1. (a) (b) (c) (d) (e) 3. There is no change. A solid has an activity of 1 whether there is a little or a lot. 5. The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve. 7. CaF 2 , MnCO 3 , and ZnS 9. (a) (b) (c) (d) 11. (a)1.77 10 –7 ; (b) 1.6 10 –6 ; (c) 2.2 10 –9 ; (d) 7.91 10 –22 13. (a) 2 10 –2 M ; (b) 1.5 10 –3 M ; (c) 2.27 10 –9 M ; (d) 2.2 10 –10 M 15. (a) 6.4 10 −9 M = [Ag + ], [Cl − ] = 0.025 M . Check: an insignificant change; (b) 2.2 10 −5 M = [Ca 2 + ], [F − ] = 0.0013 M . Check: This value is less than 5% and can be ignored. (c) 0.2238 M = [Ag + ] = 7.4 10 –3 M . Check: the condition is satisfied. (d) [OH – ] = 2.8 10 –3 M ; 5.7 10 −12 M = [Zn 2+ ]. Check: x is less than 5% of [OH – ] and is, therefore, negligible. 17. (a) [Cl – ] = 7.6 10 −3 M Check: This value is too large to drop x . Therefore solve by using the quadratic equation: [Ti + ] = 3.1 10 –2 M [Cl – ] = 6.1 10 –3 (b) [Ba 2 + ] = 7.7 10 –4 M Check: Therefore, the condition is satisfied. [Ba 2 + ] = 7.7 10 –4 M [F – ] = 0.0321 M ; (c) Mg(NO 3 ) 2 = 0.02444 M Check: The condition is satisfied; the above value is less than 5%. [Mg 2 + ] = 0.0244 M (d) [OH – ] = 0.0501 M [Ca 2 + ] = 3.15 10 –3 Check: This value is greater than 5%, so a more exact method, such as successive approximations, must be used. [Ca 2+ ] = 2.8 10 –3 M [OH – ] = 0.053 10 –2 M 19. The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change. 21. CaSO 4 ∙2H 2 O is the most soluble Ca salt in mol/L, and it is also the most soluble Ca salt in g/L. 23. 4.8 10 –3 M = = [Ca 2+ ]; Since this concentration is higher than 2.60 10 –3 M , “gyp” water does not meet the standards. 25. Mass (CaSO 4 ·2H 2 O) = 0.72 g/L 27. (a) [Ag + ] = [I – ] = 1.3 10 –5 M ; (b) [Ag + ] = 2.88 10 –2 M , = 1.44 10 –2 M ; (c) [Mn 2+ ] = 3.7 10 –5 M , [OH – ] = 7.4 10 –5 M ; (d) [Sr 2+ ] = 4.3 10 –2 M , [OH – ] = 8.6 10 –2 M ; (e) [Mg 2+ ] = 1.3 10 –4 M , [OH – ] = 2.6 10 –4 M . 29. (a) 1.45 10 –4 ; (b) 8.2 10 –55 ; (c) 1.35 10 –4 ; (d) 1.18 10 –5 ; (e) 1.08 10 –10 31. (a) CaCO 3 does precipitate. (b) The compound does not precipitate. (c) The compound does not precipitate. (d) The compound precipitates. 33. 3.03 10 −7 M 35. 9.2 10 −13 M 37. [Ag + ] = 1.8 10 –3 M 39. 6.3 10 –4 41. (a) 2.25 L; (b) 7.2 10 –7 g 43. 100% of it is dissolved 45. (a) and Cu 2+ : Add (b) and Cl – : Add Ba 2+ . (c) Hg 2+ and Co 2+ : Add S 2– . (d) Zn 2+ and Sr 2+ : Add OH – until [OH – ] = 0.050 M . (e) Ba 2+ and Mg 2+ : Add (f) and OH – : Add Ba 2+ . 47. AgI will precipitate first. 49. 1.5 10 −12 M 51. 3.99 kg 53. (a) 3.1 10 –11 ; (b) [Cu 2+ ] = 2.6 10 –3 ; = 5.3 10 –3 55. 1.8 10 –5 g Pb(OH) 2 57. 1.23 10 −3 g Mg(OH) 2 59. MnCO 3 will form first since it has the smallest K sp value among these homologous compounds and is therefore the least soluble. MgCO 3 •3H 2 O will be the last to precipitate since it has the largest K_sp value and is the most soluble. K sp value. 62. when the amount of solid is so small that a saturated solution is not produced 64. 1.8 10 –5 M 66. 5 10 23 68. [Cd 2+ ] = 9.5 10 –5 M ; [CN – ] = 3.8 10 –4 M 70. [Co 3+ ] = 3.0 10 –6 M ; [NH 3 ] = 1.8 10 –5 M 72. 1.3 g 74. 0.79 g 76. (a) (b) (c) (d) (e) 78. (a) (b) (c) (d) 80. 0.0281 g 82. 84. (a) (b) The electronic and molecular shapes are the same—both tetrahedral. (c) The tetrahedral structure is consistent with sp 3 hybridization. 86. 0.014 M 88. 7.2 10 –15 M 90. 4.4 10 −22 M 93. [OH − ] = 4.5 10 −6 ; [Al 3+ ] = 2 10 –16 (molar solubility) 95. ; [Ba 2+ ] = 4.7 10 –7 (molar solubility) 97. [OH – ] = 7.6 10 −3 M ; [Pb 2+ ] = 2.1 10 –11 (molar solubility) 99. 7.66 101. (a) K sp = [Mg 2+ ][F – ] 2 = (1.21 10 –3 )(2 1.21 10 –3 ) 2 = 7.09 10 –9 (b) 7.09 10 –7 M (c) Determine the concentration of Mg 2+ and F – that will be present in the final volume. Compare the value of the ion product [Mg 2+ ][F – ] 2 with K sp . If this value is larger than K sp , precipitation will occur. 0.1000 L 3.00 10 –3 M Mg(NO 3 ) 2 = 0.3000 L M Mg(NO 3 ) 2 M Mg(NO 3 ) 2 = 1.00 10 –3 M 0.2000 L 2.00 10 –3 M NaF = 0.3000 L M NaF M NaF = 1.33 10 –3 M ion product = (1.00 10 –3 )(1.33 10 –3 ) 2 = 1.77 10 –9 This value is smaller than K sp , so no precipitation will occur. (d) MgF 2 is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Chatelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic. 103. BaF 2 , Ca 3 (PO 4 ) 2 , ZnS; each is a salt of a weak acid, and the from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations 105. Effect on amount of solid CaHPO 4 , [Ca 2+ ], [OH – ]: (a) increase, increase, decrease; (b) decrease, increase, decrease; (c) no effect, no effect, no effect; (d) decrease, increase, decrease; (e) increase, no effect, no effect