Iron(II) is oxidized to iron(III) by bromate ion. In this reaction, bromate ion is reduced to bromide ion as shown by the net ionic equation: 6Fe^2+(aq) + BrO3^-(aq) + 6H+(aq) -> 6Fe^3+(aq) + Br^-(aq) + 3H2O(l). In a titration experiment, 50.00 mL of an iron(II) solution is titrated with a standard solution of 0.220 M bromate [BrO3^-]. The equivalence point is found to occur at the addition of 32.50 mL of the standard bromate solution. What is the molarity of the iron(II) solution?
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The Correct Answer and Explanation is:
Correct Answer:
The molarity of the iron(II) solution is 0.286 M.
Step-by-Step Solution:
Given the balanced net ionic equation:6Fe2+(aq)+BrO3−(aq)+6H+(aq)→6Fe3+(aq)+Br−(aq)+3H2O(l)6\text{Fe}^{2+}(aq) + \text{BrO}_3^-(aq) + 6\text{H}^+(aq) \rightarrow 6\text{Fe}^{3+}(aq) + \text{Br}^-(aq) + 3\text{H}_2\text{O}(l)6Fe2+(aq)+BrO3−(aq)+6H+(aq)→6Fe3+(aq)+Br−(aq)+3H2O(l)
From the stoichiometry of the equation, 1 mole of BrO₃⁻ reacts with 6 moles of Fe²⁺.
Step 1: Calculate moles of bromate used
Given:
- Volume of BrO₃⁻ solution = 32.50 mL = 0.03250 L
- Molarity of BrO₃⁻ solution = 0.220 M
mol BrO3−=0.220 mol/L×0.03250 L=0.00715 mol\text{mol BrO}_3^- = 0.220 \, \text{mol/L} \times 0.03250 \, \text{L} = 0.00715 \, \text{mol}mol BrO3−=0.220mol/L×0.03250L=0.00715mol
Step 2: Use stoichiometry to find moles of Fe²⁺
From the balanced equation:1 mol BrO3−→6 mol Fe2+1 \, \text{mol BrO}_3^- \rightarrow 6 \, \text{mol Fe}^{2+}1mol BrO3−→6mol Fe2+mol Fe2+=0.00715 mol BrO3−×6=0.0429 mol Fe2+\text{mol Fe}^{2+} = 0.00715 \, \text{mol BrO}_3^- \times 6 = 0.0429 \, \text{mol Fe}^{2+}mol Fe2+=0.00715mol BrO3−×6=0.0429mol Fe2+
Step 3: Calculate molarity of Fe²⁺ solution
Given:
- Volume of Fe²⁺ solution = 50.00 mL = 0.05000 L
Molarity of Fe2+=0.0429 mol0.05000 L=0.858 M\text{Molarity of Fe}^{2+} = \frac{0.0429 \, \text{mol}}{0.05000 \, \text{L}} = 0.858 \, \text{M}Molarity of Fe2+=0.05000L0.0429mol=0.858M
However, this value contradicts earlier logic. A reevaluation of the mole of bromate:mol BrO3−=0.220×0.03250=0.00715 mol\text{mol BrO}_3^- = 0.220 \times 0.03250 = 0.00715 \, \text{mol}mol BrO3−=0.220×0.03250=0.00715molmol Fe2+=0.00715×6=0.0429 mol\text{mol Fe}^{2+} = 0.00715 \times 6 = 0.0429 \, \text{mol}mol Fe2+=0.00715×6=0.0429molMolarity Fe2+=0.04290.05000=0.858 M\text{Molarity Fe}^{2+} = \frac{0.0429}{0.05000} = 0.858 \, \text{M}Molarity Fe2+=0.050000.0429=0.858M
Final Answer: 0.858 M
The earlier value of 0.286 M is incorrect; the correct molarity of the iron(II) solution is 0.858 M.
Explanation
In redox titrations involving iron(II) and bromate ions, the stoichiometric relationship governs the reaction. Iron(II) undergoes oxidation to iron(III), while bromate is reduced to bromide. The balanced net ionic equation confirms the molar ratio between the two species: 1 mole of bromate reacts with 6 moles of iron(II). This stoichiometric proportion is critical in determining the concentration of the unknown solution.
The titration setup involves a known concentration and volume of bromate solution, which reacts completely with the iron(II) in the flask. The number of moles of bromate is calculated from its volume and molarity. Multiplying this amount by 6 yields the number of moles of iron(II) present in the reaction at the equivalence point.
Once the total moles of iron(II) are determined, dividing this value by the volume of the iron(II) solution gives its molarity. This process is a standard approach in volumetric analysis, relying on mole-to-mole ratios and the concept of equivalence in chemical reactions.
The calculation of the molarity of the iron(II) solution results in 0.858 mol/L, reflecting the actual concentration needed to completely react with the provided volume and concentration of bromate. Each value in the problem is crucial, and misinterpreting volume units or reaction stoichiometry would lead to incorrect results. The final value represents the concentration of iron(II) that exactly reacts with the given quantity of bromate under the described experimental conditions.
