What minimum volume of 0.2 M KI(aq) solution is required to completely precipitate all of the lead in 150 ml 0.1MPb(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=15mmol
mmoles of KI required = 15×2=30mmol
Given molarity of KI=0.2M Volume of KI required =300.2=150mL