Question

You titrate 25.00 mL of the lemon-lime Kool-Aid (with KI, HCl, and starch) with 0.001000 M KIO3(aq) solution. The titration requires 10.19 mL of KIO3(aq). Determine the number of moles of ascorbic acid in 25.00 mL of your Kool-Aid solution.

Answer 1

moles = 3.057x10⁻⁵ moles of C₆H₈O₆

Step-by-step explanation:

In order to do this, we need to write the balanced reaction that is ocurring here.

In the titration we have:

KIO₃ + 6H⁺ + 5I⁻ --------> 3I₂ + 3H₂O + K⁺

As for the ascorbic acid solution:

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

From these two reactions, we can see that the KIO₃ solution reacts always with 3 moles of I₂, so, if the moles of C₆H₈O₆ are the same moles of I₂, we can assume that KIO₃ reacts with 3 moles of C₆H₈O₆. Therefore, the expression to use will be:

3n₁ = n₂

3M₁V₁ = M₂V₂

M₂ = 3M₁V₁ / V₂

Replacing we have:

M₂ = 3 * 0.001 * 10.19 / 25

M₂ = 0.001223 M

Finally the moles of ascorbic acid:

moles = 0.001223 * 0.025

moles = 3.057x10⁻⁵ moles of C₆H₈O₆

Hope this helps