Deducing the quantity of acid in a solution

Authors Avatar

Deducing the quantity of acid in a solution

INTRODUCTION:

To find the accurate concentration of a solution of sulphuric acid of which concentration is known to be between 0.05 and 0.15moldm-3, using solid anhydrous sodium carbonate and a range of indicators. I have to decide which indicator is most appropriate to use, then plan and carry out an experiment to determine the accurate concentration of the sulphuric acid.

BASIC PLAN:

  • I will make a standard solution of sodium carbonate from the anhydrous sodium carbonate and calculate its concentration.
  • After making the sodium carbonate solution of known concentration, I will use it to determine the concentration of the sulphuric acid. This will be possible through an acid/ alkali titration. Sodium hydroxide (the alkali), will neutralise the acid in the reaction to produce a salt (Sodium Sulphate – Na2SO4) water and carbon dioxide. Add the acid until solution is neutral (the alkali has been used up), then take a measurement to find out how much acid was needed to neutralise the solution, and then can work out how much acid was needed for neutralisation. I can then use this information to find out the concentration of the acid.

                                         

        

        

        


MAKING A STANDARD SOLUTION OF SODIUM CARBONATE

APPARATUS:

  • Safety goggles
  • Spatula
  • Solid anhydrous sodium carbonate
  • A balance with a 0.01g accuracy
  • 2 x 250cm3 beaker
  • Wash bottle of distilled water
  • Glass rod
  • 250cm3 Volumetric flask with stopper and label  

Plan

  1. I will weigh the first 250cm3 beaker, and reset the balance to 0.

  1. I will transfer about 2.65g of solid anhydrous Na2CO3 into this beaker (weighed to the nearest 0.01g).

 

  1. I will put about 50cm3 of distilled water into the second beaker and then carefully transfer the bulk of Na2CO3 from the first beaker into the beaker of water.
  2. I will then reweigh the first beaker to find out the amount of remaining Na2CO3.
  3. Using a glass rod, I will stir the solid and water in the beaker to dissolve, adding more water as is necessary.

 

  1. I will carefully transfer the solution to the 250cm3 volumetric flask.

  1. I will rinse both beakers three times and the glass rod once with distilled water and transfer it to the flask, to make sure all the solution goes into the volumetric flask.

  1. I will carefully make up the solution to about 2cm below the mark on the neck of the flask using distilled water. I will insert the stopper and invert it 20 times to mix the contents. Using an accurate pipette, I will add enough distilled water to bring the bottom of the meniscus to the mark.

 

  1. I will then invert it a further ten times to ensure complete mixing. I will then label the flask as “Na2CO3 solution”, inform that it is an irritant, and leave a space for the concentration to be filled in after I have calculated it. 

Equations:

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

1                    :      1           

1 mole           :      1 mole  

In the titration, I will be using a 25cm3 pipette; therefore the volume of the Na2CO3 used will be 25.00cm3. From the chemical equation above, it can be seen that the ratio of the reaction is 1:1 (it takes 1 mole of Na2CO3 to react with 1 mole of H2SO4); therefore the volume of the H2SO4 will also be around 25.00 cm3.

The concentration of the H2SO4 is between 0.05 and 0.15 moldm-3. Assuming then, that the concentration of the acid is 0.05 moldm-3, the concentration of Na2CO3 can be found as follows:

Join now!

 

Conc. of H2SO4 (C)                  = 0.05 mol/dm-3 

Volume of H2SO4 (V)               = 25cm3 = 0.025 dm3 

Moles of H2SO4 (n)               = CV = 0. 05 x 0.025 = 0.00125 mol 

0.025dm3 of Na2CO3 solution contains 0.0025 moles of H2SO4 

Concentration of Na2CO3     =        = 0.00125   = 0.05 mol/dm-3

                                                      0.025 

n =               mass

      Relative molecular mass

Number of moles (n) of Na2CO3          = 0.00125mols 

...

This is a preview of the whole essay