Coupon Accepted Successfully!


Back Titration

Let us assume that we have an impure solid substance C, weighing w gram, 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 (N1) and that of A is unknown.
For the back titration to work, following conditions are to be satisfied:
  1. Compounds A, B, and C should be such that A and B react with each other.
  2. A and pure C also react with each other, but the impurity present in C does not react with A.
  3. 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 V1 liter.
Equivalents of B reacted with A = N1V1
∴ Equivalents of A initially = N1V1
In another flask, we again take the same volume of A, but now C is added to this flask. The pure part of C reacts with A, and excess of A is back titrated with B. Let the volume of B consumed be V2 liter.
Equivalents of B reacted with excess of A = N1V2
= Equivalents of A in excess
Equivalents of A reacted with pure C = (N1V1N1V2)
Equivalents of pure C = (N1V1N1V2)
Let the n-factor of C in its reaction with A be x. Then
Moles of pure C Description: 11883.png
∴ Mass of pure Description: 11896.png
∴ Percentage purity of Description: 11904.png

Test Your Skills Now!
Take a Quiz now
Reviewer Name