In order to determine the concentration of the limewater, it is necessary to react hydrochloric acid with the limewater.
Metal Hydroxide + Acid Salt + Water
The products involved in the reaction are below:
Limewater + Hydrochloric acid Calcium Chloride + Water
The balanced equation including the state symbols is below:
Ca(OH)2 (aq) + 2HCl (aq) CaCl2 (aq) + 2H2O (l)
The ratio of the reactants Ca(OH)2 and HCl is 1:2.
This means that one mole of Ca(OH)2 reacts with two moles of HCl to give 1 mole of CaCl2 and two moles of water.
The approximate number of moles and concentration of Ca(OH)2
Moles = mass / mr
Mr of Ca(OH)2 = 40 + (16 + 1) * 2 = 74
Mass of Ca(OH)2 used in 1000cm3 = 1g.
Number of moles of Ca(OH)2 used in 1000cm3 of limewater = 1/74 = 0.0135 moles
The approximate concentration of the limewater therefore is 0.0135 mol dm-3
Number of moles used in 250cm3 of limewater = 0.0135 * 250/1000 = 0.003375 moles approximately
Approximate number of moles of HCl used in the reaction
Mole ratio of Ca(OH)2 and HCl = 1 : 2
Number of moles of Ca(OH)2 = 0.003375 moles
Approximate of the number of moles of hydrochloric acid required to neutralise the limewater =
0.003375 * 2 = 0.00675
Approximate volume of HCl required to neutralise the Ca(OH)2
Two moles are present in 1000 cm3 of the HCl. Therefore volume of the acid that contains 0.00675 moles =
1000 * 0.00675 / 2 = 3.375cm3
This means that 3.375 cm3 of hydrochloric acid will neutralise the limewater and finish the reaction.
The burette available can only read up to 1 decimal place therefore we can round up the approximate number of moles of HCl up to 3.4cm3 which is more appropriate for the burette used.
Dilution of HCl
A titre volume 3.4cm3 is too low and there would be more margin for error. In order to get reasonable titre volume I would need to dilute the acid. This will enable me to produce reliable results.
I can use the following dilution factors which would give me a variety of choices:
Approximate titre volume of 2.00 mol dm-3 HCl = 3.4cm3
Approximate Titre Volume = Dilution factor * 3.4
17cm3 is a reasonable titre volume therefore I will dilute the acid by a dilution factor of 5.
In order to dilute the acid five times I will pipette 10cm3 of hydrochloric acid two times into a 100cm3 volumetric flask. I will then fill this volumetric flask up to the mark by adding distilled water. This will give me 100cm3 of dilute hydrochloric acid. This acid would be diluted 5 times because only 20cm3 of acid was used and it was put into a volume of 100cm3. Therefore 20 100 = 5 times dilution.
In order to calculate the actual concentration of limewater we need to perform titrations. Then we can use the average titration volume to work out the actual concentration of the limewater.
Below is how this would be done:
The concentration of HCl after dilution would be: 2.00 / 5 = 0.4M
Number of moles ‘x’ cm3 of HCl = (x / 1000) * 0.4 = y
Where x = titration volume and y = moles of HCl used in titration
Molar ratio of Ca(OH)2 and HCl = 1 : 2
2 moles of HCl = 1 mole of Ca(OH)2
So, 1 mole of HCl = 1/2 mole of Ca(OH)2
Hence, y mole of HCl = y / 2 mole of Ca(OH)2
Therefore number of moles of Ca(OH)2 = y / 2
This is for the number of moles of Ca(OH)2 in a volume 250cm3.
The number of moles in 1000cm3 = (y / 2) * (1000 / 250)
Therefore actual concentration = (y / 2) * 4
These calculations come from the using the formula:
Actual concentration = number of moles used in the reaction / titration volume (dm-3)
In order for the experiment to be accurate and reliable it will be necessary to dilute the hydrochloric acid significantly. I think that a concentration of 0.08 mol dm-3 would enable me to achieve reliable results.. The dilution factor can be calculated by dividing the volume of the dilute hydrochloric acid by the volume of the hydrochloric acid used:
In this case the volume of the dilute HCl is 250cm3
The volume of HCl used is 10cm3, so:
Dilution factor = 250 / 10 = 25
This acid is too concentrated to be used for this experiment because the titre volume will be too low, hence more error margin. Therefore I will have to dilute the acid.
The limewater used will be a relatively weak base so it will be appropriate to use methyl orange as it has an end point on the PH scale between 5-8. Whereas phenolphthalein indicator has an end point much higher up in the PH scale. So if I was to use the phenolphthalein indicator then the colour change would be reached before the acid-base breaks up which would not give us proper results.
Once I have obtained the correct concentration of hydrochloric acid I can precede to the rest of the experiment. First I must set up the apparatus as I have indicated in my diagram. I will be using 25.0cm3 of limewater and I will transfer it into the conical flask using a 25.0cm3 pipette. It is also important to add the indicator into the conical flask so that the neutralisation point can be obtained. Five drops of methyl orange would be appropriate. When the indicator is added the colour of the solution is light, clear, orange and this should turn to pink by the end of the titration. In order to obtain consistent and reliable results it is necessary to repeat the experiment several times.
Results
I worked out the volume of acid used by subtracting the initial burette reading from the final burette reading:
E.g.: it can be seen above that in titration 2 the initial burette reading was 8.30cm3, and the final burette reading was 16.40cm3. I worked out the volume of acid used by using the method that I exemplified above. Below is the simple calculation:
16.10 – 8.10 = 8.00cm3
Average titre
8.10 + 8.00 + 8.20 = 24.30
24.30 / 3 = 8.10cm3
Number of moles of HCl:
1dm3 = 10*10*10 = 1000cm3
In order to convert the volume into dm3 the volume must be divided by 1000:
8.10 / 1000 = 0.0081dm3
Number of moles = concentration * volume:
0.08 * 0.0081 = 0.000648 moles
Number of moles of Ca (OH)2 (Calcium hydroxide):
Reacting ratio = 1 Ca (OH)2 : 2 HCl
0.000648 / 2 = 0.000324 moles
Concentration of Limewater in mol dm3:
0.000324 / 0.025 = 0.01296 mol dm3
Concentration of Limewater in g dm3:
Formula mass (Ca(OH)2) = 40.1+ (16+1) * 2 = 74.1g
Concentration of Ca(OH)2 = 74.1* 0.01296
= 0.960336 g dm3
= 0.960 g dm3 to 3 d.p
1 – 0.960 = 0.04
Above it can be seen that my results were quite accurate as they are near to the 1g dm3 estimate given to us. I was only 0.04 g dm3 off which shows very high precision. I am satisfied with my results but will need to go through any source of error Also the fact that my results are consistent with each other proves that they are reliable. I think that I would improve my results even more if I had another opportunity to carry out the experiment.
