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

Step-by-step 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* {\text {Volume in L}}=0.440* 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=
`(1)/(2)* 0.011=5.5* 10^(-3)` moles of
`I_3^-`

Thus Molarity of
`I_3^-=(5.5* 10^(-3))/(0.025L)=0.22M`

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