Food Sample/Unknown containing Vitamin C (Ascorbic Acid)
 
1. Determining the Titer of the potassium iodate solution

A sample of pure Vitamin C weighing 0.1138 g ( 113.8 mg ) is titrated with KIO₃ in the presence of excess KI and requires 25.94 mL

The titer of the KIO₃ solution is 113.8 mg / 25.94 mL= 4.387 mg / mL.  I.e., each mL of the KIO₃ used in the titration corresponds to 4.387 mg of Vitamin C.

As usual, the determination of the titer of the KIO₃ is tantamount to standardizing the solution. Determination of the titer of KIO₃ affects the accuracy of every subsequent procedure in this exercise.

Suppose a 150.0 mL sample of orange juice is titrated with the potassium iodate solution from part 1.1, ( again in the presence of excess postassium iodide) and it requires 20.75 mL

The number of mg of Vitamin C contained in the 150.0 mL sample of the juice is 20.75 mL * 4.387 mg/mL = 91.03 mg

• I.e., 150.0 mL of the juice contains 91.03 mg of vitamin C

Therefore, 100.0 mL of the juice contains ( 100.0 / 150.0 ) x 91.03 mg = 60.70 mg of Vitamin C, or, what is equivalent:

The concentration of vitamin C in the juice is 60.70 mg / 100.0 ml

 

Suppose a tablet containing vitamin C weighs 0.5358 g.  The tablet is ground up to a uniform powder, and 0.2436 g of the powder is titrated with the potassium iodate from part 1 with titer 4.387 mg/mL (again, in the presence of excess KI ).  15.88 mL of the KIO₃ solution is required.

The number of mg of Vitamin C contained in the 0.2436 g sample of the tablet is 15.88 mL * 4.387 mg/mL  =  69.68 mg

0.2436 g of the tablet                            contain 69.68 mg of Vitamin C
0.5358 g of the tablet (the whole tablet) contain     ?    mg of Vitamin C

Therefore, the entire tablet contained ( 0.5358 / 0.2436 ) x 69.68 mg = 153.3 mg

Suppose a sample of the unknown weighing 0.1654 g ( 165.4 mg ) requires 22.38 mL of the KIO₃ solution.

The number of mg of Vitamin C contained in the 165.4 mg sample is 22.38 mL *4.387 mg/mL  =  98.20 mg

Therefore, the percentage of vitamin C in the unknown is 100 x 98.20 / 165.4 = 59.37 %

Suppose the label of an orange juice container reads as follows:

"An 8 fl oz serving of this juice provides 100 % of the recommended daily allowance (RDA) of vitamin C."

The RDA for vitamin C is 60 mg/day

An 8 fl oz serving of the juice provides 60 mg.  I.e., the concentration of the juice is 60 mg / 8 fl oz.= 7.5 mg / fl oz

We wish to provide 300 mg of juice.  To do so will require  300 mg / (7.5 mg / fl oz ) = 40 fl oz

Since one quart = 32 fl oz, we will need well over a quart of this particular juice for the exercise.