Help with pH TITRATION INQUIRY CHALLENGE #1: Use the equivalence volume from you
ID: 964271 • Letter: H
Question
Help with pH TITRATION INQUIRY
CHALLENGE #1: Use the equivalence volume from your graphs to calculate the formula mass of the unknown acid.
CHALLENGE #2: Use the half-equivalence volume and corresponding pH from your graphs to calculate the Ka of the unknown acid.
CHALLENGE #3: Calculate the formula mass of the unknown acid using the manual titration volume of NaOH used to reach an endpoint (procedure step #3) and compare it with the formula mass calculated from the pH meter titration.
Data points
mL vs pH
0
3.84
2
4.11
4
4.37
5
4.46
6
4.56
7.5
4.71
8
4.75
8.5
4.81
9
4.86
9.5
4.89
10
4.94
10.5
4.99
11
5.02
11.5
5.07
12
5.12
12.5
5.16
13
5.2
13.5
5.25
14
5.3
14.5
5.35
15
5.4
15.5
5.45
16
5.51
16.5
5.56
17
5.63
17.5
5.71
18
5.77
18.5
5.86
19
5.94
19.5
6.06
20
6.22
20.5
6.42
21
6.71
21.2
7.03
21.5
8.97
22
10.18
23
10.68
23.5
10.81
24
10.89
24.5
10.97
25
11.02
mL vs pH
0
3.84
2
4.11
4
4.37
5
4.46
6
4.56
7.5
4.71
8
4.75
8.5
4.81
9
4.86
9.5
4.89
10
4.94
10.5
4.99
11
5.02
11.5
5.07
12
5.12
12.5
5.16
13
5.2
13.5
5.25
14
5.3
14.5
5.35
15
5.4
15.5
5.45
16
5.51
16.5
5.56
17
5.63
17.5
5.71
18
5.77
18.5
5.86
19
5.94
19.5
6.06
20
6.22
20.5
6.42
21
6.71
21.2
7.03
21.5
8.97
22
10.18
23
10.68
23.5
10.81
24
10.89
24.5
10.97
25
11.02
mL vs pH 12 10 4 0 0 1 23 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 mL addedExplanation / Answer
1. The sharp rise in the curve is the equivalence point of the titration. Look at the graph, it is about 21.5 mL NaOH used to reach to the equivalence point of the titration. At the equivalence point, the number of moles of NaOH is required is exactly the same as you had the no of moles of weak acid in your solution.
You need to calculate the no of moles of base required
= concentration of base * volume of base
equals the number of moles of acid in the sample.
Divide the mass of the sample taken (of acid) by the number of moles to get the formula mass.
2. The flat portion of the curve is where the acid is getting neutralized by the base. The middle of the flat part of the curve is where the acid is half-neutralized. The pH at that point is the pKa of the weak acid.
The volume of base at the half-neutralization point will be around 21.5/2 or 11 mL . The pH of the solution is about 5.02 at that point . So, the pKa of the weak acid = 5.02
3. The manual titration volume of NaOH is required to answer the question.
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