Question

3) Use the mole ratios contained in both equation 3 and equation 4 to calculate the

volume (in mL) of 0.0100 M potassium iodate solution needed to completely oxidize all

of the ascorbic acid contained in 20.00 mL of 0.0200 M ascorbic acid solution.

concentration of the titrant can be used along with the mole ratios from the reaction equations to determine the amount of ascorbic acid in the solution Chemical Reactions involved with the redox titration: Equation 1 shows the reduction of an iodate ion in the presence of acid to form iodine: 10 .6H5e - %12 + 3H,0 Equation 2 shows the oxidation of two iodide ions to form iodine: 51 52 5e Equation 3 shows the result of combining the reduction and oxidation reactions shown above in equations 1 and 2 10,6H51 - 31.3 HO (3) Notice that the reduction of the lodate ion in equation 1 requires the addition of five electrons. As a result, equation was scaled by 5/2 so that this equation produces the five electrons needed to reduce the lodate ion Equation 3 shows the summation of equations 1 and 2 and represents the balanced oxidation-reduction equation that we can use to measure the amount of vitamin C in a sample. The lodine formed in equation number 3 will be reduced immediately back into two iodide ions by the ascorbic acid, which is simultaneously oxidized into dehydroascorbic acid, as shown below in equation 4 HOM - NOO + 21+ 2H 000 cac OO Detyrascorbic Acid The reaction shown in equation 4 will continue as long as there is ascorbic acid present to reduce the lodine that forms. When all of the ascorbic acid has been oxidized, any additional lodine that forms cannot be reduced back into iodide ions. Instead, this iodine will persist and react with the starch indicator resulting in a blue starch-kodine complex When this complex forms, the end point of the bitration has been reached and the volume of titrant that was added up to this point can then be used to determine the amount of ascorbic acid that was present in the original solution Solutions used in this experiment: • Potassium iodate (KIOS: 0.0100 M): The potassium iodate solution will be used as the titrant. It will be dispensed from a burette 0.50 mL at a time until the end points reached. The volume (and molarity) of KIO, solution added to reach the end point is used to calculate the amount of ascorbic acid in the sample • Potassium indide K: 0.60 M: The potassium iodide solution will be added to the sample of juice drink and provides the lodide ions that are needed to react with the KIO, to form the lodine molecules that will react with the ascorbic acid • Starch indicator solution (1.0% wiv) Starch molecules form a complex with iodine that has a deep blue color of the solutions are otherwise colorless) Once all of the ascorbic acid has been oxidized, lodine will accumulate in the solution and react with the starch producing a color change that reveals the end point of the titration. It can be assumed that the ascorbic acid in the sample has been completely oxidized at the instant the starch indicator changes color • Hydrochloric acid (1.0 M): The reduction of the 10,"ion also requires six H ions These hydrogen ions will be provided by the hydrochloric acid solution

Answer 1

Moles of ascorbic acid present in solution = 0.0200mol/L × (20.00×10^-3L ) = 4.00×10^-4 mol = moles of I₂ produced

The number of moles of potassium iodate needed = moles of I₂ ÷ 3 = (4.00×10^-4/3)mol

V( in L) × 0.0100mol/L = 1.333×10^-4 mol

V = 1.333×10^-2 L = 1.333×10^-2×10³ ml

Volume of potassium iodate solution required = 13.33 ml. (Answer)