Find the solubility of a sodium hydroxide solution (lime water).

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 Chemistry coursework assessment

 November2002

AN EXPERIMENT TO DETERMINE OF THE SOLUBILITY OF CALCIUM HYDROXIDE

Giles Greenwell

Aim

To find the solubility of a sodium hydroxide solution (lime water).

In this investigation, I intend to find the solubility of calcium hydroxide in water.  Calcium hydroxide dissolves in water to form an alkaline solution.  To find the solubility, I need to find the concentration of a saturated solution.  To do this, I will titrate an acid of a known concentration, until I have a neutral solution.  I can then find the number of moles of acid that reacted with the alkali, and so find the number of moles of calcium hydroxide.  As the volume is known, the concentration, and therefore the solubility, can be found.

Ca(OH)2        +        2HCl                        CaCl2                +        2H2O

        1                2                        1                        2

You can see from the equation above that reacting with each mole of calcium hydroxide are two moles of hydrochloric acid.

Therefore if the calcium hydroxide concentration was 1 mole dm-3, and the volume was 25 cm3, 50 cm3 of 1 mol dm-3 hydrochloric acid would be needed to react with it.  However, It is stated that the concentration of the hydrochloric acid is 0.3 mol dm-3, and the concentration of the lime water is approximately 0.015 mol dm-3.  It will therefore be necessary to dilute the hydrochloric acid, so that a particularly large amount is not needed.  I will give further details of this dilution in the plan.

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While I am conducting the titration, I will put the conical flask on a white tile, so that I can see the colour change more easily.

        It is suggested that any indicator is suitable when a strong alkali reacts with a strong acid, so I have chosen methyl orange for the titration.

Prediction

I have included calculations to show the approximate amount of acid needed to neutralise the alkali.  From these calculations, you can see that if the concentration of the hydrochloric acid is 0.3 mol dm-3, there will only be approximately 2.5cm3 needed.  Smaller measurements allow a larger percentage ...

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