CHhmwk4 - moore (jwm2685) Homework 4 vanden bout (51640) 1...

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Unformatted text preview: moore (jwm2685) Homework 4 vanden bout (51640) 1 This print-out should have 42 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The relationship between the solubility s in water and K sp for the ionic solid M 2 A 3 is 1. K sp = s 5 . 2. K sp = 108 s 5 . correct 3. K sp = 6 s 2 . 4. K sp = 5 s . Explanation: 002 10.0 points The compound M 2 X 3 has a molar solubility of 0.0330 M. What is the value of K sp for M 2 X 3 ? 1. 3 . 2 10 5 2. 1 . 41 10 6 3. 3 . 91 10 8 4. . 000216 5. 4 . 23 10 6 correct Explanation: 003 10.0 points Which of the following salts would have ap- proximately the same molar solubility as AlPO 4 , K sp = 10 20 ? 1. HgI 2 , K sp = 10 30 correct 2. Co(OH) 2 , K sp = 10 15 3. BaSO 4 , K sp = 10 10 4. Cd 3 (PO 4 ) 2 , K sp = 10 33 Explanation: Both AlPO 4 and HgI 2 will approximate mo- lar solubilities of 10 10 . 004 10.0 points Rank following salts from least to most solu- ble: AgI K sp = 1 . 5 10 16 Ag 2 CO 3 K sp = 6 . 2 10 12 Ag 2 S K sp = 1 . 6 10 49 AgSCN K sp = 1 . 2 10 12 1. Ag C O 3 < Ag 2 S < AgI < AgSCN 2. AgI < AgSCN < Ag C O 3 < Ag 2 S 3. Ag 2 S < AgI < AgSCN < Ag C O 3 cor- rect 4. AgSCN < Ag C O 3 < Ag 2 S < AgI Explanation: Molar solubility can be approximated by taking the n th root of the K sp where n is the number of ions in the salt. Do- ing so results in approximate molar solubil- ities of 10 8 , 10 4 , 10 16 and 10 6 M for silver iodide, silver carbonate, silver sulfide and silver thiocyanate, respectively. Ar- ranging from least to most soluble produces: Ag 2 S < AgI < AgSCN < Ag C O 3 005 10.0 points The term solubility refers to 1. the ability of a substance to dissolve in solution. correct 2. the ability of a substance to react with acids or bases in solution. 3. the ability of a substance to form ions in solution. 4. the ability of a substance to gain electrons in solution. 5. the ability of a substance to gain or lose protons in solution. Explanation: 006 10.0 points moore (jwm2685) Homework 4 vanden bout (51640) 2 M 2+ is added to a solution of 0 . 0058 mol / L X and 0 . 0017 mol / L Z 3 . MX 2 has K sp = 5 . 2 10 12 , while M 3 Z 2 has K sp = 6 . 4 10 29 . What percentage of the Z 3 is still in solution (not precipitated) when the MX 2 just starts to precipitate? Correct answer: 7 . 74%. Explanation: K sp MX 2 = 5 . 2 10 12 [X ] = 0 . 0058 mol / L K sp M 3 Z 2 = 6 . 4 10 29 [Z 3 ] = 0 . 0017 mol / L The two relevant equilibria here are MX 2 (s) M 2+ + 2X K sp = [M 2+ ] [X ] 2 M 3 Z 2 (s) 3M +2 + 2Z 3 K sp = [M 2+ ] 3 [Z 3 ] 2 MX 2 would precipitate when K sp = [M 2+ ] [X ] 2 5 . 2 10 12 = [M 2+ ] (0 . 0058 mol / L) 2 [M 2+ ] = 1 . 54578 10 7 mol / L This corresponds to a [Z 3 ] of K sp = [M 2+ ] 3 [Z 3...
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This note was uploaded on 05/02/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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CHhmwk4 - moore (jwm2685) Homework 4 vanden bout (51640) 1...

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