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.

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:

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

the answer is a

photosynthesis