Oxidation-Reduction Titrations We pipetted a 5 mL aliquot of unknown stock solut

Question

Oxidation-Reduction Titrations

We pipetted a 5 mL aliquot of unknown stock solution into a 250 mL conical flask and added about 50 mL of DI water, and then added 1 mL of starch indicator solution

We then titrated the sample with 0.005 mol L^-1 iodine solution

My question is how do I get the calculations. I was told that these were supposed to be easy but I can't for the life of me figure this out! Can someone please help?

Trial 1 Trial 2 Trial 3 Volume pipetted 5 mL 5 mL 5 mL Titration Final Reading 18.85 mL 41.29mL 28.03 mL Initial Reading 1.43 mL 18.85 mL 5.52 mL Calculations Mols of I_2 Mols of Vitamin C Molarity of Unknown Average Molarity

Explanation / Answer

theh reactiontaking place is

ascorbic acid + I2 2 I + dehydroascorbic acid

end point is when all the ascorbic acid is converted to dehydroascorbic acid , the next drop added I2 will adsorb on starch and give the blue colour...

moles of I2 = molarity X volume titrated = 0.005 X 17.42 = 0.0871 mol (1st trial)

0.005 X 22.44 = 0.112 mol (2nd trial)

0.005 X 22.51 = 0.113 mol (3rd trial)

gram equvalents of ascorbic acid and iodone titrated will be same in each case

so gram equvalents of I2 = mol of iodone titrated ( iodine is nuetral compund )

so moles of ascorbic acid = mols of iodine

hence molarity of vitamic C = mols /volume

= 0.0871 / 0.25 = 0.348

0.112/0.25 = 0.448

0.113/ 0.25 = 0.452

hence average molarity = (0.348 + 0.448 + 0.452 ) / 3

= 0.416

voleme of I2 17.42 22.44 22.51 Mols of I_2 0.0871 0.112 0.113 Mols of Vitamin C 0.0871 0.112 0.113 Molarity of Unknown 0.348 0.448 0.452 Average Molarity 0.416

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