Please please please answer this question fully and please answer it only if you

Question

Please please please answer this question fully and please answer it only if you're 100% of the correct answer! 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 13 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? 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". 12.67 wt%

Explanation / Answer

Answer (a)

Here first of all

mol (IO3-) formed = (1.022 g)(1 mol /214.00 g) = 0.004776 mol

Reaction

IO3- + 8I- 3I3- + 3H2O

50/500 mL taken or 0.4776 mmol

so 0.4776 x 3 = 1.433 mmol

Thus,1.433 mmol I3- is formed.

Answer (b)

Titration Reaction

I3- + 2S2O32- 3I- + S4O62-

so the the concentration of the Na2S2O3 solution is

[S2O32-] = (2)(.001433) / (0.03761L) = 0.07609 M

Answer (c)

Now overall reaction would be

A + H2O + I3- DA + 2H+ + 3I-

14.35 mmol I3- (oradded to the ascorbic acid,50 out of 500 mL)

(14.35 mL)(0.07609M) =1.082mmol S2O32-added. It will react with 0.5410mmol I3-.

Therefore, 1.433–0.5410= 0.892mmol I3- reacted with 0.892 mmol of ascorbic acid.

(0.892mmol)(176.13 mg/mmol)/1000 = = 0.157g A

Wt % = (0.157/1.223) x 100 =12.8%

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