Question

An aspirin sample was reacted with 35.25 mL of 0.2214 M of NaOH and back titrated with 30.20 mL of 0.1141 M. How many moles of Aspirin are in the sample?

Answer 1

Aspirin totally reacted with 35.25 mL 0.2214 M solution

And back titrated with 30.20 mL 0.1141 M solution

Here we can find that Aspirin as the limiting reagent which is further back titrated.  

The reaction is

Aspirin + NaOH Sodium salt of Aspirin

Wt of NaOH = MmV/1000 from molarity equation

m = molecular weight of NaOH

V = volume in mL

M = Molarity of NaOH

Wt = 40 x 0.2214 x 35.25 / 1000

= 0.312 g/mol

Wt of NaOH = 30.20 x 0.1141 x 40 / 1000

= 0.137 g/mol

Total weight of NaOH consumed = 0.137 + 0.312

= 0.629 g/mol

Total number of moles NaOH = weight/molecular weight = 0.629/40

= 0.015 moles of NaOH totally reacted.

So that the number of moles of Aspirin = the number of moles of NaOH because one mole of NaOH reacted with one mole of Aspirin.

The total number of moles of Aspirin = 0.015 moles