Experiment 5
Potentiometric titrations
10/13/2014
Instructor Prof. Harman.
Yi Shen
Lab partner Anna Chon
Introduction:
Titration refers to the procedure of adding acid or base with known concentration in
molarity to an unknown base or acid solution to determine the concentration of the unknown
solution. During the experiment, the point at which stoichiometrically equivalent quantities of
acid and base is the equivalent point of titration. According to the measured volume and pH
value of the know acidic or basic solution, the equivalent point can be determined by
graphing the titration curve containing the volume of added acid or base with known
concentration on the x-axis and the pH value of the unknown solution on the y-axis. The
equilibrium constant of concentration and the concentration of the unknown solution can be
calculated based on the experiment data.
This experiment contains four titrations. In part I, strong base NaOH is added to
neutralize strong acid HCl. The chemical equation is
NaOH  HCl  H 2O  NaCl .
In part II,
NaOH is added to acetic acid. Main chemical equations
are: CH 3COOH  NaOH  CH 3COONa  H 2O . However, due to the fact that acetic acid
partially dissociate in solution and acetate ions hydrolysis in solution,
CH 3COO   H 2O  CH 3COO H  OH  , which forms a buffer, smaller Ph range is
expecting. In part III, NaOH is added to triprotic acid H 3 PO4 ,which dissociate in three
H 3 PO4  NaOH  H 2O  NaH 2 PO4
steps, NaH 2 PO4  NaOH  H 2O  Na2 H PO4 , and again for the reason that weak acid
Na2 H PO4  NaOH  H 2O  Na3 PO4
H 3 PO4 after adding NaOH forms a buffer, a smaller pH range is expecting. More
importantly, 3 equivalent points are expecting. From the graph of previous three titrations,
experimental concentration and Ka for the corresponding acid can be calculated from the half
equivalence point. In part IV, NaOH is added to an unknown acid and after getting the pH
graph, from the half equivalence point, calculated Ka, so that the unknown acid can be
determined. Before all, NaOH is standardrized by KHP.
Material and Methods:
List of Material:
pH meter and probe.
Magnetic stirrer
400ml Beakers
Two 50ml burets
phenolphthalein
Stand and clamp
Strong base NaOH(approximate concentration 0.199M and 0.207M)
Strong acid HCl ( approximate concentration 0.1M)
Weak monoprotic acid ( approximate concentration 0.1M acetic acid)
Weak polyprotic acid ( approximate concentration 0.1M phosphoric acid)
Some solid KHP
Unknown acid B
Procedure:
Part 0:
1. Measure 0.4230 g of KHP using a beaker, dissolve it into 50ml approximately.
2. Add 3 drops of phenolphthalein into the beaker, stir the solution by magnetic stirrer
until fully dissolved.
3. Clean the buret with pure water and NaOH solution(0.199M) read the initial volume.
4. Continue adding NaOH to KHP solution until the solution gets light pink, read the pH,
calculate the exact concentration of NaOH solution.
Part I:
1. Take 50ml HCl using a buret to 400ml beaker, measure the initial pH.
2. Read the initial volume of NaOH in another buret and start adding NaOH to HCl, using
2.0ml increments for 22ml continue reading corresponding pH. Using 0.1 or 0.2 ml
increments until end points, then change back to 2.0ml imcrement until pH=12. Keep stirring
during the process.
3. Plug in data ph as y axis and volume of NaOH as x axis.
Part II:
1.Take 50ml acetic acid using a buret to 400ml beaker, measure the initial pH.
2.Read the initial volume of NaOH in another buret and start adding NaOH to acetic acid,
using 2.0ml increments for 16ml continue reading corresponding pH. Using 0.1 or 0.2 ml
increments for next 8 ml, then change back to 2.0ml imcrement until pH=12. Keep stirring
during the process.
3.Plug in data ph as y axis and volume of NaOH as x axis.
Part III:
1.Take 40ml
H 3 PO4 using a buret to 400ml beaker, measure the initial pH.
2.Read the initial volume of NaOH in another buret and start adding NaOH to
H 3 PO4 acid,
using 2.0ml increments when the pH changes rapidly, ( first 15ml) Using 0.1 or 0.2 ml
increments for next 8 ml, then change back to 2.0ml increment for 8ml, and 0.1 or 0.2 ml
increments for next 8ml. Keep stirring during the process.
3.Plug in data ph as y axis and volume of NaOH as x axis.
Part IV:
1. Dissolve 0.402g unknown acid in 50ml pure water using a 400ml beaker.
2.Read the initial volume of NaOH in another buret and start adding NaOH to the unknown
acid, using 2.0ml increments for the first 8ml. Using 0.1 or 0.2 ml increments for next 8 ml,
then change back to 2.0ml increment for 10ml, and 0.1 or 0.2 ml increments for next 8ml.
And do 2ml increment until pH is 12. Keep stirring during the process.
3.Plug in data ph as y axis and volume of NaOH as x axis.
Result and Discussion:
Part 0:
NaOH is standarized by KHP.
NaOH  KHP  H 2O  NaKP
0.4230 g 
1mol
1molNaOH
1


