Experiment 7: Comparison of Unknowns: Weak Acid Identification
October 24, 2011
Objective:
The purpose of this experiment is to become more familiar with acid-base chemistry. The
necessary equations will need to be derived for this experiment and the groups with the
same unknown need to be identified.
Procedure
Part 1:
1) Obtain a weak acid unknown
2) Titrate the acid with a strong base
3) Calculate the correct amount of KHP to use
Part 2:
1) Fill weighing bottle halfway with KHP from oven and place it in dessicator to
cool
2) Weigh approximate amount of calculated KHP and place in a 250mL Erlenmeyer
(3 samples)
3) Standardize pH meter
4) Dissolve KHP in 50mL of water and add base in 2-3mL additions, checking pH
5) Graph mL of base added vs. pH and identify equivalence point and half
equivalence point
Part 3:
1) How many mL of base were used, which is determined by the equivalence point,
is equal to the number of moles present
2) Calculate the pH before any base is added, at the equivalence point, and at the
half equivalence point
Part 4:
1) Take into account the fact that a strong acid is one that is 100% ionized and
calculate the pH and concentrations
Part 5:
1) Find the acid dissociation constant
2) Predict the shape of the curve when dealing with a weak acid, titrated by a strong
base
Part 6:
1) Titrate the weak acid
2) Find the 2 points on this titration that will help identify the unkown
Part 7:
1) Calculate the MW and Ka of the unknown acid
Data
Grams of KHP (g)
Equivalence point
(mL)
Half equivalence
point (mL)
Trial 1
0.6131
Trial 2
0.6120
Trial 3
0.6135
28.50
25.5
26.5
14.25
12.75
13.25
Average
0.6130
26.8
13.41
Trial
#1
mL
added
0
3
6
9
12
15
18
21
24
27
28
28.5
29
29.5
30
32
34
37
pH
4.02
4.33
4.57
4.77
4.95
5.12
5.30
5.52
5.82
6.32
6.90
9.20
10.69
11.03
11.17
11.55
11.73
11.89
Trial
#2
mL
added
0
5
10
15
20
25
27
28
29
29.5
30
31
33
36
39
pH
4.07
4.48
4.84
5.11
5.42
5.85
6.18
6.46
8.31
10.68
11.0
11.28
11.59
11.81
11.94
Trial
#3
mL
added
0
4
8
11
14
17
20
23
26
27
27.5
28
28.5
29
29.5
30
32
34
37
pH
4.01
4.40
4.70
4.88
5.05
5.23
5.41
5.65
6.04
6.21
6.36
6.58
6.95
9.61
10.65
11.02
11.48
11.69
11.88
Unknown Weak Acid:
Grams of KHP (g)
Equivalence point
(mL)
Half equivalence
point (mL)
Trial
#1
mL
added
0
1
3
4
5
6
7
8
9
10
11
12
pH
2.94
3.31
3.89
4.11
4.35
4.68
5.39
10.81
11.23
11.41
11.56
11.67
Trial 1
0.1008
Trial 2
0.10158
Trial 3
0.1060
8.0
9.0
7.5
4.0
4.5
3.75
Trial#2
Trial#3
mL
mL
added
pH
added
pH
0
3.01
0
3.00
1
3.40
1
3.39
2
3.69
2
3.69
3
3.94
3
3.93
4
4.18
4
4.14
5
4.43
5
4.38
6
4.74
6
4.70
6.5
4.93
6.5
4.92
7
5.38
7
5.25
7.5
10.20
7.1
5.30
8
11.03
7.2
5.41
9
11.43
7.3
5.54
10
11.61
7.4
5.75
11
11.75
7.5
6.28
12
11.84
7.6
9.42
7.8
10.13
8
11.33
9
11.54
10
11.69
11
11.79
Average
0.1028
8.17
4.08
pH
Trial #1
Trial#2
Trial#3
Average
Std. Dev
Before
Based
added
3.750
3.625
3.775
3.717
0.003215
At
equivalenc
e point
7.5
7.25
7.55
7.43
0.003215
1mL after
equivalence
point
11.11
11.13
11.10
11.11
0.000577
*pKa – pH at half-euivalence point
Calculations
Grams of KHP to produce 25 mLs of Titrant
0.1M = moles base = .003mol
.030L
moles of acid=moles of base
.003mol KHP x 204.2212g = .6127g
Molarity of NaOH
(Moles of acid=Moles of Base) – mol KHP = .003mol
Molar
Mass
(g/mol)
134.4
140.0
140.4
138.3
0.69048
pKa
4
4.06
4.04
4.03
0
Ka
1x10-4
8.71x10-5
9.12x10-5
8.92x10-5
0
.003 moles NaOH = .105M NaOH
.0285 L NaOH
pH: (Using Trial 1 for example calculations)
Before Base added:
1 mol KHP x .6131g KHP = .003mol KHP = mol HP204.2212g
.003 mol HP- = 0.060M HP.05L
I
C
E
0.0500
-x
0.0500-x
+x
+x
+x
+x
3.96 x 10-6 = x2
0.0500-x
x=4.45 x 10-4
-log(4.45 x 10-4) = 3.35
At equivalence point:
1 mol KHP x .6131g = .003mol KHP = mol HP204.2212g
.003 mol KHP = .0382M P2.0785L
I
C
E
0.0382
-x
0.0382-x
+x
+x
+x
+x
Kw= 1 x 10-14
Kb = 2.56 x 10
2.56 x 10-9 =
-9
x2
.03382-x
x=9x10-6
-log(9x10-6) = 5.04 = pOH
14-5.04 = 8.96 = pH
1 mL after equivalence point:
1 mol KHP x .6131g KHP = .003mol KHP = mol HP204.2212g
.003mol HP- = .0377M P2.0795L
I
C
E
0.0377
-x
0.0377-x
Kb = 2.53x10-9
+x
+x
+x
+x
2.53x10-9 =
x2
.0377-x
x= 9.81x10-6
= -log(9.81x10-6) = 5.01
14-5.01 = 8.99 = pH
An ice table is required for a weak acid because the formal concentration taken into
consideration is not completely dissociated. The Ka is used to determine how much is
dissociated. Strong acids completely dissociate.
A small jump is predicted in the curve for a weak acid titrated by a strong base
Molar Mass of Unknown:
7.5L x 0.1M NaOH = .75mol NaOH
.1008g = 134.4g unk
.75mol
Ka of unknown:
At half-equivalence point, pH = pKa = 4
Ka = 1x10-4
Conclusion
The unknown acid was found to be benzoic acid. This was based on the molecular
weight, pKa, and Ka values. The molecular weight was a little off at 122g as well as the
Ka value of 6.28x10-5. The pKa value, however, was relatively close at 4.202.