Chemistry Coursework

To find out the molarity of the acid using 25cm3 of acid to neutralise 0.10 mole of alkali

Analysis

We used a pipette to measure the acid and this was a measurement of 25cm3. A burette was used to measure the alkali, along with methyl orange as an indicator because this has a pH range of 3-7.

We did the experiment 5 times to get a range of results.

1

2

3

4

5

Ending reading (cm3)

29.20

28.80

28.60

28.85

28.75

Starting reading (cm3)

0.00

0.00

0.00

0.00

0.00

Volume of alkali used (cm3)

29.20

28.80

28.60

28.85

28.75

Volume of acid used (cm3)

25.00

25.00

25.00

25.00

25.00

When doing an experiment the apparatus can have a error or you could actual read it wrong. Referring to “Nuffield Advanced Chemistry” as my secondary source, pipettes and burettes have a percentage error. Pipettes have and accuracy of one drop. The volume of one drop is 0.05cm3. Burettes have graduations every 0.1cm3 and this also can be 0.05cm3 too high or too low.

The percentage error for the Pipette =

0.05cm3

X 100 = 0.2%

25cm3

The percentage error for the Burette = Smallest reading = 28.60

0.05cm3

X 100 = 0.175%

28.60cm3

The results that will be excluded are 1 and 3, which are 29.20 and 28.60.

Therefore for the average we should only use results 2, 4 and 5, which are 28.80, 28.85 and 28.75cm3.

So the average of these is 28.80cm3.

We now use 28.80cm3 to work out the molarity of the acid.

The equation for the titration can be sown as:

H2SO4 + Na2CO3 H2O + Na2SO4 + CO2

This has a ratio of:

H2SO4 + Na2CO3 H2O + Na2SO4 + CO2

1 : 1

Which means that for every 1 molecule of H2SO4, 1 molecule of Na2CO3 will react with it.

So to find out the molarity of the acid using 25cm3 of acid to neutralise 0.10 mole of alkali you need to begin by making up an accurate solution of a known molarity, of the alkali.

Na2CO3 has a relative mass of: 23+23+12+16+16+16

= 106g mol-1

This total if for 1.0 mole, I want 0.10mol.

So 1M solution 106g dissolved in 1 dm3

= 1mol dm-3

= 1M

But the volumetric flask = 250cm3

So we need 1/4 mole in 1/4 of a litre = 1mol dm-3

= 106g mol-1

= 26.5g required

4

So for 0.10M of solution you need to dissolve

= 26.5

= 2.65g in 250cm3 = 0.100M

10

Now that we have made an accurate solution we can find out the molarity of the acid by using the average result of alkali used.

The average = 28.75cm3

To get the amount of alkali used in dm3 you need to:

28.75cm3

= 0.0287dm3

1000cm3 dm-3

Then we need to:

Moles = 0.100mol dm-3 x 0.0286dm3

To get the moles of alkali used in 28.75cm3

So this can be shown as:

Moles = 0.0028moles = 0.0028 moles of acid

1 : 1

Now that we have the molarity of the acid in 28.75cm3

We need to find the molarity of acid in 25.00cm3

So molarity of acid = 0.0028 moles in 25.00cm3

= 1000cm3 dm-3

= 40

25

To get the amount of times 25.00cm3 goes into a litre.

= 40 x 0.0028 moles

= 0.112 mole dm-3

This gives you the exact molarity of the acid in dm3.

From this we can see that the acid was a slightly stronger molarity.

We used 25cm3 of acid: 28.75cm3 of alkali.

It is clearly obvious that the acid is stronger because we used less of it to neutralise more alkali.

So overall out results make sense because the acid was a stronger molarity than the alkali.

Evaluation

Looking at the results you can see that they are very good except for the two anomalies.

These anomalies are results 1 and 3, which are 29.20 and 28.60cm3. They were outside the error boundary.

The anomalies could have been caused by problems within the apparatus. For example there could have been air bubbles in the burette causing the reading to actually be lower. Liquid down the side would have caused an increase in the amount, and unclean equipment in the flask could have caused the colour of the indicator to change quicker than it should have. Finally the end point has to be determined by colour. Each experiment could have had a slightly different colour.

Making the results better would be to change some of the problems that may have happened, and to make the experiment better. First of all you could run off some liquid or tap the burette to get rid of the air bubbles. Next, to get rid of the liquid down the side, you wait and leave it to settle and then run some off if you have some excess. Using distilled water to clean the equipment will help, by also making sure the equipment is thoroughly dried. Finding the end point for the pH is very difficult so instead of using methyl orange by using a digital pH meter it will be more accurate and have a less error margin. Finally by keeping each set of results and comparing the acid colours you can keep the colour constant.

Instead of using a pipette to measure the acid a burette could have been used instead as it has a smaller percentage error: pipette = 0.200% and the burette = 0.175%. This would help improve accuracy. Also, if repetitions were increased then it would lead to an increase in accuracy because more results give a more accurate mean.

Overall, there are 3 results within the error boundary. The 3 results are 28.75, 28.85 and 28.80cm3, and there mean is 28.80cm3.

They all can have a reading mistake of 0.05cm3.

So to work out the percentage error of these three results you need to:

0.05cm3

X 100 = 0.174%

28.80cm3

They are 0.174% within each other which is well within the measurement accuracy of 0.375%.

Therefore the titration for the three results was done accurately.