 .0197 M
204.22 g 1molKHP 10.49ml
Part I:
In this part, atrong base is added to strong acid, one equivalent point is expecting, and
approximately 50ml of NaOH should be added at that point, when pH is 7. The result is
shown in figure 1.
Figure 1.
Calculation:
From figure 1, at the equivalence point the volume of NaOH added is about 33ml.
At the equivalence point, ph is considered to be 7, so the mole of acid consumed should be
equal to the more of base added in.
NaOH  HCl  H O  NaCl
2
1L
.197mol 1molHCl
1
33mlNaOH 



 0.130M
1000mL
1L
1molNaOH 50ml
The exact molarity for HCl is 0.1300M
At the initial point
The half equivalence point in this graph is 16.5ml, ph = 1.45 approximately.
HCl   Cl  
At the half equvalence point, pH  pKa  1.45
Ka  101.45  .0355
24.9
24.7
1.72
Volume
Calculated
25
24.8
1.72
NaOH (ml)
V(NaOH) (ml) PH Level
25.12
24.92
1.73
0.2
0
1.17
25.22
25.02
1.73
2
1.8
1.47
25.34
25.14
1.74
3.99
3.79
1.31
25.5
25.3
1.75
5.98
5.78
1.31
25.6
25.4
1.75
8.03
7.83
1.32
25.8
25.6
1.76
9.99
9.79
1.33
25.9
25.7
1.77
12
11.8
1.36
26
25.8
1.77
13.98
13.78
1.4
26.1
25.9
1.78
16
15.8
1.43
26.2
26
1.78
18
17.8
1.47
26.3
26.1
1.79
20.01
19.81
1.52
26.4
26.2
1.8
22.02
21.82
1.59
26.5
26.3
1.8
22.14
21.94
1.6
26.61
26.41
1.81
22.23
22.03
1.63
26.71
26.51
1.82
22.38
22.18
1.62
26.81
26.61
1.82
22.48
22.28
1.62
27
26.8
1.83
22.59
22.39
1.63
27.15
26.95
1.84
22.7
22.5
1.63
27.25
27.05
1.85
22.8
22.6
1.63
27.4
27.2
1.86
22.87
22.67
1.63
27.5
27.3
1.86
23
22.8
1.64
27.6
27.4
1.87
23.1
22.9
1.64
27.7
27.5
1.88
23.27
23.07
1.64
27.8
27.6
1.89
23.4
23.2
1.65
27.9
27.7
1.89
23.5
23.3
1.66
28
27.8
1.9
23.6
23.4
1.66
30
29.8
2.09
23.7
23.5
1.67
32
31.8
2.56
23.81
23.61
1.67
34
33.8
11.13
23.95
23.75
1.67
35
34.8
11.45
24.05
23.85
1.68
36
35.8
11.61
24.15
23.95
1.68
37
36.8
11.73
24.38
24.18
1.69
37.95
37.75
11.8
24.4
24.2
1.69
38.98
38.78
11.87
24.5
24.3
1.7
40.01
39.81
11.92
24.6
24.4
1.7
41
40.8
11.96
24.7
24.5
1.71
42
41.8
11.99
24.8
24.6
1.71
42.9
42.7
12
Part II:
In this part, NaOH is added to acetic acid, a smaller range of pH and two curve parts are
expecting in the pH graph. By adding NaOH, the weak acid solution forms a buffer so the
equivalence point is not at pH 7. In addition, Ka can be calculated from the beginning and the
half equivalence point.
The result is shown in figure 2.
Figure 2.
Calculation:
From figure 2, at the initial point when no base added in, the pH is 3.14.
From the initial point, calculate Ka:
CH 3COOH  CH 3COO _  H 
CH COO   10
_
3
pH  Pka  log(
3.14  PKa  log
3.14
CH COO  )
_
3
CH 3COOH 
10 3.14
.104
pKa  5.3
Ka  10 5.3  5  10 6
Theoretical Ka for acetic acid is 1.8*10^5
pKa is 4.74
The equivalence point is 25.2ml, ph is 10.94
NaOH  CH 3COOH  H O  NaCH 3COO
2
1L
.207 mol 1molCH 3COOH
1
25.2mlNaOH 



