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Question

What minimum volume of 0.2 M KI(aq) solution is required to completely precipitate all of the lead in 150 ml 0.1 M Pb(NO3)2 solution? 2KI+Pb(NO3)2(aq)2KNO3(aq)+PbI2(s)

A
300 mL
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B
75 mL
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C
600 mL
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D
150 mL
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Solution

The correct option is D 150 mL
From the balanced chemical equation
1 mole of Pb(NO3)2 requires 2 mole of KI.
Number of mmoles of Pb(NO3)2 =molarity×volume(mL) =0.1×150=15 mmol

mmoles of KI required = 15×2=30 mmol

Given molarity of KI=0.2 M
Volume of KI required =300.2=150 mL

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