Asked by Komal
a capsule of vitamin C.. a weak acid, is analyzed by titrating it with 0.425 M sodium hydroxide. it is found that 6.20 Ml of base is required to react with capsule weighing 0.628 g. what is the percentage of vitamin C (C6H8O6) in the capsule?( one mole of vitamin C reacts with two moles of hydroxide ion)?
Answers
Answered by
DrBob222
C + 2NaOH ==> 2H2O + Na2C
How many mols of NaOH are used in the titration? That is M x L = mols NaOH.
Now convert mols NaOH to mols C using the coefficients in the balanced equation.
Convert mols C to grams C. mass C = mols C x molar mass C.
%C = [mass C/mass sample]*100
(Mass of sample is 0.628 grams). I should point out here that if the mass of the capsule and contents is 0.628 g then the percent you calculate will not be the percent C in the material inside the capsule. But that is a problem with the problem and not with the way it is being solved.
How many mols of NaOH are used in the titration? That is M x L = mols NaOH.
Now convert mols NaOH to mols C using the coefficients in the balanced equation.
Convert mols C to grams C. mass C = mols C x molar mass C.
%C = [mass C/mass sample]*100
(Mass of sample is 0.628 grams). I should point out here that if the mass of the capsule and contents is 0.628 g then the percent you calculate will not be the percent C in the material inside the capsule. But that is a problem with the problem and not with the way it is being solved.
Answered by
Ben
73.9%
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