[MW2121]

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 )

[chenbeier]

I think your calculation is ok, I think the error is in the datas.THe reactingtime of NaOH with the aspirin was to short. It is not only a neutralzation also  the ester group as to be split.

[mjc123]

But in that case, the amount of aspirin must be even greater...?

[AWK]

Incomplete hydrolysis leads to an underestimation of acetylsalicylic acid content (more NaOH is titrated with hydrochloric acid).
But if NaOH contains some sodium carbonate (which happens with older NaOH solutions), the titration with HCl (against phenolphthalein as an indicator) shows only the half of the NaOH converted to sodium carbonate and the acetylsalicylic acid content may be overestimated.

[MW2121]

Incomplete hydrolysis leads to an underestimation of acetylsalicylic acid content (more NaOH is titrated with hydrochloric acid).
But if NaOH contains some sodium carbonate (which happens with older NaOH solutions), the titration with HCl (against phenolphthalein as an indicator) shows only the half of the NaOH converted to sodium carbonate and the acetylsalicylic acid content may be overestimated.

ok, that sorta makes sense, but how does the conversion of NaOH to NaHCO3 mean that the acetylsalicylic acid content is overestimated? wouldn't it have the same effect as incomplete hydrolysis of underestimation?

[AWK]

This is a simple stoichiometry
Assume 1 mole of pure ASA (acetylsalicylic acid) and 3 moles of NaOH. Simulate 100 % or 50 % of hydrolysis and calculate results according to your formula.
Then convert 0.25 mole of CO₂ with 0.5 mole of NaOH and do complete hydrolysis of 1 mole ASA. During titration o.25 mole of Na₂CO₃ is converted to NaHCO3 (complete conversion to NaCl needs pH less than 7). and calculate results.

Note - reaction CO₂ with NaOH gives you Na₂CO₃ to the momet when all NaOH will be reacted. Since then more carbon dioxide forms sodium hydrogen carbonate.

[chenbeier]

How much NaOH do we realy need. Acetylsalicylic acid  is an Ester and an acid, after hydrolysis we have also a phenolic function.

1. CH₃COO-C₆H₄COOH  + NaOH => CH₃COO-C₆H₄COO^- + Na⁺ + H₂O
2. CH₃COO-C₆H₄COO^- Na⁺ + NaOH => CH3COO^-  Na+ + HOC₆H₄COO^- Na⁺

3.  HOC₆H₄COO^- Na⁺ + NaOH => Na+ +^-O-C₆H₄COO^- + Na⁺

What is right now. According this the ratio is 1:3  and not as mentioned above 1:2

The calculated result would be then 209,8 mg

[AWK]

How much NaOH do we realy need. Acetylsalicylic acid  is an Ester and an acid, after hydrolysis we have also a phenolic function.

1. CH₃COO-C₆H₄COOH  + NaOH => CH₃COO-C₆H₄COO^- + Na⁺ + H₂O
2. CH₃COO-C₆H₄COO^- Na⁺ + NaOH => CH3COO^-  Na+ + HOC₆H₄COO^- Na⁺

3.  HOC₆H₄COO^- Na⁺ + NaOH => Na+ +^-O-C₆H₄COO^- + Na⁺

What is right now. According this the ratio is 1:3  and not as mentioned above 1:2

The calculated result would be then 209,8 mg

Yes, with NaOH the phenolate can be formed.
No, during the titration, the phenolate is treated as NaOH. Hence the calculations of MW2121 are correct.

[chenbeier]

If so, why is then the ratio 1:2 used?

[AWK]

ASA during hydrolysis form salicylate and acetate (the same amount of moles). This is true for titration with phenolphthalein indicator.
Nonhydrolyzed ASA forms a salt with 1 mole of NaOH.

[chenbeier]

Yes this is 1 and 2, and I mentioned the built up of phenolat what means 3.

But nevertheless this takes place:

http://www1.lasalle.edu/~prushan/BackTitration-lab4.pdf

[AWK]

Check acidity of phenol. What happens to phenolate at pH of 8 (phenolphthalein becomes colorless)

[AWK]

Yes this is 1 and 2, and I mentioned the built up of phenolat what means 3.

But nevertheless this takes place:

http://www1.lasalle.edu/~prushan/BackTitration-lab4.pdf

The procedure given in above link is horrible in this point

Shutdown procedures
IMPORTANT NOTE: Your NaOH solution must be saved upon completion of the experiment; you
will use it again in future experiments

The diluted solution of NaOH reacts with carbon dioxide from the air and before next using should be checked for sodium carbonate content (and eventual correction). Otherwise, the ASA content will be overestimated.

[chenbeier]

In this experiment Aspirine and NaOH in excess was mixed and heated.  I think the pH will be more as 8. Later the titration took place with the remaining NaOH and HCl.
The difficulty could be an hydrolysis of the phenolat again at pH 8. Maybe then only Neutralisation and split of ester is mentioned.

[AWK]

In this experiment Aspirine and NaOH in excess was mixed and heated.  I think the pH will be more as 8. Later the titration took place with the remaining NaOH and HCl.

TRUE

The difficulty could be hydrolysis of the phenolat again at pH 8.  Maybe then only Neutralisation and split of ester is mentioned

At pH ~8 only free phenol exists (there some exceptions, eg nitrophenols, some fluorophenols). Hence for simplicity, we can neglect the formation of a phenolate.