To remove harmful lead ions (Pb2+) from water, ions can be added to precipitate

Question

To remove harmful lead ions (Pb2+) from water, ions can be added to precipitate the lead as salts. Using the data from Appendix II in your text, which of the following ions should be added to water to remove the most lead ions?

Please provide steps to calculation.

PbBr PbCO PbC Lead(I) bromide Lead(I) carbonate Lead(lI) chloride Lead(I) chromate Lead(I) fluoride Lead(II) hydroxide Lead) iodide Lead(II) phosphate Lead) sulfate Lead(Il) sulfide 4.67 X 10-6 740 X 10-14 1.17 X 10 5 2.8× 10 13 3.3 x 10-8 1.43 X 10-20 9.8 x 10- 1 x 10-54 1.82 × 10 8 9.04 x 10-29 PbCrO PbF Pb(OH)2 Pbl Pb (PO2 PbSO PbS

Explanation / Answer

Since the task here is to add a particular ion to result in a compound that precipitates lead ions best, when observed at the converse, the ions which when added gives rise to the least soluble lead based compound is preferred. The values given in the table are the solubility products of the corresponding lead compounds.

The solubility product of a sparingly soluble compound is the equilibrium constant of the equilibrium created by the compound as a precipitate and in its dissolved form. Thus the lower the value of solubility product for a compound, the lower its solubility in the solvent at that temperature.

Now from the table given we can find that lead (II) sulfide has the lowest Ksp value among the others. So by adding sulfide ions to the water containing lead, we can precipitate out lead (II) ions best.

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