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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).

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Answer:

2.41 M

Step-by-step 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⁻ *
(6mmolFe^(+2))/(1mmolBrO_(3)^(-)) = 26.55 mmol Fe²⁺

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

User John Prado
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