Finding out how much acid there is in a solution.

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AS chemistry coursework                                                                        Nick Collinson

February 2003        -  -

AS Chemistry coursework:

Finding out how much acid there is in a solution

AIM

I will be given a sample of sulphuric (VI) acid, thought to have a concentration between 0.050 and 0.150mol dm-3, and I have to find the exact concentration. I have been provided with solid anhydrous sodium carbonate and a range of indicators.

I will need to make up a standard solution of sodium carbonate, which I can then titrate with the acid.

Knowing that the concentration of the acid is between 0.050 and 0.150mol dm-3, and from the equation below, a good concentration for the standard solution would be between the two, at 0.100mol dm-3.

This is the equation of what will happen when the standard solution of sodium carbonate reacts with the sulphuric acid:

H2SO4 (aq) + Na2CO3 (aq)              →             Na2SO4 (aq) + CO2 (g) + H2O(l)

Risk assessment

 

Sulphuric (VI) acid         -         Very Corrosive

  •       Dangerous with water

-        If swallowed wash out mouth with water

-        If it gets in eyes flood the eye with water for 10 minutes                -        if spilt, cover with mineral absorbent and scoop up and

        put in a bucket. Add anhydrous sodium carbonate and

        leave to react, and then add plenty of water.

                                        

Sodium carbonate        -        Irritant (to eyes, skin and respiratory system)

-        If swallowed wash out mouth with water

-        If it gets in eyes flood the eye with water for 10 minutes

-        If solid is spilt, scoop up and rinse area with water.        

                                If in solution, cover with mineral absorbent, put in

                                bucket, and rinse the area with water.        

I will wear a lab coat and goggles to protect my skin, clothes and eyes.

To make up a standard solution of 0.100 mol dm-3 sodium carbonate.

 

I will be making up 250cm3 (0.250 dm3) of the solution, as we have 250 cm3 volumetric flasks.

I need to find out what mass of solid sodium carbonate will be needed to dissolve in 0.250 dm3 of water, to produce a concentration of 0.100mol dm-3

Using the equations below I can work out the mass.

RMM of Na2CO3 = (2 x 23) + 12 + (3x16) = 106

No. of  moles = concentration x volume

                  = 0.10 x 0.25

                  = 0.025 moles

Mass = No. of moles x RMM

         = 0.025 x 106

         = 2.650g                                   

So 2.650g of solid anhydrous sodium carbonate will be used.  

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I will weigh a weighing boat on a balance correct to three decimal places, record the result, and then add approximately 2.650g of the sodium carbonate and record the result. I will tip the powder into a beaker, and then re – weigh the weighing boat. To find the exact amount of sodium carbonate, I will minus the last weighing from the second weighing. This is because a very small amount will remain in the weighing boat, stuck to the sides.

To the solid in the beaker I will add distilled water and stir them with a glass ...

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