CH 302 HW8 - Thompson Margaret Homework 8 Due Nov 5 2007 11:00 pm Inst James Holcombe 1 This print-out should have 24 questions Multiple-choice

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Unformatted text preview: Thompson, Margaret Homework 8 Due: Nov 5 2007, 11:00 pm Inst: James Holcombe 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The solubility product constant of PbCl 2 is 1 . 7 10- 5 . What is the maximum concentra- tion of Pb 2+ that can be in ocean water that contains 0.0500 M NaCl? 1. 6 . 8 10- 3 M correct 2. 4 . 2 10- 8 M 3. 1 . 7 10- 3 M 4. 3 . 4 10- 3 M 5. 8 . 5 10- 7 M Explanation: K sp of PbCl 2 = 1.7 10- 5 PbCl 2 Pb 2+ + 2FCl- K sp = [Pb 2+ ][Cl- ] 2 1 . 7 10- 5 = [Pb 2+ ](0 . 05) 2 [Pb 2+ ] = 0 . 0068 002 (part 1 of 1) 10 points Calculate the solubility of iron(III) hydrox- ide at pH = 8 . 9. The solubility product of iron(III) hydroxide is 2 10- 39 . Correct answer: 3 . 99052 10- 24 . Explanation: K sp = 2 10- 39 pH = 8 . 9 Let S = molar solubility pOH = 14- pH = 14- 8 . 9 = 5 . 1 [OH- ] = 10- pOH = 7 . 94328 10- 6 The reaction is Fe 3+ (aq) + 3OH- (aq) * ) Fe(OH) 3 (s) K sp = [Fe 3+ ][OH- ] 3 S = K sp [OH- ] 3 = 2 10- 39 (7 . 94328 10- 6 ) 3 = 3 . 99052 10- 24 003 (part 1 of 1) 10 points The mineral strontianite (SrCO 3 ) is quite in- soluble in water. The best reagent which could be added to dissolve a sample of stron- tianite is 1. SrCl 2 solution. 2. NaOH solution. 3. NH 3 solution. 4. H 2 CO 3 solution. 5. HCl solution. correct Explanation: SrCO 3- Sr 2+ + CO 2- 3 To dissolve SrCO 3 , we need to shift the equilibrium of this reaction to the right. We can do this by decreasing the amount of CO 2- 3 . We can do that by combining the CO 2- 3 ion with H + to form the weak acid H 2 CO 3 . This decrease in CO 2- 3 will shift the equilibirum to the right causing the SrCO 3 to dissolve. Of the reagents listed, HCl and H 2 CO 3 are acids and can provide the H + ion. However, H 2 CO 3 will not only add H + , but also CO 2+ 3 and cause the equilibrium to shift back to the left. 004 (part 1 of 1) 10 points Consider the following K sp values for metal sulfides: CdS = 3 . 6 10- 29 CoS = 5 . 9 10- 21 CuS = 8 . 7 10- 36 PbS = 8 . 4 10- 28 Which pair would be most difficult to sepa- rate by fractional precipitation? Thompson, Margaret Homework 8 Due: Nov 5 2007, 11:00 pm Inst: James Holcombe 2 1. Cd +2 and Pb 2+ correct 2. Co +2 and Pb 2+ 3. Cd +2 and Co +2 4. Cu +2 and Pb 2+ Explanation: The pair that will be most difficult to sep- arate will be the pair with the most similar K sp values. These are CdS = 3 . 6 10- 29 and PbS = 8 . 4 10- 28 . This means that they will precipitate at about the same S 2- concentra- tion....
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This note was uploaded on 07/17/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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CH 302 HW8 - Thompson Margaret Homework 8 Due Nov 5 2007 11:00 pm Inst James Holcombe 1 This print-out should have 24 questions Multiple-choice

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