Question

What is the molar concentration of Fe2+ ion in an aqueous solution if 37.50 mL of 0.118 M KBrO3 is required for complete reaction with 11.00 mL of the Fe2+ solution? The net ionic equation is: 6Fe2+(aq)+BrO3−(aq)+6H+(aq)→6Fe3+(aq)+Br−(aq)+3H2O(l).

Answer 1

Answer:

2.41 M

Explanation:

6Fe⁺²(aq) + BrO₃⁻(aq) + 6H⁺(aq) → 6Fe³⁺(aq) + Br⁻(aq) + 3H₂O(l)

• The molar concentration of Fe²⁺ (or [Fe²⁺] ) is equal to the moles of Fe²⁺ divided by the volume, which in this case is 11.00 mL.
• The moles of Fe²⁺ are calculated from the moles of BrO₃⁻:

0.118 M KBrO₃ * 37.50 mL = 4.425 mmol BrO₃⁻

4.425 mmol BrO₃⁻ * $\frac{6mmolFe^{+2}}{1mmolBrO_{3}^{-}}$ = 26.55 mmol Fe²⁺

Thus [Fe²⁺] = 26.55 mmol Fe²⁺ / 11.00 mL = 2.41 M