Equation
Ca(OH)2 + 2HCl � CaCl2 + 2H2O
calcium hydroxide + hydrochloric acid � calcium chloride + water
Apparatus
Burette 50cm3
Pipette 20mls/25mls
Volumetric Flask 2000cm3
Conical Flask
Boss and Clamp
Funnel
White tile
Indicator
2000cm3 0.02 mol Hydrochloric acid
250 cm3 Limewater
Firstly, the acid that I was given is too strong for the reaction to be reliable, which means that it has to be diluted. To get a suitable concentration of the acid which will titrate into the base:
2.00 = 0.02 mol
100
20:2000
=1:100
So to make 0.02mol of hydrochloric acid; measure 20cm3 of 2mol dm3 accurately into a pipette.
Making sure the bottom of the meniscus is in line with the 20cm3 line.
Put this into a 2000cm3 volumetric flask and fill it up to the 2000cm3 line with water. This then needs to be mixed thoroughly by turning it upside down 100 times.
In order to decide on the volume of hydrochloric acid to use, firstly the concentration of the calcium chloride needs to be found: I was provided with 250cm3 of calcium chloride so I decided to use 25cm3, this will give up to 10 titrations.
However although doing the titration this many times would make the results very accurate, it would take up a lot of time. So to do the experiment 4 times would give an accurate enough result, if the first test was done as a practice. Here I have found the number of moles in 25cm3 of calcium hydroxide.
Ca(OH)2 + 2HCl � CaCl2 + 2H2O
n = M
m
n = 1
74.1
0.0135
n = cV
= 0.0135 x 25
1000
= 0.0003375 mol dm3
To find the volume of hydrochloric acid that is needed to neutralise 25cm3 of the base, the concentration of the acid needs to be found first.
This is found from looking at the ratio between the calcium hydroxide and the hydrochloric acid. In this case the concentration would be multiplied by two, because the ratio is 1:2.
Ca(OH)2 + 2HCl � CaCl2 + 2H2O
0.0003375 x 2
= 0.000675 mol
1:2
0.0003375
So to find the volume that will need to be used in the experiment, the number of moles needs to be divided by the concentration. This will give the answer in dm3, so to get it into cm3 it needs to be multiplied by 1000.
Ca(OH)2 + 2HCl � CaCl2 + 2H2O
V = n
c
= 0.000675
0.2
= 0.0375 mol dm3
= 0.0375 x 1000
= 33.75 cm3
This shows approximately the amount of hydrochloric acid that will be needed to neutralise the calcium hydroxide in the experiment.
Method
1. Firstly safety goggles must be worn at all times, along with any loose clothing or long hair tied back. If any of the solutions are spilt flood the area with water.
1. Collect the apparatus of:
Burette
Pipette 25mls
Boss and Clamp
Conical Flask
Beaker 250mls
Funnel
White tile
Indicator
Safety goggles
2. Set up the apparatus as it is shown in the diagram.
3. Collect the 250mls of calcium chloride in a clean beaker.
4. Measure 25mls of calcium hydroxide solution in the 25mls pipette, so that the bottom of the meniscus is in line with the 25mls mark. However before doing this make sure that the pipette is clean by rinsing it out with water and then with calcium hydroxide solution.
5. Run the calcium hydroxide solution into a clean conical flask making sure to get every last drop out of the pipette. Press the tip onto the inside of the conical flask to get the last drops out.
6. Collect the hydrochloric acid in a 250mls beaker.
7. Clean out the burette with water and then hydrochloric acid, making sure to rotate the burette to get everywhere inside it. Pour the acid into the burette with a funnel at the top, making sure to run some of the acid through to allow some to get into the tap. Measure it all the way to the 50mls at the top making sure that the bottom of the meniscus is at the 50mls line.
8. Add five drops of the indicator into the calcium hydroxide, and lightly swirl to mix it.
9. Slowly run in the acid swirling.
I chose to use 33.7 cm3 of hydrochloric acid that will only give one tiration out of a burette, rather than a smaller number that would give two or three tirations from the one burette because it works out as more accurate.
For example if the actual volume needed in the titration was 13.7cm3 and the tiration came out with 13.5, there would be a difference of 0.02ml:
%Diff = 0.2 x100
13.7
= 1.5%
But if the volume in the titration came out as 33.5 when the actual volume was 33.7 and there was the same percentage difference of 0.2 then:
% Diff = 0.2 x 100
33.7
= 0.6%
This clearly shows that using 33.7 as the volume for hydrochloric acid is much more accurate.
Remember! This is just a sample.
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