Aim: to find out the solubility of a substance that only partially dissolves in water.
Method: place about 100cm3 of distilled water in a flask and add about one spatula of solid calcium hydroxide. Stopper the flask and shake well for one minute. Leave to stand for at least 24 hours. Titrate 10cm3 samples against 0.05 mol dm-3 hydrochloric acid solution using methyl orange as an indicator. Obtain enough results to calculate an accurate average, and then calculate the number of moles of calcium hydroxide solution in 1 dm3 of solution.
Titration number | Final volume Ca(OH)2 (cm3) ±0.1ml| Initial volume Ca(OH)2 (cm3) ±0.1ml| Total volume added (cm3) ±0.1ml| Practice| 12.7| 0.0| 12.7|
1| 11.5| 0.0| 11.5|
2| 22.8| 11.5| 11.3|
3| 16.2| 5.0| 11.2|
4| 27.2| 16.2| 11.0|
5| 38.2| 27.2| 11.0|
Average amount of Ca(OH)2 added = 11.2cm3 (±0.1ml)
Hydrochloric acid – 10cm3, molarity is 0.05 mol dm3
Calcium Hydroxide – 11cm3
Ratio of calcium hydroxide : hydrochloric acid = 1 : 2 Ca(OH)2 + 2HCl CaCl2 + 2H2O
Molarity of Ca(OH)2
hydrochloric acid molarity,
Therefore, the molarity of calcium hydroxide is half that of the HCl =2.5×10-4
Therefore, the total estimated error = ±1.3%
1.3/100×2.5*10-4 = 3.25×10-6
If the Literature value is 1.53×10-3 mol/100g of water at 298 K then the percentage difference is 2.5*10-4/1.53*10-3 = 16.34
Problems with the experiment
1. Human error: since it was both Antonia and I taking turns to both read the markings on the burette and measure out 10cm3 of acid, we could have a different way of reading the meniscus and therefore have slightly different amounts of acid on every trial. 2. Colour of indicator: the indicator was turning from red to yellow, however there was different shades of yellow and we were...