Can anyone kindly help me with this problem?
20 tablets containing the monobasic drug phenytoin sodium with a molecular weight of 274.3 were powdered and 0.5541 g of tablet powder was extracted with 40 ml of 0.01 M sodium hydroxide and made up to a final volume of 50 ml with 0.01 M sodium hydroxide. The sample was centrifuged and 25 ml of the sample was acidified and then extracted into ether. The extract was evaporated and the residue was dissolved in 50 ml of anhydrous pyridine and titrated with 0.1024 M tetrabutylammonium hydroxide using thymol blue as an indicator.
Volume of titrant required = 4.65 ml
Weight of 20 tablets =2.1063 g
Stated content per tablet = 50 mg
Calculate the % of the stated content.
The answer is 99.3% according to the textbook but I'm just not sure how to get it
- hcbiochemLv 71 month ago
moles tetrabutylammonium hydroxide = 0.00465 L X 0.1024 mol/L = 5.766X10^-5 mol.
Assuming a 1:1 mole ratio in the titration, 25 mL of the extract contained 5.766X10^-5 mol of phenytoin. Since this was 1/2 of the original sample, the original sample contained 1.153X10^-4 mol.
Mass phenytoin = 1.153X10^-4 mol X 274.3 g/mol = 0.0316 g
0.0316 g / 0.5541 g X 2.1063 g = 0.120 g phenytoin in total sample
mg phenytoin/tablet = 0.120 g / 20 = 6.0 mg/tablet
% stated content = (6.0 / 50) X 100 = 12%
That seems awfully low to me...