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Titration of a weak acid with a strong base

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Titration of a weak acid with a strong base

Introduction: A weak acid is an acid that dissociates incompletely and does not release all of its hydrogens in a solution i.e. it does not completely donate all of its protons. These acids have higher pKa compared to strong acids, which release all of their hydrogens when dissolved in water.

A strong base is a basic chemical compound that is able to deprotonate very weak acids in an acid-base reaction. Compounds with a pKa of more than about 13 are called strong bases.

Titration: An acid-base titration is the determination of the concentration of an acid or base by exactly neutralizing the acid/base with an acid or base of known concentration.

Aim: To titrate a weak acid using NaOH(aq) with the concentration of 0.200 mol/dm3.

                HA + NaOH – – – > A-  + Na+ + H2O

                                              < ---------

Material:

  • Stand
  • Clamps
  • Burette (50 ml)
  • Voll pipette (25 ml)
  • Beaker
  • 50 ml of NaOH
  • 25 ml of weak acid
  • pH meter
  • magnet
  • magnetic stirrer

Method:

  1. The burette was attached to the stand with the help of a clamp.
  2. The burette was then closed and 50 ml of NaOH was poured in.
  3. 25 ml of weak acid was poured into a beaker with the help of a voll pipette and a magnet was dropped inside.
  4.  The beaker was then placed on the magnetic stirrer and the stirrer was turned on.
  5. The pH meter was calibrated using buffer solutions of pH = 4, and pH = 7.
  6. The electrode of the pH meter was then flushed with water and inserted in the beaker containing weak acid and the pH measured.
  7. The electrode was attached to the stand and the titration was started by opening the burette.
  8. pH and volume readings were noted down, as shown in the table below. A titration curve was then plotted.

pH (weak acid)

Volume in ml (NaOH) ±1

pH (weak acid)

Volume in ml (NaOH) ±1

1.98

0

11.12

18.8

2.45

2

11.35

19

2.63

3

11.51

19.2

2.78

4

11.61

19.4

2.90

5

11.69

19.6

3.03

6

11.73

19.8

3.13

7

11.79

20

3.24

8

11.98

21

3.34

9

12.10

22

3.43

10

12.20

23

3.55

11

12.26

24

3.65

12

12.32

25

3.78

13

12.41

26

3.89

14

12.44

27

4.05

15

12.48

28

4.27

16

12.50

29

4.38

16.5

12.53

30

4.56

17

12.59

32

4.82

17.5

12.62

34

5.26

18

12.65

36

5.68

18.2

12.68

38

8.31

18.4

12.71

40

10.78

18.6

12.74

42

image00.png

Discussion:

The equivalence point (point at which amount of base added is equal to the acid in the beaker) for a weak acid-strong base titration has a pH > 7.00. For a weak acid-strong base titration, the pH will change rapidly at the very beginning and then have a gradual slope until near the equivalence point. The gradual slope results from a buffer solution being produced by the addition of the strong base, which resists rapid change in pH until the added base exceeds the buffer's capacity and the rapid pH change occurs near the equivalence point.

Calculation: HA + NaOH – – – > A- + Na+ + H2O

                   < ------

Volume of NaOH at equivalence point: 18.3 ml (from table above, the value between 18.2 and 18.4 where the rapid change in pH is taking place)

Concentration of NaOH: 0.200moldm-3

n =?

c = n/v

n = c×v                                                                                                                                                                                                                              

n = 0.200 × 0.0183

n = 0.00366 mol

Since the molar relationship between the weak acid and NaOH is 1: 1, therefore number of moles (n) for the weak acid is the same as for NaOH.

Volume of weak acid: 25 ml

Number of moles: 0.00366 mol

Concentration =?

c = n/v

c = 0.00366/0.025

c = 0.146 moldm-3 (concentration of weak acid)

In order to find out the weak acid, its pKa value should be determined. At half titration point the amount of HA consumed is equal to the amount of A- formed:

 Ka = [H+][A-]/[HA]

So: Ka = [H+], If Ka = [H+] then pKa = pH.

At half titration point half as much NaOH is added as it would take to reach the equivalence point. The volume of NaOH at equivalence point is: 18.3 so half the volume at half titration point is: 9.15. At this volume the corresponding pH value is 3.35.

Since pKa = pH and pH = 3.35

Therefore: pKa = 3.35

The value nearest to 3.35 in the data booklet is the pKa value for methanoic acid which is 3.75, hence the acid in question is most probably methanoic acid. The difference between the two values is 0.40 units.

Conclusion and Evaluation:

Limitation: Analytical errors in reading the volume of NaOH in the burette are likely t o take place. This could particularly take place while reading minute volumes. The pH change is so rapid at the equivalence point that it is almost impossible to know the exact pH at that point. The experiment should be repeated for more accurate results

Improvement: Better techniques like Gran plot technique can be used to determine the equivalence point in a much more accurate way.

Conclusion: The weak acid titrated with NaOH turned out to be methanoic acid though the pKa value for the weak acid is not exactly same as the pKa value for methanoic acid. This difference of 0.40 units could be due to the analytical errors that could have occurred during reading the volume of NaOH. The experiment should be repeated for more accurate results.

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

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