Finding Out How Much Acid Is In A Solution

Aim

To discover the accurate concentration of a sample of sulphuric (VI) Acid. It is thought to have a concentration between 0.05 and 0.15 mol dm-3. This is to be done by neutralising the acid with sodium carbonate using a titration.

Theoretical Background

The reaction between Sulphuric Acid and Sodium Carbonate is a neutralisation. Sulphuric acid is a strong acid(1) . This means that it is a powerful proton, H+ donor and that it is completely in the form of ions when in dilute solution. Acids and bases are classified depending on the extent to which they form ions when dissolved in water. Sodium Carbonate is a weak base. It is a proton, H+ acceptor, however it is only a moderate one and ionisation is partially complete.

The formula for this reaction is as follows:

Sulphuric Acid + Sodium Carbonate = Sodium Sulphate + Water + Carbon Dioxide

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

In order to determine the concentration of the Sulphuric Acid it can be neutralised with a base of known concentration, in this case the Sodium Carbonate solution. The technique of titration is used to find out accurately how much of a chemical substance is dissolved in a given volume of a solution, that is, the concentration of the solution, (see diagram left) (2). In a titration the solution with known concentration is put in the burette, and the solution with unknown concentration is placed in the conical flask below. This means that the Sodium Carbonate solution will be in the Burette and the Sulphuric Acid will be in the Conical Flask, along with an indicator(2).

Sodium Carbonate is then added to the Sulphuric acid until neutralisation occurs; this is represented by a change in colour of the indicator. The difference between the initial and final reading of the burette will give a titre in cm3.

The indicator used in this experiment will be methyl orange indicator as a strong acid and a weak base are used(3). An indicator is needed to show the end point of the titration, when neutralisation has occurred. In an alkaline solution methyl orange is yellow and in acidic it is red. The colour change in this experiment will be red to yellow. Methyl orange gives a clear, sharp end point so it can be more accurate as other indicators such as universal indicator, which has a spectrum of colours. A white tiles is placed under the titration so that the colour change of the indicator can be clearly seen.

To make this experiment a fair test, before using all equipment, wash it with first distilled water, then the solution that will be used in the apparatus. This ensures no contamination with foreign substances that could affect the results in my experiment. Always read scales at the bottom of the meniscus with the eye level with the scale. If it is difficult to read, put a piece of white paper behind. I will make sure that each titre is fair by going to the same end point, as soon as the methyl orange turns red.

The sodium carbonate is in the form of a solid, anhydrous sodium carbonate that will be used to make a standard solution. To make this solution the sodium carbonate is mixed with water. When the Sodium carbonate solution is made, the beaker and funnel used should be rinsed several times with distilled water, and the washings added to the volumetric flask. This ensures that all traces of sodium carbonate are added to the volumetric flask and the concentration of the solution is exact.

Concentration on Sodium Carbonate Solution

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

From looking at the formula the ratio of Sodium Carbonate to Sulphuric Acid is 1 to 1. This means that during the neutralisation, 1 mole of Sodium Carbonate is used for each mole of Sulphuric Acid.

The concentration of the Sulphuric acid is between 0.05 and 0.15 mol dm-3.

0.05+0.15 = 0.1

2

The mean concentration of sulphuric acid is 0.1 moldm-3 so I am going to use this as a concentration value for the Sodium Carbonate Solution. However, in order to get this concentration I need to calculate the mass of Solid, anhydrous sodium carbonate I need to combine with water.

Concentration = 1000 x Number of Moles

Volume

Number of Moles = Concentration x Volume

1000

Number of Moles = 0.1 x 250 = 0.025

1000

Mass = Number of Moles x Mr

Mr= (23 x 2) + 12 + (3x16) = 106 moles

Mass = 0.025 x 106 = 2.650g

From these calculations I can tell I need to use 2.650g of Anhydrous Sodium Carbonate to make to make the solution with concentration 0.1 moldm-3

Apparatus

Ð'* Sulphuric (VI) Acid

Ð'* 0.1 moldm-3 Sodium Carbonate Solution

Ð'* Distilled Water

Ð'* Methyl Orange Indicator

Ð'* 100cm3 Beaker x 2

Ð'* 250cm3 volumetric flask

Ð'* Funnel x 2

Ð'* Glass Rod

Ð'* Burette

Ð'* White Tile

Ð'* Retort Stand

Ð'* Clamp

Ð'* 25.0 cm3 Pipette

Ð'* Pipette Filler

Ð'* 100cm3 Conical Flask

Ð'* Digital Balance able to measure to the accuracy of 0.001g

Ð'* Glass Rod

Accuracy-

o Burette- Scale can be read to the accuracy of 0.05cm3

o Pipette- An accurate volume of 25cm3

o Graduated Flask- An accurate volume of 250cm3

Making