Question: (a) Potassium iodate solution was prepared by dissolving 1.022 g of KIO3 (FM 214.00) in a 500-mL volumetric flask. Then 50.00 mL of the solution

(a) Potassium iodate solution was prepared by dissolving 1.022 g of KIO3 (FM 214.00) in a 500-mL volumetric flask. Then 50.00 mL of the solution were pipetted into a flask and treated with excess KI (2 g) and acid (10 mL of 0.5 M H2SO4). How many moles of I-3 are created by the reaction?
(b) The triiodide from part (a) reacted with 37.66 mL of Na2S2O3 solution. What is the concentration of the Na2S2O3 solution?
(c) A 1.223-g sample of solid containing ascorbic acid and inert ingredients was dissolved in dilute H2SO4 and treated with 2 g of KI and 50.00 mL of KIO3 solution from part (a). Excess triiodide required 14.22 mL of Na2S2O3 solution from part (b). Find the weight percent of ascorbic acid (FM 176.13) in the unknown.
(d) Does it matter whether starch indicator is added at the beginning or near the end point in the titration in part (c)?

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a b c d 5000 mL contains exactly 110 of the KIO3 01022 g 0477 57 mmol KIO3 Each mol of iodate ... View full answer

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