Determine the Concentration of Limewater Solution

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Determine the Concentration of a Limewater Solution

Aim:

The aim of this experiment is to determine the concentration of a limewater solution using hydrochloric acid.

Introduction:

To carry out this experiment, the method that will need to be used will be a titration. In this experiment, 250 cm3 of limewater with a concentration of 1g dm-3 will already be provided. Also, 2.00 mol dm-3 of hydrochloric acid is also provided, however, this acid is too concentrated and it will need to be diluted, before it can be used. I will be planning an experiment which will allow me to accurately determine the concentration, in g dm-3, of the limewater. I will be using an indicator to determine the point where the solution becomes neutral and when all of the limewater in the flask has reacted with the acid in the burette. A suitable indicator to use for this reaction would be methyl orange as a distinct colour (more obvious change than using methyl orange) change can be seen going from orange to pink, in this reaction.

Diluting the Hydrochloric Acid

As I stated in the introduction, I will need to dilute the acid as it is too concentrated. Before I do this, I will need to carry out a few calculations to determine how concentrated the HCl will be.

This balanced equation shows the reaction that takes place when calcium hydroxide (limewater) is added to hydrochloric acid:

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

Concentration of Calcium Hydroxide

I will now need to work out the number of moles per decimetre cubed of the calcium hydroxide required in this reaction using the following formula:

Concentration in mol dm-3 = concentration in g dm-3 / RMM

I have already been given the approximate concentration of limewater which is 1 g dm-3

The RMM of Calcium Hydroxide (Ca(OH)2) is the following:

RMM of Ca = 40.1

RMM of O2 = 32.0

RMM of H2 = 2.0

= 40.1+32.0+2.0 = 74.1

Concentration in mol dm-3 = concentration in g dm-3 / RMM

                                   = 1/74

                                    = 0.0135 mol dm-3 of Calcium Hydroxide used in this equation.

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Concentration of Hydrochloric Acid needed following dilution:

I will now work out the concentration of the hydrochloric acid needed to be the diluted value using the above information:

I have already been given the original concentration of the Hydrochloric acid which is 2.00 mol dm-3. Also, according to the balanced equation above, the ratio of Ca(OH)2 to HCl is 1:2. So, to make sure that 250 cm³ of calcium hydroxide can react with 250cm³ of diluted HCl, I can multiply the number of moles of calcium hydroxide by 2 (0.01×2) and be given 0.02 for the concentration of HCl ...

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