Last updated: May 20, 2019
Topic: Essay database
Sample donated:

Purpose: To find the concentration of acetic acid by titrating against a standard solution of NaOH ( 0.1M ) . Using this to happen the equality point, half the equality point and the neutralizing pH that will give the PKA value of acetic acid.

Data aggregation:

Table 1.1: The alteration in pH of 10 milliliter of acetic acid ( unknown concentration ) when Na hydrated oxide ( 0.1M ) was titrated over 3 tests.

pH ± 0.01

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Volume of NaOH added ( milliliter ) ± 0.1ml

Trial 1

Trial 2

Trial 3

1.0

3.93

2.15

2.92

2.0

3.72

2.76

4.20

3.0

4.55

3.47

4.05

4.0

4.37

3.69

4.48

5.0

4.63

3.73

4.40

6.0

5.08

3.85

4.60

7.0

5.48

3.65

4.59

8.0

5.64

3.94

4.34

9.0

5.48

3.86

4.42

10.0

5.57

4.12

4.54

11.0

5.76

4.33

4.85

12.0

6.63

4.51

5.28

13.0

9.06

4.77

5.32

14.0

11.34

5.91

5.38

15.0

11.61

6.27

5.45

16.0

12.16

8.21

6.11

17.0

12.43

8.78

6.34

18.0

12.39

10.13

6.58

19.0

12.44

10.86

7.13

20.0

12.66

11.28

10.26

21.0

12.65

11.15

11.26

22.0

12.69

11.29

11.53

23.0

12.59

11.42

11.80

24.0

12.61

11.32

12.22

25.0

12.62

11.68

12.56

Qualitative observations: The solution turned somewhat milklike ab initio, but so turned colourless after about 12-13 milliliter of NaOH was added.

Datas processing:

Graph 1.1: The volume of NaOH titrated into 10 ml acetic acid vs. the pH for test 1.

Using the above graph, we can generalize the center of the steep rise ( 11ml to 15 milliliter ) . The center comes to 13 milliliter of NaOH ( 0.1 M ) being titrated into the acetic acid and the neutralization pH comes to about 9.06 ( verified from table 1.1 ) .

pH = 8.21, volume = 16 mlGraph 1.2: The volume of NaOH titrated into 10 ml acetic acid vs. the pH for test 2.

Similarly we can make the same with test 2.

pH = 7.13, volume = 19 mlGraph 1.3: The volume of NaOH titrated into 10 ml acetic acid vs. the pH for test 3.

Similarly we can make the same for test 3.

Table 1.2: Summary of consequences of the neutralizing volume of NaOH ( 0.1M ) and the pH of neutralisation for tests 1,2 and 3.

Volume of NaOH ( milliliter ) ± 0.1ml

pH ± 0.01

Trial 1

13

9.06

Trial 2

16

8.21

Trial 3

19

7.13

Average ( ±0.3 milliliter, 0.03 )

16

8.13

Therefore, from table 1.2 we can see that 8.13 is the equality point for this reaction and hence half the equality point would be 4.065.

Calculations and mistake extension:

Average volume:

( Trial 1 + Trial 2 + Trial 3 ) /3

= ( 12 + 16 + 19 ) /3

= 48/3 = 16 milliliter

Mistake:

?Avg. = ( ?Trial 1 + ?Trial 2 + ?Trial 3 ) ten Avg.

Trial 1 Trial 2 Trial 3

= ( 0.1/13 + 0.1/16 + 0.1/19 ) x 16

= ±0.3 milliliter ( 1 sf )

Average pH:

( Trial 1 + Trial 2 + Trial 3 ) /3

= ( 9.06 + 8.21 + 7.13 ) /3

= 8.13 ( 3 sf )

Mistake:

?Avg. = ( ?Trial 1 + ?Trial 2 + ?Trial 3 ) ten Avg.

Trial 1 Trial 2 Trial 3

= ( 0.01/9.06 + 0.01/8.21 + 0.01/7.13 ) x 8.13

= ±0.03 ( 1 sf )

Determining the concentration of the Acetic acid:

Acid + Base i? Salt + Water

CH3COOH + NaOH i? CH3COONa + H20

1: 1

Concentration of NaOH = no. of moles/ volume in dm3 ( utilizing informations from table 1.2 )

= x/ 0.016

ten = 1.6 ten 10-3 moles of NaOH

No. of moles of NaOH = No. of moles of CH3COOH

Therefore, concentration of CH3COOH = 1.6 x 10-3 / 0.01

= 0.16 M

Uncertainties:

? No. of moles of NaOH = ( ?volume/volume ) x No. of moles of NaOH

( note: the uncertainness in the concentration of NaOH was unknown since it was pre-made )

? No. of moles of NaOH = ( 0.1/ 0.016 ) x 1.6 ten 10-3

± 0.01 moles

? Concentration of Acetic Acid = [ ( ?No. Of moles/No. of moles ) + ( ?volume/volume ) ] ten Concentration of Acetic Acid

