Aqueous calcium hydroxide, also known as lime water is used to verify the presence of carbon dioxide gas, (carbon dioxide reacts with the calcium hydroxide to produce calcium carbonate) this is achieved by bubbling the gas through the solution, if the solution turns cloudy then the precipitate calcium carbonate has formed, thus carbon dioxide is present. It is very important for the accuracy of this test to have a saturated aqueous solution of calcium hydroxide and the Ksp of this substance allows us to determine exactly how much calcium hydroxide to dissolve until the saturation point is reached.
The purpose of this experiment is to determine the Ksp value of aqueous calcium hydroxide by mixing it with a solution of hydrochloric acid with known molarity until equilibrium is reached. This is done using the titration method.
- 50 mL of saturated calcium hydroxide solution
- 50 mL of 0.100M hydrochloric acid
- phenolphthalein indicator
- 1 burette
- 1 pipette
- three 250 mL beakers
- one graduated cylinder
- retort stand
- burette clamp
- put on all necessary safety equipment
- thoroughly rinse out all containers and measuring devices before starting
- obtain 50 mL of calcium hydroxide solution from the teacher
- obtain 50 mL of 0.100M hydrochloric acid from the teacher
- set up retort stand and burette clamp
- measure out 10 mL of hydrochloric acid in the pipette and deposit in a clean beaker
- add 2 drops of indicator
- fill the burette with calcium hydroxide
- slowly add calcium hydroxide solution via the burette to the hydrochloric acid
- when the solution turns a very pale pink stop adding the calcium hydroxide
- observe and record results
- repeat experiment twice more
- clean all used equipment thoroughly
Through four successful titrations, we determined the Ksp of calcium hydroxide to be 1.2611 x 10 -5. The accepted Ksp value for this substance is 7.9 x 10 -6.
This resulted in a 59.62% error in the experiment.
When the solution contained some solids, it was known that it is saturated. Since this solution was saturated it was implied that it was at equilibrium. Being at equilibrium the forward reaction had to equal the reverse reaction. This implied that the dissolving and crystallization reactions happened at an equal rate and simultaneously. For these reactions to be equal, some precipitate would have been present, indicating the saturated solution was at equilibrium.
It was necessary to remove all of the solid Ca(OH)2 before titrating because it would have affected the rate of reaction. When the HCl was titrated with Ca(OH)2, some of the solid would go down with the liquid because of a minuscule amount of filtering. Since the solids were really concentrated, it would have many reactions with the HCl. Therefore, the amount of Ca(OH)2 had to be only liquid, with no solid pieces going into the burette, which would have neutralized the acid too quickly.
Filtering the Ca(OH)2 can’t be done perfectly with the equipment used in this lab. Without a pure solution (of Ca(OH)2) the amount of the solution would change that was needed to titrate the HCl. The value of the solution went up thus giving a different value than was first assumed for Ksp. The less that was filtered lead to a higher value for the Ksp than first thought.
Another source of error would have been the measuring of the HCl not being exactly .15 g. The changing masses of HCl would change the results a fair amount since the values theorized in the prelab calculations were for .15 g and nothing else. The more HCl added the higher the Ksp, the less added the lower the Ksp values would have been.
One way to minimize the errors in this lab would have been to maintain the temperature of the surroundings at SATP. A temperature of 25 Celsius would have guaranteed that the assumed Ksp value was the correct one that was assumed in the prelab calculation. This value would have given the most accurate results for theorizing other values.
Another thing that could have minimized the error would be having the exact same mass of HCl. Since values were theorized to be with exactly .15 g of HCl, a slight change would skew the results a lot.
CO2 in the air reacts with Ca(OH)2 to form Calcium carbonate and water. This can be shown using the equation: CO2 + Ca(OH)2 ——> CaCO3 + H2O.
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