Question

The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq) (thiosulfate ion). The determination is based on the net ionic equation 2S2O2−3(aq)+I−3(aq)⟶S4O2−6(aq)+3I−(aq) Given that it requires 27.7 mL of 0.440 M Na2S2O3(aq) to titrate a 15.0 mL sample of I−3(aq), calculate the molarity of I−3(aq) in the solution.

Answer 1

milimoles (mmol) - concentration (M) X volume (ml) mmol of Na2S2O3 = 0.440 M X 2717ML = 12.188 mmol N026, 03 dissociate completely in aqueous solution. Na2S₂ Os call anot (aq) + S. Os lag) After elmociation 2X12.128 mmol 12.188 mmol = 24,376 mmol a mmol of So032- present = 12:188 mmol 25.03 ² lag) + Ihr (aq) - S40% loal t. 2 mmol of 52 03² - react completely with Immol of Is- - 12.188 mmol of S₂O3²- react with Is = Immol x 12.188 mol ammol - 6.094 mmol - mmol of volume of I5 present 6.094 mmol Is-sample : 1500 mL molanty of Is-solution = mmol volume(mil 6.094 mmol = 0.406M 15.0mL Anwer Please Rate