? Concentration of Acetic Acid = [ ( 0.01/1.6 x 10-3 ) + ( 0.1/0.016 ) ] x 0.16

= ± 2M

Determining the PKA value:

PKA = pH at half equality point

Trail 1:

Equivalence pH = 9.06 at 13 milliliter

Half equality pH= 5.28 at 6.5 milliliters ( from graph 1.1 )

PKA = 5.28 ( As per Henderson-Haselbalch equation )

Trial 2:

Equivalence pH = 8.21 at 16 milliliter

Half equality pH= 3.94 at 8 milliliters ( from graph 1.2 )

PKA = 3.94 ( As per Henderson-Haselbalch equation )

Trial 3:

Equivalence pH = 7.13 at 19 milliliter

Half equality pH= 4.48 at 9.5 milliliters ( from graph 1.3 )

PKA = 4.48 ( As per Henderson-Haselbalch equation )

Average:

( 5.28 + 3.94 + 4.48 ) /3

= 5.57 ( 3 sf. )

Consequences:

Table 1.3: The tabular array shows the PKA value obtained from half the equality point each test and the norm.

Trial Number

PKA Value

1 ( ± 0.01 )

5.28

2 ( ± 0.01 )

3.94

3 ( ± 0.01 )

4.48

Average ( ± 0.04 )

5.57

Decision:

A weak acid merely partly dissociates from its salt. The pH will lift usually at first, but as it reaches a zone where the solution seems to be buffered, the incline degrees out. After this zone, the pH rises aggressively through its equality point and degrees out once more like the strong acid/strong base reaction.

There are two chief points to detect in the curves we have obtained in graph 1.1, 1.2 and 1.3. The first is the half-equivalence point. This point occurs midway through a buffered part where the pH hardly changes for a batch of base added. For test 1 this occurs near a point where 6.5 milliliter of NaOH is added, for test 2 it occurs nigh 8 milliliters and for test 3 it occurs nigh 9.5 milliliter. The half-equivalence point is when merely plenty base is added for half of the acid to be converted to the conjugate base. When this happens, the concentration of H+ ions equals the KA value of the acid. This can be farther extended to give, pH = pKa. The 2nd point is the higher equality point ; this point is normally above a pH of 7 as seen by the tendency in the information obtained.

The finding of the KA, of a weak acid can be hard. However, its PKA can be easy estimated by analysis of its titration curve. The, PKA is the pH value at the half equality point, that is, the point at which merely half of the volume of base needed to make the equality point has been added. The ground for this is that, at the half-equivalence point, the concentrations of the conjugate base, A- , and that of the non-dissociated acid, HA, are about equal. Therefore, they cancel out in the look for KA.

We observed the PKA value to be 5.57 ( norm ) and the scope for PKA values in an aqueous medium normally range from -2 ( strong acid ) to 12 ( strong base ) . Our value of 5.57 indicates that acetic acid is a weak acid as it has non wholly disassociated into ions. The fluctuations between the three tests could be accounted for by the mistakes mentioned subsequently, but we are certain that acetic acid is a weak acid and our experiment proved it. The theoretical value for PKA of acetic acid as per the IB informations brochure is 4.76. This is 0.81 less than the experimental value we calculated intending that there was about 17.02 % experimental mistake in this value obtained. The systematic mistake accounted for merely 0.84 % of the entire mistake this we can impute the remainder of the divergence to the random mistakes, which are subsequently, discussed in the rating. We can see that our 3rd test was the most accurate out of all the three as it is closest to the literature value.

Evaluation:

I think this lab was conducted with equal proficiency, nevertheless some beginnings of mistake could be:

It was rather hard to pull off the pH investigation, magnetic scaremonger, flow of NaOH from the burette and record the readings at the same clip as we did our pattern labs in groups. Therefore at that place could hold been mistakes in what we recorded.

There might hold been a parallax mistake while taking the readings off the burette.

The burette could hold dripped the base straight onto the pH investigation and therefore at that place may hold been excess basicity detected by it giving inaccurate readings.

The acid was non used in equal sums for every test, as the pipette holders were non working decently. Therefore there were varied sums of acid during every test.

The beakers we used were non clean and had some residue stuck at the underside, which could hold contributed for some errors in our informations.

I dripped 1 milliliter of base into the acid at a clip ; this was excessively big an interval since acetic acid is a weak acid. This means I did non enter some of the alterations in the pH, which would hold occurred with somewhat less sums of acid.

Some betterments that I can propose to this lab are:

Make the lab in braces as it becomes easier to manage all the equipment that is required to finish the lab.

Take more readings to govern out the consequence of random mistakes.

Wash all equipment exhaustively before utilizing and guarantee they all function right.

Use smaller intervals for adding the base ( possibly 0.5 milliliter ) to acquire a more gradual and accurate graph.

Keep the burette at oculus degree to cut down parallax mistake.

Use 2 pH investigations to look into the alteration in pH to be wholly certain of the readings obtained.