Answer:
0.75M Fe²⁺
Explanation:
First, we need to balance the redox reaction in acidic medium. Then, we can obtain moles of KMnO4 and with the reaction moles and molarity of the Fe²⁺ solution:
Redox Balance:
Fe²⁺ → Fe³⁺ + 1e⁻
5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
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5Fe²⁺ + 5e⁻ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
Moles of KMnO₄:
70.0mL = 0.0700L * (0.150mol / L) = 0.0105 moles KMnO₄
Moles and molarity Fe²⁺:
0.0105 moles KMnO₄ * (5 moles Fe²⁺ / 1mol KMnO₄) = 0.0525 moles Fe²⁺
In 70.0mL = 0.0700L:
0.0525 moles Fe²⁺ / 0.0700L =