Question

A student found that his titration had taken 10 ml of .1002 M NaOH to titrate .132 g of aspirin. Calculate his percent purity. Give a possible explanation of what might have affected his percent purity.

Answer 1

no of m6les of NaOH = 0.1002 x 0.01 = 0.001002 moles

no of moles of aspirin in sample = 0.00102 moles

no of moles of aspirin (theoretical ) = 0.132/ 180 = 0.00073 moles

as the aspirin OH group is free that is why no of moles of NaOH used is more than one equivalent.

So if the acetyl group was totally removed the no of moles of NaOH required = 0.00073 x 2 = 0.00146 moles.

thus no of moles of aspirin remaining intact in sample = 0.00146 - 0.001002 = 0.00044

Mass of aspirin in sample = 0.00044 x 180 = 0.0792 g

% purity = 0.0792 x 100 / 0.132 = 60%

the acetyl group present in the molecule gets cleaved due to the action of moisture in the air, thus aspirin tablets start smelling like acetic acid. the reaction involved