Back Titration (or Indirect titration) to determine concentration of acetylsalicylic acid in one brand of aspirin

[MW21]

For school, we performed a back titration involving one brand of aspirin and were told to determine the quantity of acetylsalicylic acid in a single tablet. sodium hydroxide was added in excess then the solution was heated. the excess unreacted NaOH was then reacted with HCl.  my calculations are as follows: 

V(NaOH) = 0.04085L

c(NaOH) = 0.111M        hence n(NaOH) total = 0.00435 moles

v(HCl) = 0.01778L

c(HCl) = 0.0582M          hence n(HCl) = 0.001035 moles

number of moles excess = number of moles titrated of HCl

hence 0.001035 moles didn't react with acetylsalicylic acid

 

the number of moles of NaOH that reacted with acetylsalicylic acid is 0.00435-0.001035 = 0.003495

therefore 0.003495 moles of NaOH reacted with the acetylsalicylic acid

the mole ration between NaOH and acetylsalicylic acid is 1:2

meaning the number of moles of acetylsalicylic acid is 1/2 x 0.003495

which is = 0.0017475 moles

 

Molar mass acetylsalicylic acid = 180.158g/mol

0.0017475 x 180.158 = 0.314g or 314mg

this does not make sense considering I only added 300mg of crushed aspirin powder in the first place, meaning that I would have a percentage purity of 105%

if anyone can see any obvious errors on my part or could suggest any reason for this ridiculous result, I would be extremely appreciative :))

 

 

 

[Bufofrog]

5 hours ago, MW21 said:

For school, we performed a back titration involving one brand of aspirin and were told to determine the quantity of acetylsalicylic acid in a single tablet. sodium hydroxide was added in excess then the solution was heated. the excess unreacted NaOH was then reacted with HCl.  my calculations are as follows: 

V(NaOH) = 0.04085L

c(NaOH) = 0.111M        hence n(NaOH) total = 0.00435 moles

v(HCl) = 0.01778L

c(HCl) = 0.0582M          hence n(HCl) = 0.001035 moles

number of moles excess = number of moles titrated of HCl

hence 0.001035 moles didn't react with acetylsalicylic acid

 

the number of moles of NaOH that reacted with acetylsalicylic acid is 0.00435-0.001035 = 0.003495

therefore 0.003495 moles of NaOH reacted with the acetylsalicylic acid

the mole ration between NaOH and acetylsalicylic acid is 1:2

meaning the number of moles of acetylsalicylic acid is 1/2 x 0.003495

which is = 0.0017475 moles

 

Molar mass acetylsalicylic acid = 180.158g/mol

0.0017475 x 180.158 = 0.314g or 314mg

this does not make sense considering I only added 300mg of crushed aspirin powder in the first place, meaning that I would have a percentage purity of 105%

if anyone can see any obvious errors on my part or could suggest any reason for this ridiculous result, I would be extremely appreciative :))

 

 

 

It has been a long time since I have done a titration, but I see no obvious mistake in your calculations.  Could you have missed the end point or messed up on you preparation of  the NaOH solution or HCL solutions?  Not very helpful....  Hopefully someone will respond with more information - I may have missed something in your calculation that others will pick up on.

[John Cuthber]

Boiling aspirin in alkali hydrolyses it.

I think that's covered where you say " mole ration between NaOH and acetylsalicylic acid is 1:2"
But it's always a good idea to start by writing down the equation.

[MW21]

3 hours ago, John Cuthber said:

Boiling aspirin in alkali hydrolyses it.

I think that's covered where you say " mole ration between NaOH and acetylsalicylic acid is 1:2"
But it's always a good idea to start by writing down the equation.

thanks John, ill keep that in mind for next time 

[MW21]

11 hours ago, Bufofrog said:

It has been a long time since I have done a titration, but I see no obvious mistake in your calculations.  Could you have missed the end point or messed up on you preparation of  the NaOH solution or HCL solutions?  Not very helpful....  Hopefully someone will respond with more information - I may have missed something in your calculation that others will pick up on.

I may have missed the endpoint for some of them, but I'm not sure if a difference in 0.1ml could result in a purity difference of more than 25%, especially since we were using reasonably dilute solutions. and I didn't actually prepare the standard solution for my practical as we were going to be short of time. so I'm not entirely sure about if they were prepared with precision. but i was thinking, sodium hydroxide hydrates in air so could this have been a major source of error if, for example, the standard solution was left out overnight or something like that?