Find the concentration of limewater solution Titration

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Finding the Concentration of Limewater Solution

The aim of the experiment is to find the concentration of a limewater solution (calcium hydroxide) as accurately as possible. The limewater solution is approximately 1gdm-3 but to get a more accurate concentration I will use an acid based titration. A neutralisation reaction will take place between the acid (hydrochloric acid) and the base (calcium hydroxide) during the titration. I will use the base to neutralise the acid and make a salt (calcium chloride) and water. This reaction can be written as:

2HCl + Ca(OH)2 → CaCl2 + 2H2O

        However I need to find the exact concentration of the reagent, Calcium Hydroxide solution. To overcome this problem I will need to perform an acid based titration. The other problem is that the acid reagent is too concentrated as it is 2mol/dm³. This means the reaction will occur faster because there is a greater chance of a successful collision, which means it would be harder for me to acquire an accurate result, as the end point will occur too quickly. I will need to reduce the concentration of the hydrochloric acid, thus prolonging the time taken for the reaction to take place so I can get a more accurate result*.

So I can make accurate calculations with my results I will need to convert the concentration of the limewater solution, which is in g/dm³, to mol/dm³. This will then mean all the concentrations needed for the calculations are in the same units:-

 1 g/dm³            mol/dm³.

To work out the concentration of Ca(OH) 2, I will use:      

Concentration (mol) = Mass(g) / Molecular mass of Ca(OH) 2 

        = 1g / (40 + (16X2) + (1X2))

= 1/74

= 0.01 mol/dm³ of Ca(OH)2

This means we have a 2.00 mol/dm³ solution of hydrochloric acid and a 0.01 mol/dm³ solution of calcium hydroxide.  

Dilution Factor

I will use 25cm³ of limewater solution to give me a reasonable result, which I can accurately measure out using a 25cm³ pipette. I will also use 25cm³ of HCl as a burette can dispense up to 50cm³. Also I know from the equation for the reaction that the ratio of hydrochloric acid reacting with the limewater is 2:1. so using this and the volume I can work out the concentration of acid I need.

            Moles = Volume x mol/dm³

         = (25cm³/1000)dm³ x 0.01mol/dm³

         = 0.00025 mol

Therefore the moles of the acid required  = 0.00025 x 2 (because of the 2:1 ratio) ratio

                                                = 0.0005 moles

From this, I can work out the concentration of the acid used to neutralise the base:

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mol/dm³ = moles / Volume

           = 0.0005 / (25/1000)

           = 0.02 mol/dm³

        Now that I know the Acid will be approximately 0.02mol/dm³, I can say I will need approximately 25cm³ of acid to neutralise 25cm³ of limewater solution. The dilution factor of the acid is (concentration I began with/ concentration I am going to use) 2/ 0.02 = 100, this means the dilution factor by which I have diluted the acid is 100. I will need to make 250 cm³ of 0.02mol/dm³ hydrochloric acid. The amount of 2mol/dm³ I will need to add is 250/100 = 2.5cm³. However ...

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