Chapter 4, Problem 53QAP

The molarity of iodine in solution can be determined by titration with arsenious acid, H₃AsO₄. The unbalanced equation for the reaction is

${\text{H}}_3{\text{AsO}}_3\left(aq\right)+{\text{I}}_2\left(aq\right)+{\text{H}}_2\text{O}\to 2{\text{I}}^{-}\left(aq\right)+{\text{H}}_3{\text{AsO}}_4\left(aq\right)+2{\text{H}}^{+}\left(aq\right)$A 243-mL solution of aqueous iodine is prepared by dissolving iodine crystals in water. A fifty-mL portion of the solution requires 15.42 mL of 0.134 M H₃AsO₃ for complete reaction. What is the molarity of the solution? How many grams of iodine were added to the solution?