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

Answer:

1.0190 x 10⁻⁵ mol

Explanation:

We know the titration required 10.19 mL of 0.001000 M KIO₃, from this information we can calculate the number of moles KIO₃ reacted and from there the number of moles of ascorbic acid since it is a monoprotic acid ( 1 equivalent of ascorbic acid to one equivalent KIO₃).

Molarity = mol/V

V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L

⇒ mol KIO₃ = V x M = 0.01019 L x 0.0010 mol / L =  1.0190 x 10⁻⁵ mol KIO₃

# mol ascorbic acid = # mol KIO₃ = 1.0190 x 10⁻⁵ mol