HW 8 Solutions

# HW 8 Solutions - Wright Jacqueline – Homework 8 – Due...

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Unformatted text preview: Wright, Jacqueline – 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. 8 . 5 × 10- 7 M 2. 6 . 8 × 10- 3 M correct 3. 4 . 2 × 10- 8 M 4. 1 . 7 × 10- 3 M 5. 3 . 4 × 10- 3 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 = 6 . 5. The solubility product of iron(III) hydroxide is 2 × 10- 39 . Correct answer: 6 . 32456 × 10- 17 . Explanation: K sp = 2 × 10- 39 pH = 6 . 5 Let S = molar solubility pOH = 14- pH = 14- 6 . 5 = 7 . 5 [OH- ] = 10- pOH = 3 . 16228 × 10- 8 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 (3 . 16228 × 10- 8 ) 3 = 6 . 32456 × 10- 17 003 (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. H 2 CO 3 solution. 3. NH 3 solution. 4. NaOH 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? Wright, Jacqueline – 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 12/08/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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HW 8 Solutions - Wright Jacqueline – Homework 8 – Due...

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