What minimum volume of 0.2 M $K I (a q)$ solution is required to completely precipitate all of the lead in 150 ml $0.1 M P b (N O_{3})_{2}$ solution? $2 K I + P b (N O_{3})_{2} (a q) \to 2 K N O_{3} (a q) + P b I_{2} (s)$

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Solution

From the balanced chemical equation
1 mole of $P b (N O_{3})_{2}$ requires 2 mole of $K I$ .
Number of mmoles of $P b (N O_{3})_{2}$ $= molarity \times volume(mL) = 0.1 \times 150 = 15 mmol$

mmoles of $K I$ required = $15 \times 2 = 30 mmol$

Given molarity of $K I = 0.2 M$
Volume of KI required $= \frac{30}{0.2} = 150 mL$

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