 0.104 M
1000mL
1L
1molNaOH
50ml
1L
.207 mol 1molCH 3COONa
1
25.2mlNaOH 



 0.069 MCH 3COONa
1000mL
1L
1molNaOH
50  25.2ml
The exact molarity for acetic acid is 0.1040M
0.069M of sodium acetate formed at equivalence point
CH 3COO   H 2O  CH 3COOH  OH 
From the equivalence point calculate Ka pOH  14  10.94  3.06
Kb 
CH 3COOH OH    (10 3.06 ) 2
CH COO 

3
0.069
 1.1 10 5
The half equivalence point is around 12.62 ml NaOH and the corresponding ph is around
CH 3COOH   CH 3COO  
4.86. At the half equivalence point, pH  pKa  4.86
Ka  104.86  1.4 10 5
Data:
Volume
Calculated
NaOH
V(NaOH)
7.75
7.75
4.49
15
15
5.04
(ml)
(ml)
PH
7.9
7.9
4.51
15.92
15.92
5.11
8.09
8.09
4.52
17
17
5.2
0
0
3.14
8.4
8.4
4.54
17.92
17.92
5.28
0.3
0.3
3.15
8.5
8.5
4.55
19
19
5.39
0.5
0.5
3.26
8.6
8.6
4.56
20.05
20.05
5.51
0.79
0.79
3.39
8.78
8.78
4.57
21.03
21.03
5.65
1.05
1.05
3.52
8.9
8.9
4.59
21.2
21.2
5.68
1.21
1.21
3.58
9.1
9.1
4.6
21.3
21.3
5.69
1.58
1.58
3.67
9.38
9.38
4.61
21.5
21.5
5.73
1.88
1.88
3.76
9.4
9.4
4.62
21.63
21.63
5.75
2.2
2.2
3.82
9.5
9.5
4.63
21.8
21.8
5.78
2.47
2.47
3.87
9.63
9.63
4.64
21.9
21.9
5.8
2.78
2.78
3.92
9.8
9.8
4.65
22.03
22.03
5.83
3
3
3.97
9.9
9.9
4.66
22.12
22.12
5.86
3.29
3.29
4.02
10.03
10.03
4.67
22.34
22.34
5.88
3.54
3.54
4.06
10.19
10.19
4.68
22.35
22.35
5.91
3.89
3.89
4.1
10.3
10.3
4.69
22.5
22.5
5.94
4.2
4.2
4.14
10.49
10.49
4.7
22.6
22.6
5.97
4.5
4.5
4.18
10.61
10.61
4.72
22.7
22.7
6
4.8
4.8
4.22
10.72
10.72
4.72
22.9
22.9
6.06
5.05
5.05
4.24
10.9
10.9
4.73
23.01
23.01
6.1
5.17
5.17
4.25
11
11
4.74
23.15
23.15
6.14
5.28
5.28
4.26
11.1
11.1
4.75
23.3
23.3
6.18
5.35
5.35
4.27
11.21
11.21
4.76
23.4
23.4
6.23
5.5
5.5
4.29
11.32
11.32
4.77
23.5
23.5
6.28
5.64
5.64
4.3
11.42
11.42
4.77
23.63
23.63
6.38
5.8
5.8
4.31
11.6
11.6
4.79
23.8
23.8
6.46
6
6
4.34
11.7
11.7
4.79
23.9
23.9
6.55
6.2
6.2
4.36
11.8
11.8
4.8
24
24
6.65
6.3
6.3
4.37
12
12
4.82
24.13
24.13
6.79
6.41
6.41
4.37
12.12
12.12
4.82
24.18
24.18
6.97
6.52
6.52
4.39
12.22
12.22
4.83
24.3
24.3
7.11
6.69
6.69
4.4
12.32
12.32
4.84
24.4
24.4
7.61
6.81
6.81
4.41
12.51
12.51
4.85
24.5
24.5
10.02
6.99
6.99
4.43
12.62
12.62
4.86
26.1
26.1
11.85
7.2
7.2
4.45
12.75
12.75
4.87
26.43
26.43
11.95
7.34
7.34
4.46
12.78
12.78
4.88
26.58
26.58
11.98
7.44
7.44
4.47
13
13
4.89
26.61
26.61
11.99
7.59
7.59
4.48
14
14
4.96
26.7
26.7
12
Part III:
In this part, NaOH is added to H 3 PO4 acid, which has 3 Ka, so 3 equivalence points are
expecting. And the curve should be more smooth, because buffer is made by adding NaOH.
However, the graph only shows two obvious equivalence points. The result show as figure 3.
Figure 3.
The initial point ph is 1.73.
The first equivalent point is when 19.98 ml base is added, ph is 5.99
H 3 PO4  NaOH  NaH 2 PO4  H 2O
0.207 mol 1molH 3 PO4
1


