Please answer it correctly please don\'t answer if you aren\'t 100% sure of the

Question

Please answer it correctly please don't answer if you aren't 100% sure of the answer please answer it fully
Thank you (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 was pipetted into a flask and treated with excess KI (2 g) and acid (10 mL of 0.5 M H2SO4). How many millimoles of I3 are created by the reaction? 4776 mmol (b) The triodide from part (a) reacted with 37.61 mL of Na2s203 solution. What is the concentration of the Na2s203 solution? x M 077 (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 14.35 mL of Na2s203 solution from part (b). Find the weight percent of ascorbic acid (FM 176.13) in the unknown. HINT: Think "back-titration". wt% 12.67

Explanation / Answer

(a) 1 mole of KIO3 forms 3 moles of I3-

molarity of KIO3 solution = 1.022/214 = 9.55 mM

taken 50 ml of this solution

moles of KIO3 = 9.55 mM x 50 ml = 0.48 mmol

moles of I3- formed = 3 x 0.48 = 1.44 mmol

(b) 1 I3- reacts with 2 S2O3^2-

moles of Na2S2O3 required = 2 x 1.44 = 2.88 mmol

molarity of Na2S2O3 solution = 2.88/37.61 = 0.0766 M

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