Tecl CHM 311 practice problems Ch 16-01 a. Balance the following reactions: Base
ID: 588649 • Letter: T
Question
Tecl CHM 311 practice problems Ch 16-01 a. Balance the following reactions: Based on the stoichiometry of the reactions, answer the followving questions. b. Abo ut 1 g of KI and 10 mL of 0.5 M H SO4 were added to 25.00 mL of 0.0103 M potassium iodate solution to produce iodine (actually the triodide ion). The iodine was titrated to the starch endpoint, requiring 29.82 mL of sodium thiosulfate. What is the molarity of the thiosulfate? Next, a 1.298 g vitamin C tablet containing a nominal 1000 mg of ascorbic acid (FW 176.12) was ground to a fine powder. A 0.109 g subsample of the tablet was dissolved in 30 mL of 0.3 M H2SO4, then 1 g KI plus 25.00 mL of the standard potassium iodate were added. The resulting solution was titrated to the starch endpoint, requiring 9.05 mlL of the standardized thiosulfate. What is the experimentally determined vitamin C content (mg) of the tablet? Stoichiometry of the ascorbic acid-iodine reaction is 1:1 c.Explanation / Answer
a) IO3-(aq) + 8I-(aq) + 6H+(aq) ------->3I3- (aq) + 3H2O(l)
I3- (aq) + 2S2O32-(aq) -------> S4O62-(aq) + 3I-(aq)
b) No of mole of KI = 1g/166.003g/mol = 0.006024
No of mole of H+ = (0.5mol/1000ml)×10ml = 0.005mol
No of mole of IO3- =( 0.0103mol/1000ml)×25ml = 0.0002575
IO3- is limiting reagent
0.0002575mole of IO3- produce 3× 0.0002575 = 0.0007725 mole of I3-
0.0007725mole of I3- indicate 2× 0.0007725=0.001545mole of S2O32-
Volume of thiosulphate consumed = 29.82ml
Molarity of thiosulphate = (0.001545mol/29.82mol)×1000ml =0.0518M
c) No of mole of IO3- = (0.0103mol/1000ml)×25ml = 0.0002575mol
No of mole of I3- produced = 0.0007725
No of mole of S2O32- reacted =(0.0518mol/1000ml)×9.05ml = 0.0004688
No of mole of I3- reacted with S2O32- = 0.0002344
No of mole of I3- reacted with ascorbic acid = 0.0007725 - 0.0002344= 0.0005381
No of mole of ascorbic acid present = 0.0005381
Mass of ascorbic acid = 0.0005381mol × 176.12g/mol =0.09477g = 94.77mg
Mass of ascorbic acid in Vitamin C tablet = (94.77mg/0.109g)×1.298g = 1129mg
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