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In Part 4 of the lab 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 10.40 mL of KIO3(aq). Determine the number of moles of ascorbic acid in 25.00 mL of your Kool-Aid solution.

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Answer:

3,12x10⁻⁵ moles

Explanation:

In this reaction, KIO₃ reacts with KI and HCl (H⁺) producing:

KIO₃ + 5KI + 6H⁺→ 3I₂ + 6K⁺ + 3H₂O

That means 1 mole of KIO₃ produces 3 moles of I₂.

Thus, moles of I₂ producing per with 10,40mL of 0,001000M KIO₃ are:

0,01040L×(0,001000mol / L) × 3= 3,12x10⁻⁵ moles of I₂

Then, I₂ reacts with ascorbic acid in a 1:1 reaction. That means moles of ascorbic acid in 25,00 mL of the Kool-Aid solution are also 3,12x10⁻⁵ moles

I hope it helps!

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