Question

a) Calculate the volume of 0.5M NaOH that will be required to titrate a 0.15g sample of aspirin. b) If the student reached the end point of the titration 0.76 mL prior to the calculated value in (part a). What was the purity of the sample? The student required 0.40mL more NaOH than she calculated she would need in (part a). Explain how this might happen.

Answer 1

a) the reaction is

C9H8O4 + NaOH >> C9H7O4Na + H2O

we know that

moles = mass / molar mass

so

moles of aspirin = 0.15 / 180

moles of aspirin = 8.333 x 10-4

now

consider the reaction

C9H8O4 + NaOH >> C9H7O4Na + H2O

moles of NaOH required = moles of aspirin = 8.333 x 10-4

now

we know that

moles = molarity x volume (L)

so

volume (L) = moles / molarity

volume = 8.333 x 10-4 / 0.5

volume = 1.666 x 10-3 L

volume = 1.66 ml

so

1.66 ml of NaOH is required

b)

given 0.76 ml prior to the calculated value

so

volume of NaOH = 1.66 - 0.76 = 0.9 ml

now

moles of NaOH = conc x volume (L)

moles of NaoH = 0.5 x 0.9 x 10-3

moles of NaOH = 4.5 x 10-4

now

moles of aspirin present = moles of NaOH added = 4.5 x 10-4

now

mass = moles x molar mass

so

mass of aspirin present = 4.5 x 10-4 x 180 = 0.081

now

% percent purity = ( mass of aspirin / total mass of sample ) x 100

= ( 0.081 / 0.15) x 100

= 54

so

purity of the sample is 54%

c)

NaOH is not 100% pure