Using the given data and pH at the half equivalence point, what is the pKa and K
ID: 952954 • Letter: U
Question
Using the given data and pH at the half equivalence point, what is the pKa and Ka of the reaction between acetic acid and NaOH? I tried to attach the titration Curve but couldnt. sorry:(
please show all steps and units
Acetic acid (0.1 M) and NaOH (0.1 M) were used.
Volume of NaOH delivered to Acetic Acid
pH
1
3.39
2
3.6
3
3.84
4
4
5
4.2
6
4.36
7
4.4
8
4.45
9
4.58
10
4.75
11
4.67
12
4.78
13
4.8
14
5
15
5.03
15.5
5.03
16
5.07
16.5
5.1
17
5.13
17.5
5.16
18
5.16
18.5
5.19
19
5.24
19.5
5.28
20
5.33
20.5
5.39
21
5.46
21.5
5.46
22
5.54
22.5
5.62
23
5.72
23.5
5.87
24
5.99
24.5
6.35
25
6.8
25.5
10.61
26
11.2
27
10.48
28
11.88
29
12.01
30
12.08
Volume of NaOH delivered to Acetic Acid
pH
1
3.39
2
3.6
3
3.84
4
4
5
4.2
6
4.36
7
4.4
8
4.45
9
4.58
10
4.75
11
4.67
12
4.78
13
4.8
14
5
15
5.03
15.5
5.03
16
5.07
16.5
5.1
17
5.13
17.5
5.16
18
5.16
18.5
5.19
19
5.24
19.5
5.28
20
5.33
20.5
5.39
21
5.46
21.5
5.46
22
5.54
22.5
5.62
23
5.72
23.5
5.87
24
5.99
24.5
6.35
25
6.8
25.5
10.61
26
11.2
27
10.48
28
11.88
29
12.01
30
12.08
Explanation / Answer
form the data :
at 9 ml of NaOH ------------> pH = 4.58
at 10 ml of NaOH ------------> pH = 4.75
at 11 ml of NaOH ------------> pH = 4.67
at 12 ml of NaoH -------------> pH = 4.78
from these readings the changes statered at 10 mL NaOH so . it is half-equivalence point.
at half-equivalence point. pH = pka
pH = 4.75 = pKa
pKa = 4.75
Ka = 10^-pKa
Ka = 10^-4.75
Ka = 1.8 x 10^-5
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