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I don’t understand 2b and all of C Experiment 43 Advance Study Assignment: Analy

ID: 1043543 • Letter: I

Question

I don’t understand 2b and all of C

Experiment 43 Advance Study Assignment: Analysis for Vitamin C 1. Write a balanced equation for the reaction between I, and ascorbic acid. I oxigzrg redua the reducing agent. 2. A solution of I2 was standardized with ascorbic acid (Asc). Using a 0.1137-g sample of pure ascorbic acid, 26.75 mL of 1, were required to reach the starch end point. a. What is the molarity of the iodine solution? O2ulM b. What is the titer of the iodine solution? mg AschnL 12 3. A sample of fresh grapefruit juice was filtered and titrated with the above 1, solution. A 100-mL sample of the juice took 10.26 mL of the iodine solution to reach the starch end point. What is the concentration of vitamin C in the juice in mg vitamin C/100 mL of juice? mg/100 mL b. What quantity of juice will provide the RDA amount of vitamin C?

Explanation / Answer

Q1)C6H8O6 +I2--->C6H8O6 +2H+ +2I-

Q2) Given:

a0Mass of pure Ascorbic acid=0.1137g

molar mass of Ascorbi acid(C6H8O6)=176.12g/mol

Thus, mol of ascorbic acid added=0.1137g/(176.12g/mol)=6.456*10^-4 mol

mol of I2 required for ascorbic acid to react completely=6.456*10^-4 mol (1:1 molar ratio)

So, Molarity of I2*Volume of I2=mol of I2

Molarity of I2=mol of I2/Volume of I2=(6.456*10^-4 mol)/0.02675L=0.0241M

b)titrant =I2 soutio

titre= ascorbic acid solution

you need to find the concentration of ascorbic acid/ml of I2

mass of ascorbic acid=0.1137 g=0.1137g*(1000mg/1g)=113.7 mg

volume of I2=26.75ml

[ ascorbic acid]=113.7 mg/26.75ml=4.250 mg Asc/ml I2

q3)Let V(I2)=10.26m=0.01026L

M(I2)=0.0241M

mol of I2 used up=0.01026L*0.0241mol/L=2.473*10^-4 mol

mol of ascorbic acid in juice=mol of I2 used up=2.473*10^-4 mol

mass of Ascorbic acid=(2.473*10^-4 mol)*(176.12g/mol)=0.0435 g=0.0435*1000mg=43.5mg

Concentration of asc acid=43.5mg/100 ml

b)RDA or recommended dietary allowance of ascorbic acid= 60mg daily

Volume of juice to meet RDA=(60mg/day)/(43.5mg/100ml)=137.9 ml