Question

Iodine reacts with aqueous thiosulfate ion in neutral solution according to the balanced equation 12(aq) + 25,03 -(aq) S4062(aq) + 27- (aq) How many grams of 12 are present in a solution if 35.20 mL of 0.150 M Na2S2O, solution is needed to titrate the la solution?

Answer 1

Answer: 0.67 g

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{moles}{\text {Volume in L}}$

$moles of [tex]Na_2S_2O_3=Molarity\times {\text {Volume in L}}=0.150\times 0.0352=5.28\times 10^{-3}moles$

$I_2(aq)+2S_2O_3^{2-}(aq) S_4O_6^{2-}(aq)+2I^-(aq)$

According to stoichiometry:  

2 moles of $2S_2O_3^{2-}(aq)$ require 1 mole of $I_2$

Thus $5.28\times 10^{-3}moles$ require=$\frac{1}{2}\times 5.28\times10^{-3}=2.64\times 10^{-3}$ moles of $I_2$

Mass of $I_2=moles\times {\text {molar mass}}=2.64\times 10^{-3}\times 254=0.67g$

Thus 0.67 g of iodine are present in the solution.