A-Level H2 Chemistry: Back Titration Question on Percentage Purity

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In this video, i will share with you how to solve Back Titration questions in Volumetric Analysis using 5 simple steps.

Before you continue reading, do you know when do we need to use back titration? Would a direct titration like acid-base neutralisation be sufficient?

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Back titrations are usually used when a direct titration is not possible.

e.g. when it involves solid substances such as CaCO3, where the end-point is difficult to detect.

e.g. when it involves volatile substances such as ammonia and iodine, where inaccuracy arises due to loss of substance during titration.

In back titration, a known excess of one reagent A is allowed to react with an unknown amount of B. The amount of unreacted A is then determined by titration with a reagent C of known concentration. From the titration results, the amount of unreacted A and the amount of B can be found by simple stoichiometric calculations.

5 Simple Steps in Back Titration Calculations:

1. Determine the amount of C required in the titration

2. Using stoichiometry, find the amount of A that reacted with C in the titration

3. Note that amount of A that reacted with C in the titration = amount of A that did not react with B in the earlier reaction

4. Therefore, amount of A that reacted with B = total amount (known excess) of A added – amount of A that did not react with B earlier

5. Knowing now, the amount of A that reacted with B, using stoichiometry, the amount of B can be found

Let us take a look at the question to be be discussed:

A sample of impure magnesium was analysed by allowing it to react with excess HCl solution. After 1.32 g of the impure metal was treated with 100 cm3 of 0.750 moldm-3 HCl, the remaining solution required 50 cm3 of a 0.250 moldm-3 aqueous sodium hydroxide for complete neutralisation. Assuming the impurities do not react with the acid, what is the percentage by mass of Mg in the sample? [Ans: 57.5%]


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