A sample of KIO3 weighting 0.1040 grams was added to an HCl solution containing excess KI....

A sample of KIO3 weighting 0.1040 grams was added to an HCl solution containing excess KI. The liberated I2 requried 34.6 mL of Na2S2O3 solution in the titration. calculate the molarity of the thiosulfate solution, using the overall equation ------> KIO3+6HCl+6Na2S2O3------>3S4O62-+I-+3H2O IKIO3:6Na2s2o3

Expert Solution

mass of KIO3 = 0.1040 g

molar mass of KIO3 = 214 g/mol

moles of KIO3 = mass / molar mass

= 0.1040 / 214

= 4.86 x 10^-4

KIO3 + 6HCl + 6Na2S2O3 -------------->3S4O62- + I- +3H2O IKIO3:6Na2s2o3

for 1 mol of KIO3 ----------------- 6 mol of Na2S2O3

4.86 x 10^-4 mol ----------------- ?? Na2S2O3

Na2S2O3 moles = 2.92 x 10^-3

volume of Na2S2O3 = 34.6 mL = 0.0346 L

molarity = moles / volume

= 2.92 x 10^-3 / 0.0346

= 0.084 M

molarity of Na2S2O3 = 0.084 M

queen_honey_blossom answered 7 months ago

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