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.

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

queen_honey_blossom answered 2 years ago

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