EJ Habina
Quantitative Anal. Lab
MW 2:00
Experiment 7: Comparison of Unknowns: Weak Acid Identification
Purpose:
The purpose of this experiment is to help us as students to better understand the
workings of acid-base chemistry through a guided laboratory procedure. The lab will help
us not only to understand how acids and bases react during a particular situation, but also
how to derive the calculations and how to complete such reactions as well.
Procedure:
1. Part 1
a. An unknown sample of weak acid was obtained. We know that weak acids
are titrated by a strong base, as well as potassium acid phthalate (KHP) is a
good primary standard to standardize the base. The amount of KHP required
in order to react with all the moles of base was then calculated
2. Part 2
a. Three amounts of KHP are massed based on the calculations in part 1 and
were each placed in a 250 mL Erlenmeyer flask. To each flask approx. 50 mL
of water is added and used to dissolve the amount of KHP for that sample.
For each sample’s titration, NaOH was added slowly in 2 mL additions with
pH readings taken at each increment. As the pH change began to become
greater, the additions of base were lowered to 0.5 – 1 mL until the change in
pH is minimal again. A spreadsheet and graphs of the data were created in
order to determine the equivalence and half-equivalence points of each
titration.
3. Part 3
a. Knowing that this weak acid – strong base titration and is a monoprotic
system, the equivalence point data is used to determine the amount of acid
present, or the pH. From the titrations in step 2 the pH corresponding to
before any base is added, at the equivalence point, and at 1 mL past the
equivalence point was calculated.
4. Part 6
a. After the questions in part 4 and part 5 were answered, the unknown weak
acid was then titrated just as in part 2. A spreadsheet and graphs of the data
were created in order to determine the equivalence and half-equivalence
points of each titration.
Data:
Trial # 1
Trial # 2
Trial # 3
Average
Mass of KHP (g)
0.6155
0.6061
0.6088
0.6101
1/2 Equivalence Point
(mL)
14.1
14.65
14.73
14.49
Equivalence Point (mL)
28.2
29.3
29.45
28.98
Trial # 1:
Trial # 2:
Trial # 3:
NaOH added
(mL)
pH
NaOH added
(mL)
pH
NaOH added
(mL)
pH
2
4.24
2
4.21
2
4.22
4
4.50
4
4.38
4
4.42
6
4.62
6
4.53
6
4.55
8
4.75
8
4.68
8
4.71
10
4.88
10
4.79
10
4.83
12
4.97
12
4.92
12
4.93
14
5.10
14
5.01
14
5.04
16
5.21
16
5.14
16
5.16
18
5.34
18
5.26
18
5.26
20
5.47
20
5.38
20
5.41
22
5.62
22
5.53
22
5.53
24
5.84
24
5.71
24
5.71
26
6.14
26
5.95
26
5.93
27
6.40
27
6.12
27
6.11
27.5
6.62
28
6.39
28
6.35
28
6.94
28.5
6.60
28.5
6.52
28.5
9.93
29
7.02
29
6.87
29
10.29
29.5
10.21
29.5
9.36
29.5
11.08
30
10.81
30
10.66
30
11.24
30.5
11.11
30.5
11.00
30.5
11.36
31
11.25
31
11.20
31
11.43
33
11.58
31.5
11.32
33
11.64
32
11.42
34
11.64
Trial #1: mL NaOH vs. pH
14
pH of solution
12
10
8
6
pH
4
2
0
0
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
mL NaOH Added
Trial #2: mL NaOH vs. pH
14
pH of solution
12
10
8
6
pH
4
2
0
0
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
