Chemistry planning exercise

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Finding out how much Acid there is in a Solution.

The aim of my investigation is to find the concentration of sulphuric (VI) acid from a sample. The solution is thought to have a concentration between 0.05 and 0.15 moldm-3.

To do this I will perform a titration involving sulphuric (VI) acid and sodium carbonate. A titration involves reacting a standard solution with another who’s concentration is unknown. I am given anhydrous sodium carbonate so will have to prepare a standard solution of this before attempting the titration. The indicator I will be using to indicate when the reaction is fully completed is methyl orange.(6) 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-base titration.The standard solution must be of a certain concentration for the titration to be a success. This is dependant upon the concentration of sulphuric (VI) acid. As this value ranges from 0.05 and 0.15 moldm-3 I can assume an average value of 0.10 moldm-3 as its concentration.

The reaction taking place will be

H2S04 + Na2C03                    H2CO3 + Na2S04

As this equation is in a 1:1 molar ratio, the same concentrations of both sulphuric (VI) acid and sodium carbonate should be used.

Preparing a Standard Solution

A standard solution of sodium carbonate needs to be prepared.

Sodium Carbonate, Na2CO3, is a primary standard because it meets certain requirements: -

  • It is available in a highly pure state.
  • It must be stable in air.
  • It must be easily soluble in water.
  • It should have a high molar mass.
  • In solution, when used in volumetric analysis, it must undergo complete and rapid reaction.

The process involves weighing accurately a sample of sodium carbonate and use it to make a solution close to 0.10 mol dm-3. This will be used to determine the concentration of the sulphuric (VI) acid sample.

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However, the weight of sodium carbonate I need to use is still unknown, so this will have to be found.

Using the formula n= Vc, where, n= moles of solute (mol.)

                                          V= volume of solution (dm 3)

                                        c= concentration of solution (mol dm-3)

The volume of solution to be made will be 250 cm3 as this will give me enough solution to complete the desired amount of titrations with but not too much as to have a surplus.

n=Vc

n= 0.25 x 0.1

n= 0.025 mol. (5)

Using this value I can ...

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