Experiment 7: Comparison of Unknowns- Weak Acid Identification
Purpose:
The purpose of this seven part experiment is to introduce you to a deeper understanding of acid-base
chemistry.
Procedure:




Standardize a pH meter
Obtain an unknown weak acid
Weigh out 0.5100 g KHP for three trials and dissolve in 50mL distilled water
Make 0.1M NaOH to use for titrations
Weak Acid – Strong Base Titration




Titrate KHP solution for three good trials:
- Add 2-3mL of base and record pH after each addition
- When there is a dramatic change in pH, add base in 0.5-1mL increments. This was
continued until the change in pH slowed
- Base increments were increased back to 2mL to finish out the curve
The procedure was repeated until three good trials were obtained
Calculate the pH before base was added, at the equivalence point, and after the equivalence
point
Create graph of mL of base added vs. pH
Unknown Acid – Strong Base Titration




Measure out approximately 0.25g of unknown weak acid for three trials
Combine unknown weak acid with 50mL distilled water into a 250mL Erlenmeyer flask
Titrate with base using same procedure as above
Calculate the molecular weight and Ka of unknown acid
Data:
Part 2
Trial
1
2
3
Avg.
Std
Mass KHP
0.5104g
0.5106g
0.5100g
0.5103g
.0003
Equivalence Point
25 mL
25 mL
25 mL
25 mL
0
½ Equivalence Point
12.5 mL
12.5 mL
12.5 mL
12.5 mL
0
Molarity NaOH
0.09997 M
0.1000 M
0.09989 M
0.09995 M
Titration of KHP with NaOH
Trial Two
mL NaOH
mL NaOH
0
2.5
5
7.5
10
12.5
15
17.5
20
22.5
25
26.5
27
27.5
28
4.14
4.41
4.76
4.95
5.05
5.20
5.37
5.53
5.71
5.95
6.42
11.02
11.28
11.42
11.53
11.64
11.68
11.76
11.91
12.03
12.12
12.20
12.24
0
2.5
5
7.5
10
12.5
15
17.5
20
22.5
25
26.5
27
27.5
28
mL of base added vs. pH
Trial #1
14
12
10
pH
Trial One
pH
0
2.5
5
7.5
9
11.5
14
16.5
19
21.5
24
26.5
27
27.5
28
28.5
29
29.5
31
33
35
37
39
8
6
4
2
0
0
10
20
30
40
50
40
50
mL of base added
mL of base added vs. pH
Trial #2
14
12
10
pH
mL NaOH
8
6
4
2
0
0
10
20
30
mL of base added
Trial Three
mL NaOH
pH
0
3
6
9
12
15
18
21
24
26
26.5
27
27.5
28
28.5
29
31
33
35
37
Trial #1
Trial#2
Trial#3
Average
Std. Dev
28.5
29
29.5
30
32
34
36
38
12.28
4.23
4.51
4.81
5.05
5.22
5.41
5.60
5.90
6.71
10.84
11.30
11.38
11.41
11.53
11.58
11.73
11.89
12.04
12.13
12.18
mL of base added vs. pH
Trial #3
pH
28.5
29
29.5
30
32
34
36
38
40
14
12
10
8
6
4
2
0
0
10
20
30
mL of base added
pH Before Based
added
3.35
3.35
3.36
3.35
.007
pH At equivalence
point
8.97
8.97
8.97
8.97
0
pH 1mL after
equivalence point
12.53
12.53
12.53
12.53
0
40
Part 3
How is this calculation complicated by a weak acid?
The weak acid was used with a strong base, so you have to use the I.C.E. table.
Part 4
An acid is considered weak if it does not ionize completely remember that a strong acid is one that is
100% ionized. How is this taken into account in calculating pH and concentrations?
When calculating the pH and concentration, an I.C.E. table is needed to calculate the
concentrations, which will be used to determine the pH. A strong acid has the same pH as molarity so
fewer calculations are required.
Part 5
Predict the shape of the curve when dealing with a weak acid titrated by a strong base.
The shape of the curve will be s-shaped, but it will have a different starting point and
equivalence point than the KHP curve.
Part 6
Titration of Unknown with NaOH
1
2
3
Avg.
Std
Unknown
Mass
0.5063 g
0.5095 g
0.5093 g
0.5084 g
.002
Trial One
mL NaOH
pH
0
2.5
5
7.5
10
12.5
15
17.5
20
22.5
25
27
29
29.5
30
Equivalence
Point
28.2
28.1
28.3
28.2
0.1
2.86
3.18
3.52
3.76
3.82
3.97
4.11
4.25
4.41
4.61
4.91
5.36
11.04
11.24
11.41
½ Equivalence
Point
14.1
14.05
14.15
14.1
0.05
Mol mass
pKa
Ka
180.912
182.055
181.984
181.650
0.64
4.1
4.1
4.1
4.1
0
7.94×10-5
7.94×10-5
7.94×10-5
7.94×10-5
0
mL of base added vs. pH
