LAB 2 TITRATION

Introduction

An acid/base titration is used to determine the concentration of an acid or base by

neutralizing it with either acid or base depending on the analyst. But in this case, the goal is
to determine the equivalence of the titration by using a pH meter. The equivalent point is the
point where there is enough titrant added to the analyte to neutralize it exactly.

The information gathered from this experiment, which is the equivalence point will be use to
calculate the acid dissociation constant, also known as Ka, of the acid which was titrated.
When an indicator is used in a titration, the color of the analyte will change, which is known
as the endpoint. This endpoint will be the same as the equivalent point when the indicator is
properly used. The graph for this experiment will be the pH against the volume of titrant
added. What will be observed from the graph made will be called a titration curve where the
equivalence point occurs when very small additions of titrant causes a very rapid increase in
pH.

Objectives

1. To investigate the property of weak and strong acid with pH changes

2. To determine the pKa and Ka of the weak and strong acid using titration curve

Apparatus and materials:

Phenolphthalein indicator, burette, pipette, 250 mL beakers, pH meter, white paper towel, 0.1
M NaOH, 25mL 0.1M CH3COOH, 25mL 0.1M H3PO4

Methods

A) Titration of monoprotic acid with NaOH

1. Burette with 0.1M NaOH was filled. Pipette 25.0 mL of 0.1 M M CH3COOH into a 250
mL beaker and add 3-4 drops of phenolphthalein indicator. Place the beaker on a white
paper towel to best observe the colour changes.
2. The solution was titrated by using NaOH titrant in 2 mL increments. The beaker was
carefully swirled with each addition.
3. The coloured form of the phenolphthalein stayed for a while before disappearing. NaOH
was added dropwise until the acetic acid was a light colour. This was the indication of the
endpoint for phenolphthalein.
4. The pH of the solution in the beaker was measured and recorded at this end point. The pH
probe was then rinsed with distilled water and the probe tip was replaced into its vial.
5. The colour change that occurred during titration was recorded. The pH and added NaOH
volume at that indicator’s endpoint was used to estimate the target point using the graph.
6. Once the volume of the equivalence point was known, the volume of the equivalence
point was found using the data. The pH of this volume was determined using the titration
curve, Thus pKa and by extension Ka for acetic acid was calculated.
B) Titration of polyprotic acid with NaOH
1. Burette with 0.1M NaOH was filled. Pipette 25.0 mL of 0.1 M M H3PO4 into a 250 mL
beaker and add 3-4 drops of phenolphthalein indicator. Place the beaker on a white paper
towel to best observe the colour changes.
2. The solution was titrated by using NaOH titrant in 2 mL increments. The beaker was
carefully swirled with each addition.
3. The coloured form of the phenolphthalein stayed for a while before disappearing. NaOH
was added dropwise until the phosphoric acid was a light colour. This was the indication
of the endpoint for phenolphthalein.
4. The pH of the solution in the beaker was measured and recorded at this end point. The pH
probe was then rinsed with distilled water and the probe tip was replaced into its vial.
5. The colour change that occurred during titration was recorded. The pH and added NaOH
volume at that indicator’s endpoint was used to estimate the target point using the graph.
6. Once the volume of the equivalence point was known, the volume of the equivalence
point was found using the data. The pH of this volume was determined using the titration
curve. Since there was two equivalence point, there will be two value of pKa. Thus pKa
and by extension Ka for phosphoric acid was calculated.

Results

A) Titration of monoprotic acid

Volume of NaoH added(mL) pH

0.00 2.64
2.00 4.35
4.50 4.77
6.10 4.89
8.00 5.05
10.00 5.20
12.00 5.30
14.00 5.54
16.00 5.67
18.00 5.77
20.00 6.04
22.00 6.41
24.00 7.56
24.20 10.83
26.00 12.07
28.00 12.27
30.00 12.40
 pH against volume of NaOH added
pH Series 1

14

12

10

Volume of NaOH added (mL)

