Question

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)

Answer 1

Answer:

$\large \boxed{\text{0.028 24 mol/L}}$

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

$\text{Moles of As _{4} O}_{6} =\text{ 0.1021 g As _{4} O}_{6} \times \dfrac{\text{1 mol As _{4} O}_{6}}{\text{395.68 g As _{4} O}_{6}} = 2.580 \times 10^{-4} \text{ mol As _{4} O}_{6}\\\\=\text{0.2580 mmol As _{4} O}_{6}$

3. Moles of I₃⁻

$\text{Moles of I}_{3}^{-} = \text{0.2580 mmol As _{4} O}_{6} \times \dfrac{\text{4 mmol I}_{3}^{-}}{\text{1 mmol As _{4} O}_{6}} =\text{1.032 mmol I}_{3}^{-}$

4. [I₃⁻]

$c = \dfrac{\text{1.032 mmol }}{\text{36.55 mL }} = \textbf{0.028 24 mol/L}\\\text{The concentration of I _{3}^{-} is \large \boxed{\textbf{0.028 24 mol/L}} }$