An iodine solution is standardized using pure ascorbic acid. A 0.450 g sample of
ID: 870862 • Letter: A
Question
An iodine solution is standardized using pure ascorbic acid. A 0.450 g sample of ascorbic acid requires 35.50 mL of iodine to reach a starch endpoint.A. What is the concentration of the iodine solution?
B. What is the titer? (mg ascorbic/mL iodine)
C. If 125.0mL sample of juice requires 12.50mL of the iodine solution to reach a starch endpoint what Is the concentration of the juice?
D. What is the mass of ascorbic acid I the sample?
E. If your RDA for ascorbic acid is 60mg per day how much juice would you drink to meet this requirement? An iodine solution is standardized using pure ascorbic acid. A 0.450 g sample of ascorbic acid requires 35.50 mL of iodine to reach a starch endpoint.
A. What is the concentration of the iodine solution?
B. What is the titer? (mg ascorbic/mL iodine)
C. If 125.0mL sample of juice requires 12.50mL of the iodine solution to reach a starch endpoint what Is the concentration of the juice?
D. What is the mass of ascorbic acid I the sample?
E. If your RDA for ascorbic acid is 60mg per day how much juice would you drink to meet this requirement?
A. What is the concentration of the iodine solution?
B. What is the titer? (mg ascorbic/mL iodine)
C. If 125.0mL sample of juice requires 12.50mL of the iodine solution to reach a starch endpoint what Is the concentration of the juice?
D. What is the mass of ascorbic acid I the sample?
E. If your RDA for ascorbic acid is 60mg per day how much juice would you drink to meet this requirement?
Explanation / Answer
The equation between ascobic acid and iodine is
C6H8O6 + I3- + H2o -------> C6H6O6 + 3I- + 2H+
Q1)The change in oxidation state of both species is 2.
thus equivalents of acid = equivalents of iodine
0.450/176.12 = 35.5x N/1000
normality of iodine = 0.07197
Q2) The titer is 450mg of ascorbic/ 35.5 mL of iodine -=12.67mg/mL of iodine
Q3) equivalents of juice = equivalents of iodine
125XN = 12.5 x 0.07197
therefore concentration of juice = 7.197 x 10-3N
Q4) mass of ascorbic acid in the juice sample = Normality x equivalent weight
= 7.197x 10-3 x 176.12 x125/(2 x 1000)= 0.0828 g
Q5) Since the juice has 0.0828 g in 125 mL to get 60x10-3 gg per day
we need to take 90.579 mL of juice per day
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