mL of NaOH Added
Trial #3: mL NaOH vs. pH
14
pH of solution
12
10
8
6
pH
4
2
0
0
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
mL of NaOH Added
Mass of Unknown (g)
1/2 Equivalence
Point (mL)
Equivalence Point
(mL)
Trial # 1
NaOH Added
(mL)
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
35
35.5
36
36.5
37
37.5
38
38.5
39
39.5
40.5
42.5
pH
1.66
1.79
1.96
2.09
2.23
2.36
2.48
2.59
2.71
2.81
2.93
3.04
3.16
3.32
3.46
3.65
3.95
4.20
4.35
4.60
5.22
10.12
10.63
10.83
10.99
11.11
11.20
11.33
11.49
Trial # 1
0.501
Trial # 2
0.5045
Trial # 3
0.5017
Average
0.5024
18.425
18.10
22.25
19.592
36.85
36.20
44.50
39.183
Trial # 2
NaOH Added
(mL)
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
33
33.5
34
34.5
35
35.5
36
36.5
37
37.5
38
38.5
39.5
41.5
pH
1.83
1.98
2.17
2.34
2.47
2.60
2.73
2.83
2.95
3.04
3.14
3.24
3.35
3.47
3.60
3.84
3.99
4.09
4.19
4.31
4.58
4.80
5.62
10.31
10.74
10.95
11.07
11.17
11.33
11.53
Trial # 3
NaOH Added
(mL)
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
45
45.5
46
48
pH
1.66
1.85
2.04
2.29
2.48
2.63
2.77
2.91
2.96
3.03
3.09
3.16
3.21
3.25
3.32
3.42
3.53
3.63
3.79
3.95
4.26
5.50
10.57
10.77
10.91
11.29
pH of solution
Trial #1: mL NaOH Added vs. pH
14
12
10
8
6
4
2
0
pH
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46
mL NaOH Added
Trial #2: mL NaOH Added vs. pH
pH of solution
15
10
5
pH
0
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44
mL NaOH Added
Trial #3: mL NaOH Added vs. pH
pH of solution
12
10
8
6
pH
4
2
0
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52
mL NaOH Added
pH
1 mL after
Equivalence
Point
MW
(g/mol)
pKa
Ka
Trial #
Before NaOH added
At Equivalence
Point
1
3.320
8.997
8.9995
135.96
2.73
1.86E-03
2
3.319
8.98
8.988
139.36
3
1.00E-03
3
3.318
8.98
8.992
113.9
3.18
6.60E-04
Average
3.319
8.986
8.993
129.74
2.97
1.17E-03
Calculations:
Grams of KHP required for 30 mL of Titrant:
0.1 𝑀 𝑁𝑎𝑂𝐻 =
0.1 𝑀 0.03 𝐿
×
= 0.003 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
1𝐿
1
𝑚𝑜𝑙𝑒𝑠 𝑎𝑐𝑖𝑑 = 𝑚𝑜𝑙𝑒𝑠 𝑏𝑎𝑠𝑒
0.003 𝑚𝑜𝑙𝑒𝑠 𝐾𝐻𝑃 ×
pH:
204.2212 𝑔
= 0.6127 𝑔 𝐾𝐻𝑃
1 𝑚𝑜𝑙𝑒
Before adding NaOH:
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐾𝐻𝑃 𝑢𝑠𝑒𝑑 ×
1 𝑚𝑜𝑙 𝐾𝐻𝑃
= 𝑚𝑜𝑙𝑒𝑠 𝐾𝐻𝑃 = 𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
204.2212 𝑔
𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
= 𝑀 𝐻𝑃−
𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒
𝐾𝑎 =
[𝑥 2 ]
→ 𝑥 = [𝐻]
[𝐻𝑃 − 𝑥]
𝑝𝐻 = −log[𝐻]
Ex:
0.6155 𝑔 ×
1 𝑚𝑜𝑙 𝐾𝐻𝑃
= 0.003013 𝑚𝑜𝑙𝑒𝑠 𝐾𝐻𝑃 = 0.003013 𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
204.2212 𝑔
0.003013 𝑚𝑜𝑙𝑒𝑠 𝐻𝑃
= 0.0603 𝑀 𝐻𝑃−
0.050 𝐿
3.9 × 10−6 =
[𝑥 2 ]
→ 𝑥 = [𝐻] = 0.000483 𝑀
[0.0603 − 𝑥]
𝑝𝐻 = −log[0.000483] = 3.32
At the Equivalence Point
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐾𝐻𝑃 𝑢𝑠𝑒𝑑 ×
1 𝑚𝑜𝑙 𝐾𝐻𝑃
= 𝑚𝑜𝑙𝑒𝑠 𝐾𝐻𝑃 = 𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
204.2212 𝑔
𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
= 𝑀 𝑃−2
𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒
𝐾𝑤
10−14
𝐾𝑏 =
=
𝐾𝑎 3.9 × 10−6
𝐾𝑏 =
[𝑥 2 ]
→ 𝑥 = [𝑂𝐻]
[𝑃−2 − 𝑥]
𝑝𝑂𝐻 = −log[𝑂𝐻] → 14 − 𝑝𝑂𝐻 = 𝑝𝐻
Ex:
0.6155 𝑔 ×