 0.103M
l
1molNaOH 40ml
0.207 mol 1molNaH 2 PO4
1
19.98ml 


 0.071MNaH 2 PO4
l
1molNaOH
40  18.45ml
19.98ml 
The exact concentration of
H 3 PO4 is 0.103M

H 2 PO4  H 2O  H 3 PO4  OH 
pOH  14  5.99  8.01

From first equivalent point calculate Ka: Kb  H 3 PO4  OH
H 2 PO4
Ka 

  (10
10 14
 .0735
1.36  10 13
8.01 2
) / 0.07  1.36 10 13
The first half equivalent point is around 10ml, ph is around 2.3
At this point
pH  Pka
Ka  5 10 3
Theoretically, Ka1=7.5*10^-3
The second equivalence point is when 38ml NaOH is added, pH is 10.43.
2
HPO4  H 2O  H 2 PO4  OH 
pOH  14  10.43  3.57

H 2 PO4 OH  
Kb 
 (10 3.57 ) 2 / 10 8.01  0.74
H PO4
Ka 
10 14
 1.35 10 13
0.74
The second half equivalence point is when 29 ml NaOH is added, Ph is 7.2
At this point
pH  Pka  7.2
Ka  10 7.2  6.3  10 8
Theoretically, Ka2=6.2*10^-8
Ka3=4.2*10^-13
Data:
Calculated
37.6
37.6
9.93
Volume
NaOh
37.8
37.8
10.31
NaOH
added
38
38
10.43
added(ml) (ml)
PH
38.3
38.3
10.57
0
0
1.73
38.5
38.5
10.69
2.1
2.1
1.8
38.8
38.8
10.79
4.1
4.1
1.89
39
39
10.93
5.82
5.82
2
39.2
39.2
10.96
7.5
7.5
2.1
39.5
39.5
11.04
9.18
9.18
2.21
39.72
39.72
11.1
11
11
2.37
39.95
39.95
11.11
12.98
12.98
2.52
40.15
40.15
11.19
14.8
14.8
2.78
43.34
43.34
11.23
16.35
16.35
3.07
40.5
40.5
11.25
17.98
17.98
3.86
40.8
40.8
11.31
18.91
18.91
5.5
19.98
19.98
5.99
21
21
6.26
22
22
6.43
23.91
23.91
6.57
24.03
24.03
6.71
25
25
6.8
26.15
26.15
6.92
27
27
6.99
27.9
27.9
7.01
29
29
7.2
30
30
7.3
30.48
30.48
7.35
31.5
31.5
7.45
31.9
31.9
7.49
33
33
7.61
33.5
33.5
7.71
34
34
7.78
34.5
34.5
7.88
35
35
8.02
35.2
35.2
8.04
36.2
36.2
8.28
36.9
36.9
8.77
37
37
8.9
37.15
37.15
9.07
37.38
37.38
9.46
Part IV:
In this part, NaOH is added to unknown acid, which is made by adding 0.402g to 50ml water.
Since it is known that the unknown acid has two Ka and molecular weight around 130 g, two
equivalent points are expecting. However, one of them cannot be really tell. The result is
shown in figure 4.
Figure 4.
The initial ph is 2.28
The second equivalence point is when 36.8ml NaOH is added, ph is 9.31.
NaOH  H 2 A  NaHA  H 2O
NaOH  H A  Na2 A  H 2O
.207 mol 1molH 2 A

