Question

Vitamin C in a titration with potassium iodate

References Mailings Review View AaBbcode Abccdee AaBbcc No Spacing Heading 1 Normal 3. A suitable method for the determination of vitamin C (C.H.O.) is a titration with potassium iodate (KIO). Potassium iodate is used as a titrant and is added to an ascorbic acid solution that contains strong acid and potassium iodide (KI). Potassium iodate reacts with excess potassium iodide, liberating molecular iodine (12): [1] KIO, + 5KI + 6H 31, +6K + 3H-0 As long as the solution contains ascorbic acid, the le produced in equation 1 is used up in a rapid reaction with ascorbic acid (equation 2), during which dehydroascorbic acid (C.H.O.) and iodide ion (I-) are formed: [2] CH.06 +12CHO.+ 21+ 2H The excess iodine can be determined by reacting it with thiosulfate S.O, until the solution turns from blue color (iodine starch complex) to colorless shown in equation 3. [3] 1. + 2 S.O.2. - 21 + .0.2- A student used 0.500 grams of potassium iodate to generate excess iodine to react with 15.00 ml vitamin C solution. The excess iodine was titrated with 15.00 mL of 0.0100 M thiosulfate solution to reach the end point. (a) Determine the molar concentration of vitamin C. En unted States 30 8884 .

Answer 1

Given, 0.5g of KIO broduce excess iodine, to react with is mL vit c solh. Remaining iodine solh was titrated with 15 mL of 0.01M thiosulfate KIO ASKI +6H + 3 I +6K+ + 3H00 Male of KIO3 = 0.5 mal 214 6.2.336 x 10-3 mot. from above ean 1 mo of KIO, 2-336x103. gives 3 Imole of I " 3x2.336x10-3 md of Iz 7.008 × 10-3 mol. so, excess mole of I, Co Hq ft I coho +21 +24+ a scorbic acid dehydso ascorbic acid . concentration of ascoobic acid be s molar. of vit c is reacted with excess I. Let 15 mL Mole of acid in 15mL = Max 15 x xmol 1000 L - 150 mot. lood left = lexcess of I -(mole of acid in So, Mole of Remaining الم I I mol 7oo8Xlors - 15 n 1000)

- Iz +25.02- 2 I- + Suo? remaining Iz Mole of so2 = 15L x 0.01 mol/l 1000 = 15xios mole This much of so2 is reacted mole of Iz so both are equal (7.008x10 3 154 1= 15x10-5 -) is n = 0.006858 Зе mac 1000 1000 n = 0.45 72 molar of C M. vitamin 0.4572 conen Hence, molar