Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + Pbl2(s) Part A What minimum volume of 0.229 M potassium iodide solution is required to completely precipitate all of the lead in 200.0 mL of a 0.150 M lead(II) nitrate solution? O 65.5 mL O 131 mL O 524 mL O 262 mL

Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + Pbl2(s) Part A What minimum volume of 0.229 M potassium iodide solution is required to completely precipitate all of the lead in 200.0 mL of a 0.150 M lead(II) nitrate solution? O 65.5 mL O 131 mL O 524 mL O 262 mL

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter15: Equilibria Of Other Reaction Classes

Section: Chapter Questions

Problem 10E: The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities...

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