Trial #1
14
12
10
pH
Trial
8
6
4
2
0
0
10
20
30
mL of base added
40
50
Trial Two
pH
0
3
6
9
12
15
18
21
24
27
29
29.5
30
30.5
33
35.5
38
40
42
2.95
3.41
3.66
3.83
4.00
4.14
4.31
4.51
4.79
5.40
11.06
11.28
11.46
11.48
11.87
12.03
12.13
12.18
12.20
Trial Three
mL NaOH
pH
0
3
6
9
12
15
18
21
24
27
29
3.02
3.44
3.67
3.88
4.00
4.19
4.36
4.54
4.82
5.35
10.31
mL of base added vs. pH
Trial #2
14
12
10
8
pH
11.50
11.58
11.60
11.89
12.03
12.12
12.20
12.23
6
4
2
0
0
10
20
30
40
50
mL of base added
mL of base added vs. pH
Trial #3
14
12
10
pH
mL NaOH
30.5
31
31.5
34
36.5
39
42
44
8
6
4
2
0
0
10
20
30
mL of base added
40
50
29.5
30
30.5
31
31.5
32
32.5
35
37.5
39
40.5
11.12
11.33
11.48
11.58
11.65
11.71
11.75
12.00
12.10
12.14
12.14
Part 6
Does the curve look like what you might predict? What are the two points on this titration curve that
can help identify your unknown? What do they correspond to?
Yes, the curve was what I thought it was going to be. The two points on the curve are the
equivalence point and the half equivalence point. The equivalence point is when moles of acid= moles of
base, and the half equivalence point is where pH=pKa.
Calculations:
Amount of KHP Needed:
0.1 𝑀 𝑁𝑎𝑂𝐻 × 0.025 𝐿 𝑡𝑖𝑡𝑟𝑎𝑛𝑡 ×
204.2212 𝑔 𝐾𝐻𝑃
= 0.51 𝑔 𝐾𝐻𝑃 𝑛𝑒𝑒𝑑𝑒𝑑
1 𝑚𝑜𝑙
NaOH Solution:
𝑀1 𝑉1 = 𝑀2 𝑉2
(0.1 𝑀 𝑁𝑎𝑂𝐻)(250 𝑚𝐿) = (6 𝑀 𝑁𝑎𝑂𝐻)(𝑉2 )
𝑉2 = 4.17𝑚𝐿
Molarity of NaOH Using Trial 1:
0.5104 𝑔 𝐾𝐻𝑃 ×
1 𝑚𝑜𝑙 𝐾𝐻𝑃
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1
×
×
= 0.09997 𝑀 𝑁𝑎𝑂𝐻
204.2212 𝑔 𝐾𝐻𝑃 1 𝑚𝑜𝑙 𝐾𝐻𝑃 0.025 𝐿
pH: (Using Trial 1)
Before Base added:
0.5104 grams KHP ×
1 mol KHP 0.00250 mol
=
= 0.04999 M
204.2212 g
0.05L
I
C
E
Ka =
H+ +
+x
+x
HA
⇌
0.04999
-x
0.04999-x
A−
+x
+x
[A− ][H + ]
[HA]
3.9 × 10−6 =
x2
0.04999 − x
x= 4.42 × 10−4
pH = −log([H + ])
pH = − log([4.42 × 10−4 ])
pH = 3.35
At equivalence point:
𝐿 𝐻2 𝑂 + 𝐿 𝑁𝑎𝑂𝐻 = .075 𝐿
0.5104 grams KHP ×
I
C
E
1 mol KHP 0.00250 mol
=
= 0.03333 M
204.2212 g
. 075 L
B + H2 O ⇌ BH + +
0.03333
-x
+x
0.03333-x
+x
Kb =
Kw
Ka
Kb =
1 × 10−14
3.9 × 10−6
OH −
+x
+x
K b = 2.564 × 10−9
Kb =
[ BH + ][OH − ]
[B]
2.564 × 10−9 =
x2
0.03333 − x
x= 9.24 × 10−6
pOH = −log([OH − ])
pOH = − log([9.24 × 10−6 ]) = 5.034
pH = 14 − pOH
pH = 14 − 5.034
pH = 8.97
1 mL after equivalence point:
mols NaOH = .09997 M × 0.026 L = 0.0025992 mol NaOH = mol OH −
0.0025992 mol of OH −
= 0.0342 M OH −
0.50L + 0.26L
pOH = − log([OH − ])
pOH = − log([0.0342]) = 1.466
pH = 14 − pOH
pH = 14 − 1.466
pH = 12.53
MW of Unknown Acid
0.028L × 0.09995 M NaOH = 0.0027986 moles
0.5063 g unknown
= 180.912 g/mol
0.0027986 moles
Ka Calculation
pKa=pH at the half eq point
Ka = 10−pKa
Possible Identity of Unknown Acids
Benzoic Acid-
MW: 122.12g
Ka: 6.28x10-5
pKa: 4.202
2,4-Dinitrophenol-
MW: 184.11g/mol
Ka: 7.69x10-5
pKa: 4.114
2-Naphthoic acid-
MW: 172.18g/mol
Ka: 4.6x10-5
pKa: 4.16
Conclusion:
The purpose of this experiment was to determine the identity of our unknown acid by
completing a series of acid- base titrations. We were able to determine the equivalence point (28.2mL),
and the half equivalence point (14.1mL) of our unknown by looking at our titration curves. The Ka, and
pKa were determined by looking at the titration curves, and we also calculated the molecular weight
which in turn helped us identify our unknown. The Ka value was 7.94x10-5, the pKa value was 4.1 and the
average molecular weight was 181.650g/mol. The choices that were considered to be our unknown
were benzoic acid, 2,4-Dinitrophenol, and 2-Naphthoic acid. Since 2,4-Dinitrophenol’s MW was the
closest to the value of our unknown’s, we determined that 2,4-Dinitrophenol was our unknown. There
are several errors that could have occurred during experimentation there was when you were titrating
your solutions, and also losing your amount of unknown when transferring it to Erlenmeyer flasks.