1 6 9 10 9 2 1 69 10 14 3 5 2 10 19 4 1 4 10 21 5 7 6

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1. 6 . 9 × 10 - 9 2. 1 . 69 × 10 - 14 3. 5 . 2 × 10 - 19 4. 1 . 4 × 10 - 21 5. 7 . 744 × 10 - 13 6. 0.0053 7. 8 . 281 × 10 - 17 correct Explanation: S = 9 . 1 × 10 - 9 mol / L The solubility equilibrium is AgI(s) Ag + (aq) + I - (aq) [Ag + ] = [I - ] = S = 9 . 1 × 10 - 9 mol / L K sp = [Ag + ][I - ] = ( 9 . 1 × 10 - 9 ) 2 = 8 . 281 × 10 - 17 020 10.0 points What is K sp for Ni(OH) 2 , if its molar solu- bility is 4 . 1 × 10 - 6 mol / L?
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– Homework #8 – holcombe – 6 1. 4 . 7 × 10 - 30 2. 8 . 3 × 10 - 8 3. 2 . 8 × 10 - 16 correct 4. 1 . 7 × 10 - 14 5. 6 . 9 × 10 - 11 6. 5 . 2 × 10 - 25 7. 1 . 4 × 10 - 21 Explanation: S = 4 . 1 × 10 - 6 mol / L The solubility equilibrium is Ni(OH) 2 (s) Ni 2+ (aq) + 2 OH - (aq) [Ni 2+ ] = S = 4 . 1 × 10 - 6 mol / L [OH - ] = 2 S = 8 . 2 × 10 - 6 mol / L K sp = [Ni 2+ ][OH - ] 2 = (4 . 1 × 10 - 6 ) ( 8 . 2 × 10 - 6 ) 2 = 2 . 75684 × 10 - 16 021 10.0 points Rank following salts from least to most solu- ble: BiI K sp = 7 . 7 × 10 - 19 Cd 3 (AsO 4 ) 2 K sp = 2 . 2 × 10 - 33 AlPO 4 K sp = 9 . 8 × 10 - 21 CaSO 4 K sp = 4 . 9 × 10 - 5 1. Cd 3 (AsO 4 ) 2 < CaSO 4 < AlPO 4 < BiI 2. BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 < AlPO 4 3. AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 correct 4. CaSO 4 < AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 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. Doing so results in approximate molar solubilities of 10 - 10 , 10 - 7 , 10 - 11 and 10 - 3 for bismuth iodide, cadmium arsenate, aluminum phosphate and calcium sulfate, respectively. Arranging these from least to greatest produces: AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 . 022 10.0 points Rank the following salts from greatest to least molar solubility: A) BaCrO 4 K sp = 1 . 17 × 10 - 10 B) AlPO 4 K sp = 9 . 84 × 10 - 21 C) CaSO 4 K sp = 4 . 93 × 10 - 5 D) FeS K sp = 8 × 10 - 19 1. C > A > B > D 2. A > C > D > B 3. A > C > B > D 4. C > A > D > B correct Explanation: All of the salts are composed of a single cation and anion, so the solubility of each is the square root of its K sp . 023 10.0 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. 1 . 7 × 10 - 3 M 2. 3 . 4 × 10 - 3 M 3. 6 . 8 × 10 - 3 M correct 4. 8 . 5 × 10 - 7 M 5. 4 . 2 × 10 - 8 M Explanation: K sp of PbCl 2 = 1.7 × 10 - 5 PbCl 2 Pb 2+ + 2FCl -
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– Homework #8 – holcombe – 7 K sp = [Pb 2+ ] [Cl - ] 2 1 . 7 × 10 - 5 = [Pb 2+ ] (0 . 05) 2 [Pb 2+ ] = 0 . 0068 024 10.0 points AgCl would be least soluble in 1. 0.1 M HNO 3 . 2. pure water. 3. 0.1 M HCl. 4. 0.1 M CaCl 2 . correct 5. 0.1 M NH 3 . Explanation: 025 10.0 points Does a precipitate form when 100 mL of 0.00250 M AgNO 3 and 100 mL of 0.00200 M NaBr solutions are mixed? 1. AgBr precipitates. correct 2. AgBr does not precipitate. Explanation: V AgNO 3 = 0 . 1 L [AgNO 3 ] = 0 . 0025 M V NaBr = 0 . 1 L [NaBr] = 0 . 002 M AgNO 3 + NaBr -→ AgBr + NaNO 3 AgBr(s) Ag + (aq) + Br - (aq) K sp = [Ag + ] [Br - ] = 5 . 0 × 10 - 13 (0 . 1 L) (0 . 0025 M Ag + ) = 0 . 00025 mol (0 . 1 L) (0 . 002 M Br - ) = 0 . 0002 mol total volume = 0 . 1 L + 0 . 1 L = 0 . 2 L 0 . 00025 mol Ag + 0 . 2 L = 0 . 00125 mol / L Ag + 0 . 0002 mol Br - 0 . 2 L = 0 . 001 mol / L Br - [Ag + ] [Br - ] = (0 . 00125 mol / L) × (0 . 001 mol / L) = 1 . 25 × 10 - 6 > K sp AgBr precipitates.
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