Triiodide ions are generated in solution by the following reaction in acidic sol

Question

Triiodide ions are generated in solution by the following reaction in acidic solution:
6H+ +IO3-(Aq) +8I-(Aq) > 3I3-(Aq) +3H20
Triiodide ion is determined by titration with a sodium thiosulfate(Na2S2O3) solution. The products are iodide ion and tetrathionate ion (S4O6 -2)
A sample of 0.6013g of potassium iodate was dissolved in water. Solid potassium iodide was then added in excess. What is the minimum volume of 3.00M HCl required to convert all of the IO3- ions to I3- ions? Triiodide ions are generated in solution by the following reaction in acidic solution:
6H+ +IO3-(Aq) +8I-(Aq) > 3I3-(Aq) +3H20
Triiodide ion is determined by titration with a sodium thiosulfate(Na2S2O3) solution. The products are iodide ion and tetrathionate ion (S4O6 -2)
A sample of 0.6013g of potassium iodate was dissolved in water. Solid potassium iodide was then added in excess. What is the minimum volume of 3.00M HCl required to convert all of the IO3- ions to I3- ions?
6H+ +IO3-(Aq) +8I-(Aq) > 3I3-(Aq) +3H20
Triiodide ion is determined by titration with a sodium thiosulfate(Na2S2O3) solution. The products are iodide ion and tetrathionate ion (S4O6 -2)
A sample of 0.6013g of potassium iodate was dissolved in water. Solid potassium iodide was then added in excess. What is the minimum volume of 3.00M HCl required to convert all of the IO3- ions to I3- ions?

Explanation / Answer

moles of KIO3 = mass / molar mass = 0.6013 / 214

                        = 2.81 x 10^-3

6H+ +IO3-(Aq) +8I-(Aq) -------------> 3I3-(Aq) +3H20

6 mol HCl --------------------> 1 mol KIO3 or IO3-

x mol HCl -----------------> 2.81 x 10^-3 mol KIO3

x = 6 x 2.81 x 10^-3 = 0.0169

moles of HCl = 0.0169

molarity = 3.00 M

volume = moles / molarity

              = 0.0169 / 3.00

               = 5.62 x 10^-3 L

volume of HCl = 5.62 mL

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