Let us assume that we have an impure solid substance `'C'`, weighing `'w'` g and we are required to calculate the percentage purity of `'C'` in the sample. We are also provided with two solutions `'A'` and `'B'`, where the concentration of `'B'` is known (`N_1`) and that of `'A'` is unknown. For the back titration to work, following conditions are to be satisfied

(`a`) Compounds `'A'`, `'B'` and `'C'` should be such that `'A'` and `'B'` react with each other.

(b) `'A'` and pure `'C'` also react with each other but the impurity present in `'C'` does not react with `'A'`.

(c) Also the product of `'A'` and `'C'` should not react with `'B'`.

Now we take out certain volume of `'A'` in a flask (the equivalents of `'A'` taken should be equivalents of pure `'C'` in the sample) and perform a simple titration using `'B'`. Let us assume that the volume of `'B'` used be `V_1` litre.

Equivalents of `'B'` reacted with `'A'= N_1V_1`

Equivalents of `'A'` initially `= N_1 V_1`

In another flask, we again take same volume of `'A'` but now `'C'` is added to this flask. Pure part of `'C'` reacts with `'A'` and excess of `'A'` is back titrated with `'B'`. Let the volume of `'B'` consumed is `V_2` 1itre.

Equivalents of 'B' reacted with excess of `'A'= N_1 V_2`

Equivalents of `'A'` in excess `= N_1 V_2`

Equivalents of `'A'` reacted with pure `'C' = (N_1 V_1 - N_1 V_2)`

Equivalent of pure `'C'=(N_1V_1-N_1V_2)`

Let the `n -` factor of `'C'` in its reaction with `'A'` be `x`, then the moles of pure `'C' = (N_1V_1-N_1V_2)/x`

Mass of pure `'C' =(N_1V_1-N_2V_2)/x * ` Molar mass of `'C'`.

Percentage purity of `'C' = (N_1V_1-N_1V_2)/x * (text(Molar mass of C))/w * 100`