To determine the concentration of a limewater solution
To determine the concentration of a limewater solution
Aim
In this experiment I will be determining the concentration of limewater solution, in g dm-3, by titrating the calcium hydroxide solution, Ca (OH)2, with dilute hydrochloric acid, HCl.
Introduction
One of the most accurate ways of determining the concentration of a given solution is to establish how much of another solution will neutralize it. This type of experiment is called a titration. The solution in question is limewater , which is an alkali, so in order to neutralize it, we require an acid. A good acid for this experiment would be hydrochloric acid. This is because the products given off when mixed with the limewater are not harmful, and hydrochloric acid is a strong acid, so the end-point (the point where the solution has been neutralized,) will be more profound.
The equation I have looked briefly upon above is shown below. This is the balanced equation for titrating the limewater solution with hydrochloric acid is as follows:
The hydrochloric acid available is far too concentrated to be used safely and accurately in this experiment. Therefore I will need to make a standard solution of it. "A standard solution is one which is made up to have a known concentration of a known solute."
Equipments
The things that I will need are as follows:
* A volumetric flask that has 250 cm3 clearly marked
* Methyl orange indicator
* A stand to hold the burette rigidly
* I will need distilled water, which will be needed for dilution
* Four 250 cm3 conical flasks
* 250 cm3 of limewater which has been made such that it contains approximately 1g dm-3 of calcium hydroxide
* Approximately 250 cm3 of 2M hydrochloric acid
* A pipette to pipette out 25 cm3 of Ca(OH)2
* A burette, which I'll need for the titration of Ca(OH)2 with HCl
* Two 10 cm3 measuring cylinder will be used for the dilution. This will be needed to measure out the HCl.
* A funnel which will aid me in putting the HCl into the burette and I will also need a white tile, which will help me to clearly see the reaction taking place
Dilution
As I've stated above, I'll be provided with 2M hydrochloric acid. This is far too concentrated, so I will have to dilute this concentrated acid to approximately 0.1 moles dm-3.
In order to dilute the HCl to 1 molar; I will need to apply the formula shown below:
C1V1=C2V2
We are given the concentration of the HCl, which is 2.00mol dm-3, this is C1.
We are also given the volume of distilled water, which is 250cm3, this is V2.
We want to make the concentration of HCl into 0.1 ...
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Dilution
As I've stated above, I'll be provided with 2M hydrochloric acid. This is far too concentrated, so I will have to dilute this concentrated acid to approximately 0.1 moles dm-3.
In order to dilute the HCl to 1 molar; I will need to apply the formula shown below:
C1V1=C2V2
We are given the concentration of the HCl, which is 2.00mol dm-3, this is C1.
We are also given the volume of distilled water, which is 250cm3, this is V2.
We want to make the concentration of HCl into 0.1 moles dm-3, this is C2.
In order work out the volume of HCl, V1, needed to make 0.1 moles dm-3 we need to apply the above formula:
So By putting in the values
V1= C2V2 V1= 0.1? 250 C1 2
V1= 12.5cm3 (to 3.s.f)
I will measure out 12.5 cm3 of 2M HCl, before putting this in a volumetric flask. The number of moles in 12.5 cm3 of 2M HCl is as follows:
Number of moles = concentration (moles dm-3) ? volume (dm3)
= 2.0 ? (12.5/1000)
= 0.0250 moles of HCl (to 3.s.f)
At the moment I've got 12.5 cm3 of 2M HCl in the volumetric flask, which has a mark on 250 cm3. So, if I was to fill up the volumetric flask up to that mark with distilled water, including the12.5 cm3 of 2M HCl, the concentration of the acid will be 0.1 moles dm-3, as the following calculation illustrates:
Concentration (moles dm-3) = number of moles
volume (dm3)
= 0.025 = 0.100 moles dm-3 (to 3.s.f)
0.250
Therefore, in the volumetric flask I'll have 12.5 cm3 of 2M HCl and (250 -12.5) = 237.5 cm3 of distilled water to give me a concentration of 0.1 moles per dm3.
Method: Implementation of experiment
. Dilute the 2M HCl as explained above so that the concentration is 0.1 moles dm-3.
2. Get a stand for the burette and assemble it so that it is stable on the stand
3. Get four 250 cm3 conical flasks out, as one will be needed for a trial and the other three for the actual titres
4. At this point wear an apron, goggles and make sure all loose objects or things are out of the way
5. Pipette out 25 cm3 of limewater into a clean 250 cm3 conical flask. Add five drops of methyl orange indicator
6. Putting the burette under eye level, and making sure the tap is closed so no HCl falls out, fill it up until zero with 0.1M HCl via the funnel on top of the burette
7. Under the conical flask, put a white tile so as to make it easier to spot colour changes
8. Start titrating. Gently shake the conical flask and as soon as the solution turns from orange to pink, it indicates that the solution has ceased, so record the amount of HCl used clearly on a table. Make sure that the next time you do it the colour is consistent, which should be a pink tinge, and to get an accurate value of HCl used, look at the meniscus from eye level.
