Respuesta :
Answer:
[tex]\large \boxed{\text{0.028 24 mol/L}}[/tex]
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
[tex]\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}[/tex]
3. Moles of I₃⁻
[tex]\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}^{-}[/tex]
4. [I₃⁻]
[tex]c = \dfrac{\text{1.032 mmol }}{\text{36.55 mL }} = \textbf{0.028 24 mol/L}\\\text{The concentration of I$_{3}^{-}$ is $\large \boxed{\textbf{0.028 24 mol/L}}$}[/tex]
The molarity of the iodide ion in the given solution has been 0.028 M.
The balanced chemical reaction can be given as:
[tex]\rm As_4O_6\;+\;4\;I^3^-\;+\;10\;H_2O\;\rightarrow\;4\;H_3AsO_4\;+\;12\;I^-\;+\;8\;H^+[/tex]
Moles of [tex]\rm \bold{As_4O_6}[/tex] = [tex]\rm \dfrac{0.1021}{395.68}[/tex]
Moles of [tex]\rm \bold{As_4O_6}[/tex] = 0.2580mmol
Moles of [tex]\rm \bold{I^3^-}[/tex] = [tex]\rm moles\;of\;As_4O_6\;\times\;ratio\;of\;\dfrac{I^3^-}{As_4O_6}[/tex]
Moles of [tex]\rm \bold{I^3^-}[/tex] = 0.2580 [tex]\rm \times\;\dfrac{4}{1}[/tex]
Moles of [tex]\rm \bold{I^3^-}[/tex] = 1.032 mmol
Molarity = [tex]\rm \dfrac{moles}{volume\;(L)}[/tex]
Molarity = [tex]\rm \dfrac{1.032\;\times\;10^-^3}{36.55\;\times\;10^-^3}[/tex]
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:
https://brainly.com/question/2817451
