Finding out how much acid there is in a solution

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Richard Dilley

Finding out how much acid there is in a solution


The aim of this investigative experiment is to discover the accurate concentration of sulphuric acid (H2SO4), which is found in a solution. The concentration is thought to be between 0.05 mol dmˉ³ and 0.15 mol dmˉ³. I have been given access to anhydrous sodium carbonate (NaCO3) and a range of indicators.

In order to obtain the concentration of the acid in the solution I will have to titrate the known solution of sodium carbonate with the unknown sulphuric acid. The indicator I will be using to indicate when the reaction is fully completed is methyl orange. This is because I am using a strong acid and a weak alkali and methyl orange is the most appropriate indicator for this type of acid-alkali titration.

Firstly I must create a solution of sodium carbonate that has a known concentration:

Standard solution of sodium carbonate


  • Safety Goggles
  • Spatula
  • Solid anhydrous sodium carbonate
  • Balance (2 d.p.)
  • Beaker (250cm³)
  • Bottle of distilled water
  • Stirring rod
  • Volumetric flask (250cm³)
  • Pipette  (25cm³)

Before carrying out the procedure it is necessary to work out how much solid sodium carbonate is to be used in making the known concentration. Since the equation for this titration is:

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

This shows that 1 mole of sulphuric acid reacts with 1 mole of sodium carbonate. Since the concentration of the acid is thought to be between 0.05 mol dmˉ³ and 0.15 mol dmˉ³ it would seem appropriate to use concentration of sodium carbonate that is 0.10 mol dmˉ³. Therefore I must work out how much anhydrous sodium carbonate is required to make this concentration:

Relative molecular mass of Na2CO3 = 46+12+48 = 106g

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To make a concentration of 1 mol dmˉ³ you would need 106g of Na2CO3

To make the desired concentration of 0.1 mol dmˉ³ you would need 10.6g of Na2CO3 (106/10 = 10.6)

Therefore to make the concentration of our 250cm³ 0.1 mol dmˉ³ you would need 2.65g of Na2CO3 (10.6/4 = 2.65)

Therefore I will need 2.65g of Na2CO3 to be dissolved to make the solution.


  1. Using a spatula transfer exactly 2.65g of the solid sodium carbonate to 250cm³ of distilled water in the beaker.
  2. Stir to dissolve the solid using ...

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