Question

How do I calculate the concentration of ascorbic acid in (mg/100 mL) from the following data?

C₆H₈O₆ + I₂ → C₆H₆O₆ + 2I^- + 2H⁺

KIO₃ + 5KI + 6H⁺ → 3I₂ + 6K⁺ + 3H₂O

Molarity of KIO3 = 0.001987 mol/L

Volume of unknown = 10.0 mL

Volume of KIO3 used = 33.90 mL

The calculation scheme given was:
Volume of KIO₃ = mols of KIO₃ (use molarity) = mols of I₂ (1:3 ratio) = mols C₆H₈O₆/ 10.0 mL of unknown

Report in mg/100 mL

I was able to follow the scheme and ended up with the mass of ascorbic acid but im not sure how to do the last part (which is expressing in mg/ 100 mL)

Answer 1

C₆H₈O₆ + I₂ → C₆H₆O₆ + 2I^- + 2H⁺

KIO₃ + 5KI + 6H⁺ → 3I₂ + 6K⁺ + 3H₂O

Given that;

Molarity of KIO3 = 0.001987 mol/L

Volume of unknown = 10.0 mL

Volume of KIO3 used = 33.90 mL

Molar mass of ascorbic acid (C₆H₈O₆) = 176.12 g/mol

First calculate the moles of KIO3 as follows:

Number of moles = molarity * volume in L

= 0.001987 mol/L *33.90 mL *1.00 L/ 1000 ml

= 6.74*10^-5 moles KIO3

Now calculate the moles of I2 as follows:

6.74*10^-5 moles KIO3 * 3 mole I2/ 1 mole KIO3

= 2.02*10^-4 moles

According to the given reaction number of moles of I2 is equal to the number of moles of ascorbic acid.

Moles of ascorbic acid = 2.02*10^-4 moles

Molar mass of vitamin c = 176.12 g/mol

So, 1 mol of vitamin c = 176.12 g

2.02*10^-4 moles of vitamin c = 0.000234 x 176.12 g

= 0.0356 g

= 35.6 mg

Because 1000 mg = 1.00 g

So, concentration = 35.6 mg / 10.0 mL

= 3.56 mg / ml

= 3.56 mg / ml *100 ml

= 356 mg / 100 ml