So, your aim is to find out the number of moles/concentration of an unknown substance - just like a normal titration.
Let A = The unknown substance
B = The substance that will be used in both parts of the reaction. It will be in excess for the first reaction against the unknown, and will then be neutralised in the second reaction
C = The other substance of known concentration. It will be used in the second reaction to react with the leftover B that is left from the first reaction with A.
1) So, firstly you react A with B. B will be in excess, so all of A will be neutralised in the first reaction, leaving some of the B unreacted.
2) In the second part of the method, the remaining B that is left unreacted from the first reaction is reacted with C. C is used to neutralise the remaining B leftover.
You will be given the concentration of c(C), then you also will determine the V(C) that is needed to react with the B, so you have the n(C) and thus the n(B) used in that reaction. Also, you will be given the c(B) and the volume added in the beginning.
From this, you use the n(B) added in the first reaction - n(B) leftover, you are left with the amount of n(B) that reacted with A
Finally, use the n(B) reacted in the first equation to determine the n(A) - and if needed, the concentration of A
Hope this helps, feel free to question anything
