# Applications of Law of Equivalence , Simple Titration , Back Titration

### Simple titration

In this, we can find the concentration of a substance with the help of the conc. of another substance which can react with it.

For example: Let there be a solution of a substance A of unknown concentration. We are given another substance B whose concentration is known (N1).

We take a certain known volume of A in a flask (V2) and then we add B to A slowly till all the A is consumed by B. (This can be known with the help of indicators). Let us assume that the volume of B consumed is V1.

According to the Law of equivalents, the number of gm equivalents of A is equal to the number of gm equivalents of B.

∴ N1V1 = N2V2 , where N2 is the conc. of A.

From this we can calculate the value of N2

### Back titration

Back titration is used to calculate % purity of a sample. Let us assume that we are given an impure solid substance C weighing w gms and we are asked to calculate the percentage of pure C in the sample. We will assume that the impurities are inert.

We are provided with two solutions A and B, where the concentration of B is known (N1) and that of A is not known. This type of titration will work only if the following conditions are satisfied (a) A, B and C should be such compounds that A and B can react with each other, A and C can react with each other but the product of A and C should not react with B.

Now we take a certain volume of A in a flask (the A taken should be such that the gm equivalents of A taken should be ≥ gm equivalents of C in the sample. This can be done by taking A in excess). Now we perform a simple titration using B.

Let us assume that the volume of B used is V1. In another beaker, we again take the solution of A in the same volume as taken earlier. Now, C is added to this and after the reaction is complete, the solution is being titrated with B. Let us assume that the volume of B used up is V2.

Gram equivalents of B used in the first titration = N1V1

∴ gm. equivalents of A initially = N1V1

gm. equivalents of B used in the second titration is N1V2

∴ gm. equivalents of A left in excess after reacting with C = N1V2

gm. equivalents of A that reacted with C = N1V11 – N1V2

gm. equivalents of pure C = N1V1 – N1V2

If the n factor of C is x ,

The the moles of Pure C $\large = \frac{N_1 V_1 – N_1 V_2}{x}$

The Weight of C $\large = \frac{N_1 V_1 – N_1 V_2}{x} \times Molecular \; Weight \; of \; C$

Percentage Weight of C $\large = \frac{N_1 V_1 – N_1 V_2}{x} \times \frac{Molecular \; Weight \; of \; C}{W} \times 100$