The Determination of the Solubility of Calcium Hydroxide

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The Determination of the Solubility of Calcium Hydroxide

I have to plan an experiment to find the solubility of calcium

hydroxide, Ca(OH)2, in water. I have to make up a solution of calcium

hydroxide and carry out a titration using hydrochloric acid solution

of the chosen concentration.

The equipment need is as below:

· Solid calcium hydroxide

· Methyl orange indicator

· Volumetric flask (250cm3)

· Clamp and boss

· Clamp stand

· Burette (50cm3)

· Conical flask

· Pipette (25cm3)

· Pipette filler

· Distilled water

· White spotting tile

· Hydrochloric acid of chosen concentration

· Beaker x2

· Rubber bung

· Funnel x2

· Electronic scale

‘The maximum mass of calcium hydroxide needed to produce 1dm3 of

saturated solution at room temperature is 1.5g.’

I only want 250cm3 as I am using a 250cm3 volumetric flask. Therefore:

1dm3 / 4 = 250cm3

1.5g / 4 = 0.375g

The number of moles in volumetric flask:

0.375 / 74 = 0.005 moles

I need an excess of 0.5g to make sure that all the calcium hydroxide

has been fully dissolved:

0.375g + 0.5g = 0.875g

I have to now work out the concentration of hydrochloric acid I will

be using. The molar mass of calcium hydroxide is:

C = 40 O = 16 (x2) H = 1 (x2)

R.A.M = 74

The concentration of calcium hydroxide at the beginning will be:

1.5 / 74 = 0.02

So, the concentration is 0.02 mol/dm3

In the experiment I will be using 25cm3 of the solution from the

volumetric flask, so the mass of the calcium hydroxide in one

titration will be:

0.375 / 10 = 0.0375g

Therefore the number of moles of calcium hydroxide:

0.0375 / 74 = 0.0005 moles

Ca(OH)2(aq) + 2HCl à CaCl2(aq) + 2H20(l)

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