The Concentration of Limewater Solution.

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Ibrar Razaq 12PWR

                The Concentration of Limewater Solution.

Aim: The aim of this experiment is to find out the concentration of limewater solution.

I am provided with 250.00 cm3 of limewater, which has been made such that it contains approximately 1g dm3 of calcium hydroxide. Also available is hydrochloric acid, which has a concentration of 2.00mol dm3. To determine the exact concentration of the limewater, the hydrochloric acid must be titrated against it. The hydrochloric acid is very concentrated, and a very little amount will be needed to neutralise a flask full of limewater. Therefore, it will be necessary to dilute the hydrochloric acid.

I can use an appropriate value of limewater concentration to calculate the approximate value of 2.00mol dm-3 hydrochloric acid that will neutralise 25.00cm3 of the limewater. I can then calculate how much to dilute the hydrochloric acid by, and make up the volume of hydrochloric acid that will neutralise the limewater into a measurable amount.  

Calculation:

Using 25.00cm3 of limewater solution of approximately 1.00g dm-3.

Since the concentration of limewater is 1.00g dm-3 

Therefore the mass of Ca(OH)2(aq) in 1g dm-3 of the solution = 1.00g

 

 Therefore Mass Ca(OH)2 in 25.00cm3 of solution = 1x25    =    1g   = 0.025g

                                                                1000          40

Hence no. of moles of Ca(OH)2 in 25.00cm3 of solution =        Mass     =   0.025    =  3.373 x  10-4  moles                                                                                    RMM          74.1        

Since Mole Ratio of Ca(OH)2  : HCl  = 1: 2

Therefore No. of moles of HCl = 3.373 x 10-4 x 2

                                   =  6.746 x 10-4 moles

Volume of HCl = No. of Moles        = 6.746 x 10-4    = 3.373 x 10-4 dm3  = 0.337 cm3

                    Concentration                   2        

So 0.337 cm3 of 2.00mol dm-3 hydrochloric acid will neutralise 25.00cm3 of the limewater. However, 0.337 cm3 is not a measurable using a burette and is far too small and inaccurate, 30.00cm3 would be a measurable volume using a burette, so I will now calculate how much to dilute the 2.00mol dm3 hydrochloric acid by, to make the approximate volume that will neutralise the 30.00cm3 limewater.

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Concentration of diluted HCl  = no of moles  = 6.746 x10-4   = 0.0224mol dm3

                                  Volume          0.030dm3

Therefore, there would be 0.0224 moles of HCl in 1.00dm3 of diluted solution.

Volume of 2.00mol dm3 HCl   = No. Of Moles        = 0.224   =0.01124 dm3  = 11.24 cm3

in 1.00dm3 of diluted solution        Concentration               2        

As I would not be able to measure 11.24 cm3 of hydrochloric acid exactly using a bulb pipette, as they only measure certain set volumes of liquids. Therefore, I will make the volume 10.00cm3, as ...

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