Question

If a 108.3 ml sample of grapefruit juice takes 10.14 ml of a 0.01720 M standarized iodine solution to reach the starch end point, what is the concentration of Vitamin C in units of mg/ml?

Answer 1

V = 108.3 mL

V = 10.14 mL Iodine

M = 0.0172 M

find vitamine C

ascorbic acid + I2 --> 2 I- + dehydroascorbic acid

or teehcnically:

C6H8O6(aq) + I2(aq)→C6H6O6(aq) + 2H+(aq) + 2I-(aq)

therefore, raito is 1:1

so

mmol of Iodine = MV = (10.14)(0.0172) = 0.17441 mmol of Iodine

therefore, from the preivous raito

1 mol of acid = 1 mol of iodine

0.1744 mmol iodine = 0.1744 mmol of acid

then

[Concentration of acid ] = mmol/mL = 0.1744 /108.3 = 0.0016103 M

we need mg/mL so

mass of Vit C = mol*MW = 0.1744 *176.12 = 30.71532 mg

V = 108.3 mL

C = 30.71532/108.3

C = 0.28361 mg / mL

answered by: anonymous