Monday

March 30, 2015

March 30, 2015

Posted by **Sara** on Sunday, February 27, 2011 at 8:17pm.

1. Using the equations,

IO3- +5I- + 6H+ -> 3I2 + 3H2O

C6H8O6 + I2 -> C6H6O6 + 2I- + 2H+

how many equivalents of IO3- are required to produce enough I2 to oxidize one equivalent of ascorbic acid?

Answer: 1IO3- <-> 3I2 <-> 3C6H8O6

1/3 equivalents of IO3-

2. A 10 g sample of no name grape drink mix is diluted to 250 mL. 50 mL of grape drink is titrated with 35 mL of 0.005 M KIO3 solution.

a) How many moles of ascorbic acid are present in 50 mL of the grape solution?

0.005 mol/L KIO3 x 0.035 L KIO3 =

1.75 x 10^-4 mol KIO3

1.75 x 10^-4 mol KIO3 x (3 mol I2/1 mol KIO3) = 5.25 x 10^-4 mol I2.

3 KIO3 <-> 3 C6H8O6, so same amount of moles of C6H8O6.

b) What is the percent by mass of ascorbic acid in a package of grape drink mix?

5.25 x 10^-4 mol C6H8O6 x 176.124 g/mol C6H8O6 = 0.092 g C6H8O6

But this is grams in 50 mL of solution - we have 250 mL throughout which the ascorbic acid is distributed. So 0.092 g x 5 = 0.462 g C6H8O6 within 250 mL solution.

0.462 g/10.00 g x 100 = 4.62%

c) If there are 8 servings per package, calculate the %RDA of vitamin C in each serving of no name grape mix. Formula: (mg ascorbic acid/serving)/75 mg

(462 mg C6H8O6/8)/75 mg

= 0.77 % RDA

Thanks for your help!

- Chemistry -
**DrBob222**, Monday, February 28, 2011 at 3:26pmI think all of your work is good except for 1. I believe the answer is 1/6 mol IO3^- = 1 equivalent ascorbic acid.

1 mole IO3^- = 3moles I2

3 moles ascorbic acid = 3 moles I2; therefore, 1 mole IO3^- = 3 moles I2

In the next step,

C6H8O6 + I2 ==> C6H6O2 +2H^+ + 2I^-

So this step is 1 mol C6H8O = 1 mol I2 but

2 equivalents C6H8O6 = 2 equivalents I2 (since I2 goes from 0 to -2 for 1 mole, that is 2e change for 1 mol, the equivalent weight I2 is 1/2 molar mass pr 2 equivalents. Therefore,

1/3 mol IO3^- = 1 mole I2 = 2 equivalents I2 and

1/6 mol IO3^- = 1 equivalent I2.

1/6 mol IO3^- = 1 equivalent ascprbic acid. You can check that out in your percent, which you did with moles and the 4.62% is correct.

If we do that with equivalents (and milliequivalents) we should get the same answer. m.e. = # milliequivalents and m.e.w. = milliequivalent weight.

So mL = 35 mL

Normality = 0.005*6 (since there are 6 equivalents in 1 mole IO3^-).

m.e.w. ascorbic acid = molar mass/2 = 176.124/2000 =0.088062

(mL x N x m.e.w./mass sample)*100 =

35.0mL x 0.03N x 0.088062g/2g sample)*100 = 4.62%

Viola!

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