The lead(II)ion can be effectively removed from solution via the addition of phosphate ions. The lead(II) phosphate precipitates and can be filtered off. +2 Na3PO4 (aq) →>> Pb3(PO4)2 (s) + 6 NaNO3(aq) A solution is made by dissolving 6.93 grams of lead (II) nitrate into about 50 mL of water. The volume is then precisely brought up to 100 mL and the solution is saved as stock solution. A 50.0 mL aliquot of this stock solution is then titrated with 0.222 M sodium phosphate. What would be the minimum number of milliliters (mL) of the the phosphate solution that are needed to completely precipitate (knock out) the lead in this aliquot? (tolerance is ±0.1 mL) *Note: The term aliquot means a part or portion of the whole. In this case, the aliquot is exactly 50.0 mL of the solution that was mixed. Type your answer... 3 Pb(NO3)2(aq)

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter4: Types Of Chemical Reactions And Solution Stoichiometry
Section: Chapter Questions
Problem 154CP: Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution:...
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The lead(II)ion can be effectively removed from solution via the addition of phosphate ions. The lead(II) phosphate precipitates and can be filtered off.
+ 2 Na3PO4 (aq) →>> Pb3(PO4)2 (s) + 6 NaNO3(aq)
A solution is made by dissolving 6.93 grams of lead (II) nitrate into about 50 mL of water. The volume is then precisely brought up to 100 mL and the solution is saved as stock
solution. A 50.0 mL aliquot of this stock solution is then titrated with 0.222 M sodium phosphate. What would be the minimum number of milliliters (mL) of the the phosphate
solution that are needed to completely precipitate (knock out) the lead in this aliquot? (tolerance is ±0.1 mL)
*Note: The term aliquot means a part or portion of the whole. In this case, the aliquot is exactly 50.0 mL of the solution that was mixed.
Type your answer...
3 Pb(NO3)2(aq)
Transcribed Image Text:The lead(II)ion can be effectively removed from solution via the addition of phosphate ions. The lead(II) phosphate precipitates and can be filtered off. + 2 Na3PO4 (aq) →>> Pb3(PO4)2 (s) + 6 NaNO3(aq) A solution is made by dissolving 6.93 grams of lead (II) nitrate into about 50 mL of water. The volume is then precisely brought up to 100 mL and the solution is saved as stock solution. A 50.0 mL aliquot of this stock solution is then titrated with 0.222 M sodium phosphate. What would be the minimum number of milliliters (mL) of the the phosphate solution that are needed to completely precipitate (knock out) the lead in this aliquot? (tolerance is ±0.1 mL) *Note: The term aliquot means a part or portion of the whole. In this case, the aliquot is exactly 50.0 mL of the solution that was mixed. Type your answer... 3 Pb(NO3)2(aq)
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