Question

The amount of I − 3 ( aq ) in a solution can be determined by titration with a solution containing a known concentration of S 2 O 2 − 3 ( aq ) (thiosulfate ion). The determination is based on the net ionic equation 2 S 2 O 2 − 3 ( aq ) + I − 3 ( aq ) ⟶ S 4 O 2 − 6 ( aq ) + 3 I − ( aq ) Given that it requires 29.4 mL of 0.380 M Na 2 S 2 O 3 ( aq ) to titrate a 30.0 mL sample of I − 3 ( aq ) , calculate the molarity of I − 3 ( aq ) in the solution.

Answer 1

The molarity of I₃⁻ (aq) solution: M₂ = 0.186 M

Step-by-step explanation:

Given net ionic equation:

2S₂O₃²⁻ (aq) + I₃⁻ ( aq ) ⟶ S₄O₆²⁻ (aq) + 3I⁻ (aq)

Number of moles of S₂O₃²⁻: n₁ = 2, Number of moles of I₃⁻: n₂ = 1

Given- For S₂O₃²⁻ solution: Molarity: M₁ = 0.380 M, Volume: V₁ = 29.4 mL;

For I₃⁻ (aq) solution: Molarity: M₂ = ? M, Volume: V₂ = 30.0 mL

To calculate the molarity of I₃⁻ (aq) solution, we use the equation:

`(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))`

`((0.380 M)* (29.4 mL))/(2)=(M_(2)* (30.0 mL))/(1)`

`\Rightarrow M_(2) = ((0.380 M)* (29.4 mL))/((30.0 mL)* 2) = 0.186 M`

Therefore, the molarity of I₃⁻ (aq) solution: M₂ = 0.186 M