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annnoyyying
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Homework Statement
25 cm3 of 1M NaOH added to an aspirin tablet. 25cm3 of distilled water also added. Mixture heated for 10 mins. Hydrolysed solution and washings transferred to a 250 cm3 conical flask and made up to the mark (250cm3).
25 cm3 of the hydrolysed solution is titrated against 0.1M HCl. The average titration is 12.13 cm3 HCl required to reach end point.
1 mol acetyl salicylic acid reacts with 2 mols NaOH
1 mol NaOH reacts with 1 mol HCL
Mr(ASA) = 180.15 g/mol
theory:
excess NaOH is added to neutralise the acid in the tablet. the hydrolysed solution is titrated against HCl, the number of moles HCl used proportional to the moles of remaining NaOH.
Homework Equations
n=C*V/1000
The Attempt at a Solution
a) number of moles of NaOH originally added to the aspirin tablet is:
25*1/1000=0.025 mols
b) number of moles of NaOH remaining in 250 cm3 hydrolysed solution(not used to hydrolyse ASA):
=number of moles of HCl used to titrate against 25cm3 hydrolysed solution *10
=(12.13*0.1/1000)*10 =0.01213 mols
c) number of moles of NaOH used to hydrolyse the ASA
= a) - b)
=0.025-0.01213
=0.01287 mols
d) number of moles of acetyl salicylic acid hydrolysed by NaOH:
=number of moles of NaOH used to hydrolyse ASA/2
= 0.01287/2 = 0.00644mols
e) mass of ASA in the tablet:
=0.00644*180.15
=1.6g
however, the mass of ASA should be 300mg (as on the packet)...
There must be something wrong with my calculation.
Can someone please kindly check??