 3.8110 3 mol
l
2molNaOH
3
0.402 g  3.8110 mol  105.5 g / mol
36.8ml 
The sencond half equivalence point is when 29.7 ml NaOH is added, ph is 5.65
At this point
Data:
.
pH  Pka  5.65
Ka  10 5.65  2.24 10 6
Volume
added NaOH Calculated
20.3
20.3
4.62
36
36
6.88
20.5
20.5
4.66
36.2
36.2
7.02
20.7
20.7
4.69
36.41
36.41
7.21
ml
NaOH ml
PH
0
0
21.23
4.77
36.5
36.5
7.54
1
1
2.28 21.23
2.3
21.7
21.7
4.83
36.75
36.75
8.54
2
2
22.2
4.89
36.8
36.8
9.31
3
3
2.35 22.2
2.4
22.7
22.7
4.95
36.9
36.9
9.74
4
4
23
5
37
37
10.04
5
5
2.47 23
2.53 23.7
23.7
5.07
37.1
37.1
10.28
6
6
2.6
24.2
24.2
5.11
37.2
37.2
10.46
7
7
24.7
5.16
37.3
37.3
10.61
8
8
2.67 24.7
2.72 25.2
25.2
5.22
37.4
37.4
10.72
9
9
25.7
5.26
37.5
37.5
10.82
10
10
2.82 25.7
2.89 26.2
26.2
5.31
37.6
37.6
10.89
10.5
10.5
26.75
5.36
37.7
37.7
10.96
11
11
2.95 26.75
3
27.2
27.2
5.41
37.8
37.8
11.02
11.5
11.5
27.7
5.45
37.9
37.9
11.06
12
12
3.05 27.7
3.1
28.2
28.2
5.5
38
38
11.11
12.5
12.5
28.7
5.54
38.1
38.1
11.14
13
13
3.15 28.7
3.21 29.2
29.2
5.59
38.2
38.2
11.18
13.5
13.5
29.7
5.65
38.4
38.4
11.21
14
14
3.26 29.7
3.31 30.2
30.2
5.7
38.5
38.5
11.24
14.5
14.5
3.4
5.75
38.6
38.6
11.27
15
31.2
5.8
38.8
38.8
11.32
15.5
15.5
30.7
3.47 31.2
3.54 31.7
30.7
15
31.7
5.87
39.05
39.05
11.36
16
16
32.2
5.93
39.25
39.25
11.4
16.5
16.5
3.64 32.2
3.74 32.7
32.7
5.99
40
40
11.46
17
17
33.2
6.07
41
41
11.58
17.5
17.5
3.85 33.2
3.97 33.3
33.3
6.08
42
42
11.66
18.05
18.05
33.42
6.11
43
43
11.72
18.3
18.3
4.14 33.42
4.19 33.53
33.53
6.12
44
44
11.77
18.5
18.5
33.7
6.17
45
45
11.82
18.7
18.7
4.24 33.7
4.28 34.1
34.1
6.23
46
46
11.86
18.92
18.92
34.4
6.29
47
47
11.89
19.1
19.1
4.35 34.4
4.39 34.6
34.6
6.34
48
48
11.92
19.3
19.3
34.9
6.4
49
49
11.94
19.5
19.5
4.42 34.9
4.46 35.2
35.2
6.48
50
50
11.97
19.7
19.7
35.4
6.58
19.9
19.9
4.51 35.4
4.55 35.7
35.7
6.69
20.1
20.1
4.59 35.85
35.85
6.78
Conclusion:
From the molar mass and Ka2 calculated in part 4, the unknown is Malonic acid. Becasue
according to the unknown chart provided,, malonic acid has the closed molar mass and Ka2.
From the calculations done in Part I, II, and III, the best approximation of Ka can be made by
using the half equivalent point. When the titration happens between strong acid and base,
equivalence point is expected to be at pH7. However, when strong base is added to a weak
acid, the equivalence point is not suppose to be at pH7 because the formation of buffer. Also,
weak acid and strong base titration has a smaller range of pH as well as smoother curve.
Polyprotic acid should have equivalence point as many as the Ka’s but there are exceptions
like H3PO4 and the unknown in this experiment.