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

B) Titration of polyprotic acid

Volume of NaOH added (mL) pH

0.00 1.30
2.00 1.97
4.00 2.00
6.00 2.03
8.00 2.15
10.00 2.20
12.00 2.24
14.00 2.27
16.00 2.34
18.00 2.42
20.00 2.50
22.00 2.58
 24.00 2.68
26.00 2.78
28.00 2.89
30.00 3.01
32.00 3.16
34.00 3.35
36.00 3.69
38.00 4.61
40.00 5.87
42.00 6.25
44.00 6.48
46.00 6.63
48.00 6.73
50.00 6.84
52.00 6.87
54.00 7.00
56.00 7.10
58.00 7.18
60.00 7.29
62.00 7.39
64.00 7.49
66.00 7.59
68.00 7.70
70.00 7.84
72.00 8.01
74.00 8.22
76.00 8.66
78.00 9.20
80.00 9.93
82.00 10.72
84.00 11.04
86.00 11.27
88.00 11.44
90.00 11.55

pH against the volume of NaOH added

pH Series 1
12

11

10

Volume of NaOH added (mL)

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95
Discussion

A) Titration of monoprotic acid

Moles of titrant= volume of titrant (L) X concentration of titrant (M)

=0.025L X 0.1M

=2.5 X 10-3 mol NaOH

Moles of titrant = moles of analyte

Moles of analyte = 2.5 X 10-3 mol

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑒𝑠, 𝑚𝑜𝑙

𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑒𝑠 =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑒𝑠, 𝐿

2.5 × 10−3 𝑚𝑜𝑙

=
0.25𝐿
= 0.01 𝑀
(12.07+6.41)
Equivalence point = pH = (pH1 + pH2 )= =9.24
2

9.24
Half equivalence point = pKa = pH = =4.62
2

pKa = - log Ka

4.62 = - log Ka
-3.77
10 = Ka

Ka = 2.4 X 10-5

From the calculations above, the endpoint for this reaction is at pH 9.24, which means that
the mole of the titrant is equal to the mole of the analye at this point, which is 2.5 X 10-3 mol.
The pKa will then be 4.62 as well as the Ka being 2.4 X 10-5 .The titration of the acetic acid
stops at 30mL NaOH due to it being a weak acid, meaning it is easy to be neutralized and the
colour of the solution to turn light pink.

B) Titration of polyprotic acid

Moles of titrant= volume of titrant (L) X concentration of titrant (M)

=0.025L X 0.1M

=2.5 X 10-3 mol NaOH

Moles of titrant = moles of analyte

Moles of analyte = 2.5 X 10-3 mol

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑒𝑠, 𝑚𝑜𝑙

𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑒𝑠 =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑒𝑠, 𝐿

2.5 × 10−3 𝑚𝑜𝑙

=
0.25𝐿
= 0.01 𝑀

There are 2 equivalence point for this group, which are:

1st equivalence point

(6.25+3.69)
Equivalence point = pH = (pH1 + pH2 )= =4.97
2

4.97
Half equivalence point = pKa = pH = =2.49
2

pKa1 = - log Ka

2.49 = - log Ka
-2.49
10 = Ka

Ka1 = 3.24 X 10-3

2nd equivalence point

(10.72+8.22)
Equivalence point = pH = (pH1 + pH2 )= =9.47
2

9.47
Half equivalence point = pKa = pH = =4.74
2

pKa1 = - log Ka

4.74 = - log Ka
-4.74
10 = Ka

Ka1 = 1.82 X 10-5

From what can be seen, polyprotic acid is an acid which gives out 3 hydrogen protons when it
reacts in a solution. However, there will only be 2 equivalent point because the graph stops at
by the point when the second equivalence point is determined. 90 mL NaOH is added to
reach its second end point because phosphoric acid is a strong acid, meaning it takes more
volume of NaOH to neutralize it. Its first and second equivalence point is 4.97 and 9.47
respectively as well as their pKa being 2.49 and 4.74.

Conclusion

The pH of the weak and strong acid increase of more volume of NaOH is added. However,
the monoprotic acid required less volume of NaOH than polyprotic acid needed to reach the
end point since it was a weak acid. The pKa of an acid can be determined by using a pH
meter. A monoprotic acid can only give out one equivalence point since it only release one
hydrogen ion in the solution while a polyprotic acid can give out more hydrogen ion than a
monoprotic acid can. Errors will also occur if the pH meter is not used correctly.

References

1. http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Streng
th/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf
2. Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper
Saddle River, New Jersey: Pearson/Prentice Hall, 2007