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Unformatted text preview: estrada (gse77) Homework 10 Sutcliffe (52410) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points 3 . 58 g of Ag 2 SO 4 will dissolve in 1 L of water. Calculate the solubility product constant for this salt. Correct answer: 6 . 05628 10 6 . Explanation: solubility of Ag 2 SO 4 = 3 . 58 g / 1 L H 2 O solubility = parenleftbigg 3 . 58 g Ag 2 SO 4 L parenrightbigg parenleftbigg 1 mol Ag 2 SO 4 311 . 77 g Ag 2 SO 4 parenrightbigg = 0 . 0114828 mol / L Ag 2 SO 4 Ag 2 SO 4 (s) 2 Ag + (aq) + SO 2 4 (aq) [Ag + ] = 2 (0 . 0114828) = 0 . 0229656 [SO 2 4 ] = 0 . 0114828 K sp = [Ag + ] 2 [SO 2 4 ] = (0 . 0229656) 2 (0 . 0114828) = 6 . 05628 10 6 002 10.0 points A saturated solution of MgF 2 (s) is found to have a Mg 2+ concentration of 2 . 10 3 M. What is the value of K sp for MgF 2 ? 1. 4 . 10 9 2. 3 . 2 10 8 correct 3. 4 . 10 12 4. 8 . 10 9 5. 1 . 10 9 Explanation: [Mg 2+ ] = 2.0 10 3 M MgF 2 Mg 2+ + 2 F K sp = [Mg 2+ ] [F ] 2 = ( x ) (2 x ) 2 = 4 x 3 = 4 ( 2 10 3 ) 3 = 3 . 2 10 8 003 10.0 points A solution of AgI contains 2 mol/L Ag + . K sp of AgI is 8 . 3 10 17 . What is the maximum I concentration that can exist in this solution? Correct answer: 4 . 15 10 17 mol / L. Explanation: [Ag + ] = 2 [I ] = ? K sp = 8 . 3 10 17 AgI(s) Ag + (aq) + I (aq) K sp = [Ag + ] [I ] [I ] = K sp [Ag + ] = 8 . 3 10 17 2 = 4 . 15 10 17 mol / L 004 (part 1 of 2) 10.0 points It is necessary to add iodide ions to precipitate the lead(II) ions from 250 mL of 0 . 073 M Pb(NO 3 ) 2 (aq). What minimum iodide ion concentration is required for the onset of PbI 2 precipitation? The solubility product of PbI 2 is 1 . 4 10 8 . Correct answer: 0 . 000437928 mol / L. Explanation: [Pb 2+ ] = 0 . 073 M K sp = 1 . 4 10 8 The balanced equation is Pb 2+ (aq) + 2 I (aq) PbI 2 (s) Analyzing the reaction using concentrations, Pb 2+ + 2 I PbI 2 . 073 + x . 073 x estrada (gse77) Homework 10 Sutcliffe (52410) 2 K sp = [Pb 2+ ] [I ] 2 1 . 4 10 8 = (0 . 073) ( x ) 2 x = 0 . 000437928 mol / L = [I ] 005 (part 2 of 2) 10.0 points What mass of KI must be added for PbI 2 to form?...
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 Spring '07
 Holcombe

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