1 𝑚𝑜𝑙 𝐾𝐻𝑃
= 0.003013 𝐾𝐻𝑃 = 0.003013 𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
204.2212 𝑔
0.003301 𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
= 0.0385 𝑀 𝑃−2
0.050 𝐿 + 0.0282 𝐿
𝐾𝑏 =
𝐾𝑤
10−14
=
= 2.564 × 10−9
𝐾𝑎 3.9 × 10−6
2.564 × 10−9 =
[𝑥 2 ]
→ 𝑥 = [𝑂𝐻] = 9.93 × 10−6
[0.0385 − 𝑥]
𝑝𝑂𝐻 = −log[9.93 × 10−6 ] → 14 − 5.003 = 𝑝𝐻 = 8.997
At 1 mL After Equivalence Point
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐾𝐻𝑃 𝑢𝑠𝑒𝑑 ×
1 𝑚𝑜𝑙 𝐾𝐻𝑃
= 𝑚𝑜𝑙𝑒𝑠 𝐾𝐻𝑃 = 𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
204.2212 𝑔
𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
= 𝑀 𝑃−2
𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒
𝐾𝑏 =
𝐾𝑤
10−14
=
𝐾𝑎 3.9 × 10−6
𝐾𝑏 =
[𝑥 2 ]
→ 𝑥 = [𝑂𝐻]
[𝑃−2 − 𝑥]
𝑝𝑂𝐻 = −log[𝑂𝐻] → 14 − 𝑝𝑂𝐻 = 𝑝𝐻
Ex:
0.6155 𝑔 ×
1 𝑚𝑜𝑙 𝐾𝐻𝑃
= 0.003013 𝑚𝑜𝑙𝑒𝑠 𝐾𝐻𝑃 = 0.003013 𝑚𝑜𝑙𝑒𝑠 𝐻𝑃 −
204.2212 𝑔
0.003013 𝑚𝑜𝑙 𝐻𝑃 −
= 0.03804 𝑀 𝑃−2
0.05𝐿 + 0.0282𝐿 + 0.001𝐿
𝐾𝑏 =
𝐾𝑤
10−14
=
= 2.564 × 10−9
𝐾𝑎 3.9 × 10−6
2.564 × 10−9 =
[𝑥 2 ]
→ 𝑥 = [𝑂𝐻] = 9.88 × 10−6
[0.03804 − 𝑥]
𝑝𝑂𝐻 = −log[9.88 × 10−6 ] → 14 − 5.0005 = 𝑝𝐻 = 8.9995
How does the incomplete ionizing of weak acids effect the overall result when
calculating both pH and concentration values?
Since weak acids do not ionize completely, the concentrations of each species must be
calculated using a ICE table. This is due to the fact that the acid is not completely disassociated
and therefore any change in species is significant.
Predict the shape of the curve when titrating a weak acid with a strong base
The shape of the curve will be more gradual than that say with KHP. Not only will it be
more gradual, but also the “jump” which will occur when it reaches equivalence will be much
smaller.
Molar Mass of Unknown:
𝐿 𝑁𝑎𝑂𝐻 𝑎𝑑𝑑𝑒𝑑 𝑎𝑡 𝐸𝑞𝑢𝑖𝑣. 𝑃𝑡. × 𝑀 𝑁𝑎𝑂𝐻 = 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
Ex:
𝑚𝑎𝑠𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
= 𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑈𝑛𝑘𝑛𝑜𝑤𝑛
𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
0.03685 𝐿 × 0.1 𝑀 𝑁𝑎𝑂𝐻 = 0.003685 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
0.5010 𝑔
= 135.96 𝑔/𝑚𝑜𝑙
0.003685
Ka of Unknown:
𝐴𝑡 ℎ𝑎𝑙𝑓 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑐𝑒 𝑝𝑜𝑖𝑛𝑡 𝑝𝐾𝑎 = 𝑝𝐻
𝐾𝑎 = 10−𝑝𝑘𝑎
Ex:
𝐴𝑡 ℎ𝑎𝑙𝑓 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑐𝑒 𝑝𝑜𝑖𝑛𝑡 𝑝𝐾𝑎 = 𝑝𝐻 = 2.73
𝐾𝑎 = 10−𝑝𝑘𝑎 = 10−2.73 = 1.86 × 10−3
Identity of Unknown B:
Possible identity: Salicylic Acid (4-Hydroxybenzoic Acid)
Real Molar Mass: 138.12 g/mol
Real pKa: 2.80
Conclusion:
In this experimental procedure, the goal was to identify an unknown weak acid
through titration by a strong base. After completing the procedure, one possible identity of
Unknown B was found to be Salicylic Acid. Its actual molar mass of 138.12 g/mol as very
close to the calculated molar mass, which was 135.96 g/mol. Also, the actual pKa value for
salicylic acid of 2.80 is close to that of the calculate value, which was 2.73. Overall, the
proximity of the actual values for salicylic acid to that of the calculated values makes this
the most probably identity of the unknown sample.