Answer:
The molarity of I₃⁻ (aq) solution: M₂ = 0.186 M
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:
[tex]\frac{M_{1}V_{1}}{n_{1}}=\frac{M_{2}V_{2}}{n_{2}}[/tex]
[tex]\frac{(0.380 M)\times (29.4 mL)}{2}=\frac{M_{2}\times (30.0 mL)}{1}[/tex]
[tex]\Rightarrow M_{2} = \frac{(0.380 M)\times (29.4 mL)}{(30.0 mL)\times 2} = 0.186 M[/tex]
Therefore, the molarity of I₃⁻ (aq) solution: M₂ = 0.186 M
