= 1/ 74.1
= 0.013495276 mol dm-3
As I predict that, I will use 25.0 cm3 of calcium hydroxide per titration
Moles = concentration x volume
= 0.01 x (25.0 /1000)
=0.0003 moles (1 s.f)
As the equation is 2HCl + Ca(OH)2→ CaCl2 + 2H2O
The ratio of Hydrochloric acid to Calcium hydroxide is 2:1
Therefore the moles of hydrochloric acid is 0.0003 x 2 = 0.0006 moles
Volume HCl is needed is approximately 30.0cm per titration
Concentration= moles / volume
=0.0006/ (30.0/1000)
=0.02 mol dm-3
As the concentration of hydrochloric acid given is 2.00 mol dm-3 and the concentration needed is 0.02 mol dm-3.
2.00/0.02= 100
So, the hydrochloric acid needs to be diluted 100 times.
Apparatus
Overall
Safety glasses
1 graduated syringe
2 conical flasks
1 pipette
1 pipette filler
1 burette
1 small tile
Phenolphthalein indicator
A glass rod
A volumetric flask
250cm3 of calcium hydroxide
Hydrochloric acid solution
A funnel
A clamp and stand
Beaker
Diagram of titration
Risk Assessment
Calcium hydroxide can be harmful.
Hydrochloric acid is corrosive and may burn skin so you should wear an overall to protect your clothing and skin.
Phenolphthalein indicator may irritate eyes and can damage respiratory system so you should wear safety glasses to protect your eyes.
Plan
- Before starting the experiment, make a table showing the pointing and end point for each titration and showing the total volume used.
-
Fill the concentrated hydrochloric acid in the graduated syringe up to 2.5 cm3.
-
Put the hydrochloric acid from the graduated syringe into a 250cm3 beaker, and add diluted water until the 250cm3 mark and stir with a glass rod.
-
Then pour into a 250cm3 volumetric flask and fill until 250cm3 mask accurately, this is done by filling up until the bottom of the meniscus measures from the eye level.
-
Get the 250cm3 of calcium hydroxide from a beaker use a 25cm3 pipette and put it into a conical flask
- Place the conical flask on top of a tile
- Set up the clamp and stand to hold a burette
- Fill the burette with the diluted hydrochloric acid making sure the bottom of the burette is filled with hydrochloric acid.
- Put the burette on the clamp, and record the starting point of the acid
- Add a few drops of phenolphthalein indicator into the calcium hydroxide and it will turn the solution pink/red
- Put the conical flask underneath the burette and begin to titrate
- When the solution turns colourless permanently turn the tap of the burette off, and record the end point in the table
-
Begin the titration process again, with a new 25cm3 of calcium hydroxide in the conical flask and top the amount of hydrochloric acid and record the starting point. Continue to titrate until you get a titre within 0.1 of each other
Results
The volume of original hydrochloric acid used was 2.5cm3
The volume for diluted hydrochloric acid in the volumetric flask is 250.0cm3
The pipette volumes of calcium hydroxide used for each titration is 25.0cm3
Conclusion
The concentration of diluted hydrochloric acid used
As 2.5cm3 of hydrochloric acid was used and the concentration was 2 mol dm-3
Moles of diluted HCl= concentration x volume
= 2 x (2.5/1000)
= 0.005 mol
The volume of diluted hydrochloric acid was 250cm3
Concentration of diluted of HCl= moles/ volume
= 0.005/(250/1000)
= 0.02 mol dm-3
The concentration is the amount required to do the titration, which is shown above in the introduction
For the average titre, I used the 5th and 6th reading as they were within 0.1 of each other.
The average titre is 31.75+ 31.85= 31.80 cm3 (2 d.p)
2
For this I ignored any anomalous results.
Using the average titre volume, I can now calculate the accurate concentration of limewater which I titrated.
The moles used per titration = volume x concentration
= 0.02 x (31.8/1000)
= 0.000636 mol
As the chemical equation is: [2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)]
Ratio of HCl and Ca(OH)2 is 2:1
Therefore moles of calcium hydroxide: 0.000636 /2 = 0.000318
Concentration of Ca(OH)2 = moles/volume
= 0.00318/(25/1000)
= 0.01272 mol dm-3
Concentration of Ca(OH)2 in g dm-3 = concentration x mr
= 0.01272 x (40.1+(16 x 2) +2)
=0.01272 x 74.1
= 0.94 g dm-3 (2.d.p)
Evaluation
At the beginning of the experiment, you were told that the limewater given to you was just approximately 1 g dm-3. The answer calculated is 0.94 g dm-3, which is only 0.06 away from the approximated amount. The average concentration found of the limewater solution is 0.76 g dm-3. My results were 0.18 g dm-3 more concentrated than the average. This could be because the titration method as it is difficult to get the exact volume of hydrochloric acid needed to neutralise the calcium hydroxide solution that is why everyone final concentration for calcium hydroxide was slightly different.