Answer :
Approximately 49.1 mL of potassium iodide (KI) solution is necessary to react with 34.1 mL of a 0.216N solution of potassium permanganate (KMnO₄) in a basic medium. The patient balances the chemical equation and applies the stoichiometry and equivalent concentration concepts in this chemistry question.
The subject matter of the question pertains to a chemical reaction between potassium iodide (KI) and potassium permanganate (KMnO4). This is a stoichiometry problem, which is a subfield of chemistry dealing with the relationships between quantities of reactants and products. The question asks for the volume of KI solution needed to react with a given volume of a KMnO4 solution. Given that the normality (N) is the measure of equivalent concentration, it’s crucial to balance the equation first to know the ratio of reactants.
The reaction between KMnO₄ and KI in a basic medium involves a 5:2 stoichiometric ratio respectively. Hence for each mole of KMnO₄, 0.4 moles of KI are required.
By applying the formula: N₁V₁ = N₂V₂ where N₁=0.216N, V₁=34.1mL (of KMnO₄),
N₂=0.150N, we can solve for V₂, the volume of KI needed.
Therefore, V₂ = (N1V1)/N2 = (0.216N*34.1mL)/0.150N ≈ 49.1 mL
hence the correct answer should be (e) 49.1 mL.
Learn more about the topic of Chemical Reactions and Stoichiometry here:
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