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# Year 11 - Concentration

## Concentration of Solutions - Introduction

Some measure of the amount of solute dissolved in a solvent is needed when dealing with solutions (particularly aqueous solutions).

e.g. mass per unit volume : g cm-3, kg m-3 or g dm-3 amount per unit volume : mol m-3 or mol dm-3 N.B.: 1 dm3 = 1000 cm3

The amount concentration equation given above can be manipulated to give another useful equation : This equation must be used when dealing with volumes and solutions and must not be confused with another equation giving the amount of substance : back to top

## Concentration of Solutions - Calculations

### (1) Question :

Sulphuric acid of concentration 0.1 mol dm-3 reacts exactly with 10 cm3 of 0.2 mol dm-3 sodium carbonate(aq). What volume of sulphuric acid is needed?

The amount of sodium carbonate = amount concentration × volume

= 0.2 mol dm-3 × (10 cm3/1000) dm3

= 0.2 mol dm-3 × 0.01 dm3

= 0.002 mol

#### Balanced chemical equation :

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

This chemical equation states that the amount ratio of sulphuric acid to sodium carbonate is 1 : 1.

Therefore the amount of sulphuric acid in the reaction = amount of sodium carbonate

= 0.002 mol  = 0.02 dm3

### (2) General mathematical equation :

For the general reaction between acid A and base B the equation is,

naA(aq) + nbB(aq) → salt + water

where na and nb are the balancing numbers for acid and base respectively, in the chemical equation.

Now, let the concentration of A be Ca mol dm-3 and the concentration of B be Cb mol dm-3.

And, let the volume of A used be Va cm3 and the volume of B used be Vb cm3.

Therefore, the amount of A used = Va × Ca

Therefore, the amount of B used = Vb × Cb

Since na mol of A reacts with nb mol of B these two equations may be combined to give,

nb × Va × Ca = na × Vb × Cb

In an examination question you will be given five out of the six pieces of information above and by rearrangement of the mathematical expression the sixth value can be worked out.

e.g. in the above worked example between sulphuric acid and sodium carbonate,

 na = 1 nb = 1 Va = unknown Vb = 10 cm3 Ca = 0.1 mol dm-3 Cb = 0.2 mol dm-3 Va = 20 cm3 or 0.02 dm3

## Concentration of Solutions - Titration

N.B.: This method of salt preparation is used to prepare ALL types of sodium, potassium or ammonium salts.

Equipment diagram - #### Method

Using a pipette and pipette filler measure out 10.0 cm3 of the solution of base into a conical flask. Add a few drops of an indicator, e.g.

 indicator colour in acid colour in base phenolphthalein colourless purple methyl orange red yellow

Place the conical flask under a burette filled with the solution of acid.

Then slowly add the acid from the burette to the base in the conical flask. Swirl the conical flask constantly to ensure a complete reaction. Keep a careful eye on the colour of the indicator in the conical flask and when it changes stop adding the acid from the burette.

This will give a rough value for the volume of acid needed to neutralise the base in the conical flask. The experiment then needs to be repeated with a fresh sample of the base in a fresh conical flask.

The acid can now be added quickly until 1 or 2 cm3 under the volume needed the first time.  After that the acid should be added one drop at a time until the just the addition of one drop causes the indicator to change colour.

That will be the precise end-point of the reaction.

#### Results

sample of results table :

 titration number final burette reading, cm3 initial burette reading, cm3 volume used, cm3 1 ( rough ) 11.10 0.00 11.10 2 11.10 0.00 11.10 3 10.30 0.00 10.30 4 10.10 0.00 10.10 5 10.05 0.00 10.05  back to top written by Dr Richard Clarkson : © Saturday, 1 November 1997