Investigating the Concentration of Citric Acid

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Investigating the Concentration of Citric Acid

Problem

I have been set the task of determining the concentration of a sample of citric acid, using only a 1.0 M solution of sodium hydroxide and any laboratory glassware that is required. I have also been given the RMM of citric acid as 210g and I have been told that one mole of citric acid will react with 3 moles of sodium hydroxide.

My Experiment

Using this information I have determined that the best way to conduct this experiment is to do a titration. This will allow me to determine the amount of impure citric acid required to neutralise a known volume of 1.0 M sodium hydroxide. Consequently, by calculating the correct volume of pure citric acid that would be required to neutralise this volume of sodium hydroxide, I can calculate the percentage purity of the citric acid.

Calculating the Amount of Citric Acid

In order to make this experiment a good titration, I must use a reasonable amount of citric acid. This is so that the amount of citric acid needed to neutralise the sodium hydroxide is not so small that it is difficult to read off the burette, and so that it is not so large that it takes more than 50cm3, and so 2 burettes must be used. I think that I can safely assume that in this experiment the percentage purity will be greater than 50%. Therefore, if I calculate the amount of pure citric acid required to neutralise 25cm3 of sodium hydroxide when the citric acid is 100% pure, I should easily be within the confines of my 50 cm3 burette. I will therefore now calculate the required mass of sodium hydroxide:

Ratio citric acid : sodium hydroxide

1 : 3

Concentration of sodium hydroxide = 1.0 M

Therefore, require concentration of citric acid = 1.0 M /3 = 0.33 M

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I will be using a volumetric flask which can hold 250cm3 of liquid. Therefore, in order to make a 1.0 M solution of citric acid I will require 210 / 4 as I am not using a whole litre of water.

210 / 4 = 52.5g of citric acid.

However, I need to make a 0.33 M solution and therefore I will require 52.5 / 3 grams of citric acid.

52.5 / 3 = 17.5g of citric acid required to make a 0.33 M solution of citric acid, assuming that the acid is 100% pure. It is clear that the concentration of ...

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