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# Lead(II) ions can be removed from solution by

ISBN: 9780321809247 1

## Solution for problem 82E Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Problem 82E

Lead(II) ions can be removed from solution by precipitation with sulfate ions. Suppose that a solution contains lead(II) nitrate. Write complete ionic and net ionic equations to show the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead(II) sulfate and aqueous potassium nitrate.

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Solution: Step1: a) Given volume of Ba(OH) solution = 50.0 mL = 50/1000 L = 0.050 L 2 Molarity of the Ba(OH) solution2 0.101 M Therefore, number of moles of Ba(OH) = Volume of th2solution in litres X Molarity = 0.050 L X 0.101 M = 0.005 mol Now, the reaction between HCl and Ba(OH) takes place in th2following manner: 2HCl(aq) + Ba(OH) --------2-> MgCl (aq) + 2H O 2 2 In the above equation, it is seen that, 1 mole of Ba(OH) requires 2 mole2of HCl for complete neutralization. Therefore, 0.005 moles of Mg(OH) will require 2 x 0.005 = 0.01) moles of HCl for complete neutralization. Now, using the expression for molarity, we can calculate the volume of HCl required. Volume of HCl in litres = Number of moles of HCl / Molarity = 0.01 mol / 0.120 M = 0.0833 L = 0.0833 X 1000 mL = 83.3 mL Thus, required volume of HCl is 83.3 mL Step2: b) Mass of NaOH = 0.200 g Molar mass of NaOH = 40.01 g/mol Therefore, number of moles of NaOH = 0.200 / 40.01 = 0.005 mol The reaction between H2SO4 and NaOH takes place in the following manner: H2O 4q) + 2NaOH(aq) -----------> Na SO (aq) + 2 O 4 2 In the above equation, it is seen that, 1 mole of H SO requires2 mo4s of NaOH for complete neutralization. Therefore, 0.005 moles of NaOH will require (0.005 / 2 = 0.0025) moles of H SO for 2 4 complete neutralization. Now, using the expression for molarity, we can calculate the volume of H SO required. 2 4 Volume of H SO 2 lit4s = Number of moles of H SO / Molarity 2 4 = 0.0025 mol / 0.125 M = 0.02 L = 0.02 X 1000 mL = 20 mL Thus, the required volume...

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Lead(II) ions can be removed from solution by

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