# Determining the concentration of acid in a given solution

Determining the concentration of acid in a given solution

Planning 4

I have been given a sample of sulfuric (VI) acid solution with a concentration between 0.05 and 0.15 mol dm-³.  I am going to find out the accurate concentration of the sulfuric acid. To find out the concentration of the acid I will react it with a known volume and concentration of a base and see how much base was needed to neutralise the acid.

The acid is a strong acid which means that I know all the H+ ions have been disassociated and are in the solution. The H+ ions will react with the OH- ions in the alkali which will neutralise the solution.

I am provided with solid, hydrated sodium carbonate with the formula Na2CO3·10H2O.1 This is a readily available base and I can dilute it down to achieve the concentration I want to react with the acid.

The formula of the reaction that will take place is

H2SO4 (aq)    +     Na2CO3 (aq)    →             Na2SO4 (aq)    +    CO2 (g)   +    H2O (l)

So 1 mole of   H2SO4   reacts with 1 mole of   Na2CO3.

A titration will give me the most reliable and accurate results with the available equipment.

To do my titration I will need:

A Burette 7

I will need a burette to add the sodium carbonate to the sulfuric acid solution. The burettes provide me with very accurate results of volume of solution added.

The class set of burettes measure 50cm3. I want to do a titration of around 25cm3 - 35cm3 each time. This is because the larger the amount of solution I titrate, the smaller the percentage error. I don’t want to plan for each titration to exceed 35cm3 as if each titre is slightly larger then I may run off the scale of the burette. If I had to refill my burette during a titre, then the percentage error would double as I would have to read the burette at the beginning and the end of the first amount of solution and again at the beginning and end of the second solution.

I can read the burette to the nearest 0.05cm3 so the precision error is ±0.025cm3 for each reading. I will have to read the burette at the beginning and end of the titre so the precision error for each titre is ±0.05cm3.

If the titre is 30cm3 the error is   (0.05/30.00) x 100 = 0.16% which is acceptably low.

A pipette 7

I will need a very accurate, glass pipette which I can use to transfer a specific known volume of sulfuric acid added to the conical flask to be reacted.

There are class sets of pipettes of 10cm3 and 25cm3. I have a large amount of sulfuric acid, so I am not limited to using small volumes of sulfuric acid each time.

A pipette measuring 10cm3 is accurate to ±0.04cm3, so has a percentage error of 0.4%. A pipette measuring 25cm3 has a precision uncertainty of ±0.06cm3, yet because of the larger volume it measure, it has a percentage error of 0.24%.

I am going to use a pipette which measures 25cm3 of solution as I am not limited on the amount of solution I can use and it will provide me with the least percentage error and therefore mean my results will be more accurate.

I will use a volumetric flask to create the correct concentration of sodium carbonate solution. I will add the mass of solid sodium carbonate I need and then fill the volumetric flask up with distilled water, so that there is a know concentration of the solution.

The error on a 250cm3 volumetric flask is 0.2cm3 if read correctly.

This means that the percentage error is 0.08%. This is a very low percentage error for the volumetric flask and will have very little effect on the overall concentration of the sodium carbonate solution.

I know that the concentration of sulfuric acid is between 0.05 and 0.15 mol dm-3. To do some estimation calculations I will presume that my acid is 0.10 mol dm-3. I will use 25cm3 of the acid in a conical flask as this is how much I will be able to add accurately using a 25cm3 pipette.

This means that there will be            0.10 moles H2SO4                                      in 1000cm3

(0.10/1000)                                in 1cm3

((0.10/1000) x 25) = 2.5x10-3 moles            in 25cm3

In the reaction between the acid and base, the disassociated H+ ions and OH- will react to neutralise the solution. I am going to use a similar concentration of sodium carbonate and sulfuric acid so that when the reaction reaches the endpoint, all the H+ ions have reacted and the OH- ions are not in excess. I will therefore use a concentration of approximately 0.10 mol dm-3 as this is the middle value of the possible concentration of sulfuric acid.

1 mole of H2SO4 will react with 1 mole of Na2CO3

Using an approximate titre of 30cm3 I will need

As these calculations were all based on approximate values I will round 0.083 up to 0.1 mol dm-3, as 0.1 is an easier concentration to make up and will ensure that there are enough OH- ions to react with the H+ ions.

I am using titres of approximately 30cm3. I will need some sodium carbonate solution to rinse the burette, solution to do a rough titration, and I will want about 5 titres worth to get 3 accurate titres so I will need (7x30)= approximately 210cm3 of sodium carbonate solution.

I will use a volumetric flask to measure up the amount of sodium hydroxide solution needed. There are volumetric flasks of 100cm3 and 250cm3. I will use the 250cm3 flask as this will provide me with enough solution to do all my titres and will minimise the percentage error as the volume it carries is larger.

Measuring the sodium carbonate solution 1 2

I need to make 250cm3 of a sodium hydroxide solution with the concentration of 0.1 mol dm-3. I will use solid hydrated sodium carbonate with the formula Na2CO3·10H2O.1

Mr (Na2CO3·10H2O)  = (23+23+12+48+20+160) = 286

250cm3 of 0.1 mol dm-3 contains (0.1 x 250/1000) moles

I will need (0.1 x 250 x 286/1000) = 7.15g of Na2CO3·10H2O in 250cm3

The error on a 2dp balance would be ± 0.005g. I would need to tare the balance before I weigh my solid, so this error would be doubled for every weighing because the balance needs to estimate 0.00 and the actual weight of the solid.

If I measured out 7.15g the percentage error for each reading is (0.005 x 2/ 7.15) x 100= 0.139%

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#### Here's what a star student thought of this essay

Spelling, grammar and puncutation all clearly set out and demonstrated to a very accurate and high level.

Planning is adequate and sets out the scientific background and intentions of calculation for the experiment well. The use of each piece of apparatus is explained well, with the percentage errors that might arise from each piece of equipment considered before the experiment is set out, also suggesting basic improvements that may be done to improve the test and implementing them, although I would have liked the candidate to have considered a wider range of improvement strategies. Method of the experiment is explained logically and clearly with the results set out correctly and consistently within a suitably designed table, although I'm not sure why the candidate highlighted some figures in bold. Considers improvements and things that may have gone wrong in the experiment to a very high level in the evaluation, and the fact the candidate considered improvements both before and after the experiment shows a higher level of thinking and consideration. Overall an examplary piece.

Response to the question is very clearly set out, this piece of coursework is an exemplary piece for the level I would expect from A level candidates. The introduction, main text and evaluation are all carefully considered and the only way the candidate could have improved the experiment and content would have been to consider properties of the experiment that show thinking outside of the A level syllabus.