A solution of I3¯(aq) can be standardized by using it to titrate As4O6(aq). The titration of 0.1021 g of As4O6(s) (MW = 395.68) dissolved in of water requires 36.55 mL of I3¯(aq). Calculate the molarity of the I3¯(aq) (Reaction ratio is 4 mol I3¯ / 1 mol As4O6)

Explanation:

1. Balanced chemical equation.

The unbalanced equation is  

As₄O₆ + I₃⁻ ⟶ H₃AsO₄ + I⁻

The balanced equation is

1As₄O₆ + 4I₃⁻ + 10H₂O ⟶ 4H₃AsO₄ + 12I⁻ + 8H⁺

2. Moles of As₄O₆

3. Moles of I₃⁻

4. [I₃⁻]

The molarity of the iodide ion in the given solution has been 0.028 M.

The balanced chemical reaction can be given as:

Moles of =

Moles of = 0.2580mmol

Moles of =

Moles of = 0.2580

Moles of = 1.032 mmol

Molarity =

Molarity =

Molarity = 0.028 M

The molarity of the iodide ion in the given solution has been 0.028 M.

For more information about molarity, refer to the link:

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