Homework 08 Solutions - Gadhia Tejas Homework 8 Due Nov 5...

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Gadhia, Tejas – 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. 3 . 4 × 10 - 3 M 2. 6 . 8 × 10 - 3 M correct 3. 8 . 5 × 10 - 7 M 4. 1 . 7 × 10 - 3 M 5. 4 . 2 × 10 - 8 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 = 3 . 8. The solubility product of iron(III) hydroxide is 2 × 10 - 39 . Correct answer: 7 . 96214 × 10 - 9 . Explanation: K sp = 2 × 10 - 39 pH = 3 . 8 Let S = molar solubility pOH = 14 - pH = 14 - 3 . 8 = 10 . 2 [OH - ] = 10 - pOH = 6 . 30957 × 10 - 11 The reaction is Fe 3+ (aq) + 3 OH - (aq) * ) Fe(OH) 3 (s) K sp = [Fe 3+ ] [OH - ] 3 S = K sp [OH - ] 3 = 2 × 10 - 39 (6 . 30957 × 10 - 11 ) 3 = 7 . 96214 × 10 - 9 003 (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. HCl solution. correct 2. H 2 CO 3 solution. 3. SrCl 2 solution. 4. NaOH solution. 5. NH 3 solution. 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?
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Gadhia, Tejas – Homework 8 – Due: Nov 5 2007, 11:00 pm – Inst: James Holcombe 2 1. Cd +2 and Pb 2+ correct 2. Cu +2 and Pb 2+ 3. Co +2 and Pb 2+ 4. Cd +2 and Co +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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