HW 8 Solutions - Wright, Jacqueline Homework 8 Due: Nov 5...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Page1 / 6

HW 8 Solutions - Wright, Jacqueline Homework 8 Due: Nov 5...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online