Question

The amount of I₃⁻(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 2S₂O₃²⁻(aq)+I₃⁻(aq)⟶S₄O₆²⁻(aq)+3I⁻(aq). Given that it requires 38.1 mL of 0.440 M Na₂S₂O₃(aq) to titrate a 25.0 mL sample of I₃⁻(aq), calculate the molarity of I₃⁻(aq) in the solution.

Answer 1

Answer: 0.22 M

Explanation:

$2S_2O_3^{2-}(aq)+I_3^-\rightarrow S_4O_6^{2-}(aq)+3I^-(aq)$

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 $Na_2S_2O_3=Molarity\times {\text {Volume in L}}=0.440\times 0.025=0.011moles$

$Na_2S_2O_3\rightarrow 2Na^+S_2O_3^{2-}$

Thus moles of $S_2O_3^{2-}$ = 0.011

According to stoichiometry:  

2 moles of $S_2O_3^{2-}(aq)$ require 1 mole of $I_3^$

Thus 0.011 moles of $S_2O_3^{2-}(aq)$ require=$\frac{1}{2}\times 0.011=5.5\times 10^{-3}$ moles of $I_3^-$

Thus Molarity of $I_3^-=\frac{5.5\times 10^{-3}}{0.025L}=0.22M$

Therefore, the molarity of $I_3^-$ in the solution is 0.22 M