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A solution of sodium thiosulfate was standardized by dissolving 0.1668 g KIO3 (214.00 g/mol) in water,...

A solution of sodium thiosulfate was standardized by dissolving 0.1668 g KIO3 (214.00 g/mol) in water, adding a large excess of KI, and acidifying with HCl. The liberated iodine required 22.11 mL of the thiosulfate solution to decolorize the blue starch/iodine complex. Calculate the molarity of the sodium thiosulfate solution.

Solutions

Expert Solution

IO3- (aq) + 8 I- (aq)+ 6 H+ (aq) <------> 3I3- (aq) + 3 H2O (l)

In acidic solutions tri iodide oxidizes thiosulfate totetrathionate.

I3- (aq) + 2S2O3-2 (aq) <-----> 3I- (aq) + S4O4-2(aq)

The number of moles of KIO3 added = 0.1668 / 214.0g/mol

                                                         = 7.79 x 10-4 mol

The number of moels of I3- generated from this solution = 3 x 7.79 x 10-4

                                                                                        = 2.34 x 10^-3 mol

Thus the number of moles of thiosulfate = 2 x 2.34 x 10^-3

                                                               = 4.68 x 10^-3 mol

concentration of the thiosulfate solution = 4.68 x 10^-3 /0.02211 L

concentration of the thiosulfate = 0.212 M


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