Safety: HCl is high in concentration and so could be irritant.
Lab coat and safety goggles must be worn.
Apparatus List:
For dilution:
-
Graduated pipette (5cm-3)
-
250cm3 volumetric flask.
- Distilled Water.
- HCl solution.
For titration:
-
Graduated Pipette (25cm3)
- Burette, Stand and Clamp
- Funnel
-
Beaker containing 0.02moldm-3 HCl solution.
- Beaker containing limewater of unknown concentration.
- Conical Flask
- Phenolphthalein Indicator & White Tile
Preparations: To work out the number of moles we can use the ‘n=m/M’ equation.
Concentration of Ca(OH)2(aq).
1g dm-3 approximately.
M of Ca(OH)2(aq) = 40 + 2(16 + 1) = 74
Number of moles = 1/74 = 0.0135 moles.
Concentration of HCl(aq)
Original Concentration is 2.00 moldm-3.
In order to react with the Ca(OH)2 it has to be at least 0.0135 moles.
Therefore, it can be diluted by a x100 factor to obtain 0.0200 moldm-3.
Method: To dilute the HCl solution.
This dilution can be carried out by adding 2.5cm3 of HCl into a 250cm3 volumetric flask and filling to the line with water, i.e. 247.5cm3 of water.
The last few drops of water should be added with a pipette.
The flask should then be stoppered and shaken to mix the contents.
Titration Method:
- Set up the equipment as shown in the diagram below.
- Transfer 25cm3 of the lime water to the conical flask using the pipette
- Add three drops of the indicator into the conical flask.
-
Fill the burette to the max level (0) with the 0.02 moldm-3 HCl solution
- Carry out the titrations: open the tap on the burette and keep shaking the conical flask until it changes from colourless to pink/red. Read the level on the burette and take down the result.
Empty out the flask and wash it out, then repeat steps 1 to 5 until you have obtained three results within 0.10 cm3 of each other.
Record the results in a suitable table that has an initial and final volume line, which can be used to work out a volume used. Volume used is the final volume minus the initial.
Calculations:
The results can be used to calculate an average volume used.
This is the known information at this point.
V represents the average volume of HCl used.
As results were calculated in cm3 it must be divided by 1000 to get dm3.
Moles can be worked out using the equation ‘Concentration = Moles / Volume’
i.e. 0.02 x V.
The reaction shows that two moles of HCl react with one mole of Ca(OH)2 so in the neutralisation reaction there must have been twice as many moles of HCl than Ca(OH)2 . Therefore moles of Ca(OH)2 must be (0.02 x V) / 2 = 0.01 x V.
Using the same equation, Conc. of Ca(OH)2 = (0.01 x V) / 0.025 = 0.4 x V.
The mass of the Calcium Hydroxide in the solution can now be worked out using ‘n=m/M’.
M(Ca(OH)2) = 40 + 2(16 + 1) = 74
Mass of Calcium Hydroxide in dm3 = 0.4 x V x 74
Therefore the concentration of the Limewater in gdm-3 will be:
0.4 x V x 74 = ????? gdm-3
Bibliography:
T. Lister and J. Renshaw , “Understanding Chemistry for Advanced Level” Chapter 12.
T. Lister and J. Renshaw, “Essential AS Chemistry for OCR” Chapter 15.