chapter17_part2

chapter17_part2 - 798 Chapter 17: Additional Aqueous...

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Chapter 17: Additional Aqueous Equilibria 798 52. Answer: K sp = 4.23 × 10 –12 Strategy and Explanation: Adapt the method developed in Problem-Solving Example 17.10. Write the chemical equation and the equilibrium expression for the dissociation of the solute: PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq) K sp = [Pb 2+ ][Br ] 2 At equilibrium, the moles of solid that dissolve per liter are: 3.74 " 10 # 3 g PbBr 2 100.0 mL " 1 mol PbBr 2 367.0 g PbBr 2 " 1000 mL 1 L = 1.02 × 10 –4 M The stoichiometry of the equation shows that the lead ion concentration is the identical to the number of moles of solid that dissolve per liter, 1.8 × 10 –5 M, and the bromide ion concentration is twice that value, 2 × (1.02 × 10 –4 M) = 2.04 × 10 –4 M. K sp = (1.02 × 10 –4 )(2.04 × 10 –4 ) 2 = 4.23 × 10 –12 53. Answer: see equations and expressions below Strategy and Explanation: Adapt the method developed in Problem-Solving Example 17.8. (a) PbCO 3 (s) Pb 2+ (aq) + CO 3 2– (aq) K sp = [Pb 2+ ][CO 3 2– ] (b) Ni(OH) 2 (s) Ni 2+ (aq) + 2 OH (aq) K sp = [Ni 2+ ][OH ] 2 (c) Sr 3 (PO 4 ) 2 (s) 3 Sr 2+ (aq) + 2 PO 4 3– (aq) K sp = [Sr 2+ ] 3 [PO 4 3– ] 2 (d) Hg 2 SO 4 (s) Hg 2 2+ (aq) + SO 4 2– (aq) K sp = [Hg 2 2+ ][SO 4 2– ] 54 . Answer: see equations and expressions below Strategy and Explanation: Adapt the method developed in Problem-Solving Example 17.8. (a) FeCO 3 (s) Fe 2+ (aq) + CO 3 2– (aq) K sp = [Fe 2+ ][CO 3 2– ] (b) Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2– (aq) K sp = [Ag + ] 2 [SO 4 2– ] (c) Ca 3 (PO 4 ) 2 (s) 3 Ca 2+ (aq) + 2 PO 4 3– (aq) K sp = [Ca 2+ ] 3 [PO 4 3– ] 2 (d) Mn(OH) 2 (s) Mn 2+ (aq) + 2 OH (aq) K sp = [Mn 2+ ][OH ] 2 55 . Answer: K sp = 3.0 × 10 –18 Strategy and Explanation: Adapt the method developed in Problem-Solving Example 17.10. Write the chemical equation and the equilibrium expression for the dissociation of the solute: Ag 3 AsO 4 (s) 3 Ag + (aq) + AsO 4 3– (aq) K sp = [Ag + ] 3 [AsO 4 3– ] At equilibrium, the moles of solid that dissolve per liter are: 8.5 " 10 # 6 g Ag 3 AsO 4 1 mL " 1 mol Ag 3 AsO 4 462.5238 g Ag 3 AsO 4 " 1000 mL 1 L = 1.8 × 10 –5 M The stoichiometry of the equation shows that the arsenate ion concentration is the identical to the number of moles of solid that dissolve per liter, 1.8 × 10 –5 M, and the silver ion concentration is three times that value, 3 × (1.8 × 10 –5 M) = 5.4 × 10 –5 M. K sp = (5.4 × 10 –5 ) 3 (1.8 × 10 –5 ) = 3.0 × 10 –18 56 . Answer: K sp = 2.22 × 10 –4 Strategy and Explanation: Adapt the method developed in the solution to Question 55. Write the chemical equation and the equilibrium expression for the dissociation of the solute: CaSO 4 (s) Ca 2+ (aq) + SO 4 3– (aq) K sp = [Ca 2+ ][SO 4 2– ]
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Chapter 17: Additional Aqueous Equilibria 799 At equilibrium, the moles of solid that dissolve per liter is: 2.03 g CaSO 4 1 L " 1 mol CaSO 4 136.14 g CaSO 4 = 0.0149 M The stoichiometry of the equation shows that the concentrations of calcium ion and sulfate ion are both the
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This note was uploaded on 07/15/2009 for the course CHE 131 taught by Professor Kerber during the Fall '08 term at SUNY Stony Brook.

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chapter17_part2 - 798 Chapter 17: Additional Aqueous...

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