9. Repeat step 5 to 8 a further three times, as the first one was a trial. Bear in mind the amount of HCl needed in the trial, so you'll be aware of when the solution will turn colourless
0. The results will be recorded in the following way shown and the average will be the mean of the differences between the three titres, titres 1,2 and 3. Trial will not be taken into account.
Trial
Titre 1
Titre 2
Titre 3
Average
Initial volume (cm3)
Final volume (cm3)
Difference (cm3)
Method: Processing of results
. Calculate the number of moles of HCl used in the titration. Use the formula: number of moles = concentration (moles dm-3) ? volume (dm3).
2. Deduce the number of moles of Ca (OH)2 used in the titration. This can be done because we know the number of moles of HCl as well as the stoichiometric ratio from the equation between HCl and Ca (OH)2. Use this: number of moles of Ca (OH)2 = number of moles of HCl ? 1/2
3. Calculate the concentration of Ca (OH)2. This is viable, as we know the number of moles of it from above and also the volume, which is 0.25 dm3, and we could simply rearrange the formula in 1 to work out the concentration. The concentration will be given in moles per dm3.
4. However, we're required to work out the concentration in grams per dm3, so we would have to convert it from moles per dm3 into grams per dm3. First I'll
5. have to work out the molecular mass of Ca (OH)2, before applying the following formula 'mass = number of moles ? molecular mass'. The answer to that will be in grams; therefore we get the concentration of the limewater in grams dm-3.
Health and safety
Hydrochloric acid is a very corrosive chemical so at all times it should be treated with the maximum amount of caution. At all times when dealing with hydrochloric acid, you must wear goggles and aprons. When putting the hydrochloric acid into the burette, extra care must be taken. When doing the experiment make sure everything loose is out of the way. This means long hair must be tied back, bags and coats must be completely out of the way as well as the stools, which people can trip over, with catastrophic consequences a likely scenario.
Analysis
I was presented with 250 cm3 of calcium hydroxide, which I titrated with 0.1M HCl. I did four titres, out of which one was a trial. Here are the results.
Trial
Titre 1
Titre 2
Titre 3
Average
Initial volume (cm3)
0.00
0.00
0.00
0.00
Final volume (cm3)
4.80
4.80
4.80
4.75
Difference (cm3)
4.80
4.80
4.80
4.75
4.78
Mean titre = Titre 1 + Titre 2 + Titre 3
3
= 4.80 + 4.80 + 4.75 = 4.78 cm3 (to 3.s.f)
3
I know the concentration of the hydrochloric acid, which is 0.1M, and from the mean titre I also know the volume of the hydrochloric acid used. This will allow me to work out the number of moles of HCl, which is as follows:
Number of moles = concentration (moles dm-3) ? volume (dm3)
= 0.1 ? (4.78/1000)
= 0.000478 moles of HCl (to 3.s.f)
Now that I am aware of the number of moles of HCl, I look at the chemical equation to see the stoichimetric ratio between HCl and Ca (OH)2.
Ca (OH)2(aq) + 2HCl(aq) ? CaCl2 (aq) + 2H2O(l)
The equation tells me that 2 moles of HCl reacts with 1 mole of Ca (OH)2. Therefore I can now work out the number of moles of calcium hydroxide, as I know the number of moles of HCl.
Number of moles of Ca (OH)2 = number of moles of HCl ? 1/2
= (4.78 ? 10-4) ? 1/2
= 0.000239 moles of Ca (OH)2 (to 3.s.f)
Now that I know the number of moles of calcium hydroxide and the volume of it, I can work out its concentration by virtue of applying the following formulae:
Concentration (moles dm-3) = number of moles
volume (dm3)
= 2.39 ? 10-4
0.025
= 0.00956 moles dm-3 (to 3.s.f)
The criterion for this particular assessment is to work out the concentration of calcium hydroxide in grams per dm3, not in moles per dm3. So, I will have to work out the molecular mass of Ca(OH)2, but before that I need to know the relative atomic masses of the elements concerned.
Relative atomic mass of H = 1
Relative atomic mass of Ca = 40
Relative atomic mass of O = 16
Therefore molecular mass of Ca(OH)2 = 40 + (2 ? 16) + (2 ? 1) = 74g
Since the concentration of the limewater solution is 0.00956 moles dm-3, I need to know the mass of Ca(OH)2 that has 0.00956 moles of it. Since I know the molecular mass of Ca(OH)2, I can work out the mass that contains 0.00956 moles, which is as follows:
Mass = moles ? molecular mass
= 0.00956 ? 74
= 0.707g (to 3.s.f)
Therefore the concentration of the limewater solution is 0.707 grams per dm3.
Bibliography
* http://www.keot.co.uk/misc/as-che-coursework-task-p-240203.doc, from this site I got the balanced equation for my limewater investigation. I also got a definition on what is a standard solution.
* From my Chemistry OCR book 1, I used pages 21-28, which showed me how to carry out my required calculation for my investigation. It also showed me how titration is carried out.