2.0 / 3.02
2.5 / 3.08
3.0 / 3.17
3.5 / 3.25
4.0 / 3.28
4.5 / 3.30
5.0 / 3.33
5,5 / 3.36
6.0 / 3.40
6.5 / 3.42
7.0 / 3.47
7.5 / 3.50
8.0 / 3.55
9.5 / 3.58
11.0 / 3.62
12.0 / 3.66
12.5 / 3.70
13.5 / 3.74
14.0 / 3.78
14.5 / 3.82
15.0 / 3.85
15.5 / 3.89
16.0 / 3.95
16.5 / 3.98
17.0 / 4.02
17.5 / 4.05
18.0 / 4.09
18.5 / 4.18
19.0 / 4.27
19.5 / 4.38
20.0 / 4.48
21.5 / 4.60
23.0 / 4.75
24.5 / 4.92
25.0 / 5.17
25.5 / 5.38
26.0 / 5.57
26.5 / 5.96
27.0 / 7.60
28.0 / 11.70
Given that data, find the following values for BOTH determinations 1 AND 2:
Volume of NaOH solution required to reach equivalence point, mL:
Number of equivalents of NaOH required to reach equivalence point, equiv:
Number of equivalents of unknown acid titrated, equiv:
Equivalent mass of unknown acid, g/equiv:
pKa of unknown acid:
Ka of unknown acid:
Explanation / Answer
Equivalence point is the point after which there is a sharp in the pH of the titration .
So from the given information ,
Equivalence point is at pH = 7.60
So the volume of NaOH added to reach equivalence point is 27 ml
At half equivalence point
pH = pKa
half equivalnce point is at 13.5 ml
so
pH at 13.5 ml is 3.74
so
pKa = 3.74
we know that
pKa = -log Ka
3.74 = -log ka
Ka = 1.8197 x 10-4