Equilíbrio Iônico Aquoso

Prévia do material em texto

372 Chapter 17 Aqueous Ionic Equilibrium 17.91 Given: ionic compound formulas AX and and = 1.5 X 10⁻⁵ Find: higher molar solubility (S) Conceptual Plan: The expression of the solubility product constant of is = The molar solubility of a compound, can be computed directly from by solving for S in the expression = = Solution: For AX, = 1.5 X 10⁻⁵, m = 1, and n = 1; so = 1.5 X 10⁻⁵ = Rearrange to solve for S. S = 1.5 X 10⁻⁵ = 3.9 X 10⁻³ M. For = 1.5 X 10⁻⁵, m = 1, and n = 2; so = 1.5 X 10⁻⁵ = Rearrange to solve for S.S = 1.5 X 4 10⁻⁵ = 1.6 X 10⁻² M. Because 10⁻² M > 10⁻³ M, has a higher molar solubility. Check: The units (M) are correct. The more ions generated, the greater the molar solubility for the same value of the 17.93 Given: Fe(OH)₂ in 100.0 mL solution Find: grams of Fe(OH)₂ Other: = 4.87 X 10⁻¹⁷ Conceptual Plan: The expression of the solubility product constant of is = The molar solubility of a compound, can be computed directly from by solving for S in the expression = = Then solve for then mL L then 1L 1000 mL S, L mol Fe(OH)₂. 89.87 Fe(OH)₂ L 1 mol Fe(OH)₂ Solution: For Fe(OH)₂, = 4.87 10⁻¹⁷, A = Fe²⁺, m = 1,X = and n = 2; = 4.87 10⁻¹⁷ = Rearrange to solve for S.S = 3 4.87 X 4 10⁻¹⁷ = 2.30050 X 10⁻⁶ M. Then 100.0 mL X 1000 1L mL = 0.1000 L then 2.30050 X 10⁻⁶ mol Fe(OH)₂ X = 2.30050 X 10⁻⁷ X 89.87 1 Fe(OH)₂ = 2.07 X 10⁻⁵g Fe(OH)₂. Check: The units (g) are correct. The solubility rules from Chapter 4 (most hydroxides are insoluble) suggest that very little Fe(OH)₂ will dissolve; so the magnitude of the answer is not surprising. 17.95 (a) Given: BaF₂ Find: molar solubility (S) in pure water Other: = 2.45 X 10⁻⁵ Conceptual Plan: The expression of the solubility product constant of is = The molar solubility of a compound, can be computed directly from by solving for S in the expres- sion = = Solution: For BaF₂, = 2.45 X 10⁻⁵, A = Ba²⁺, m = 1,X=F, = and n = 2; so = 2.45 X 10⁻⁵ = Rearrange to solve for S.S = 2.45 4 10⁻⁵ = 1.83 X 10⁻² M (b) Given: Find: molar solubility (S) in 0.10 M Ba(NO₃)₂ Other: = 2.45 X 10⁻⁵ Conceptual Plan: M Ba(NO₃)₂ M then M S Ba(NO₃)₂(s) + ICE table Solution: Because one Ba²⁺ ion is generated for each Ba(NO₃)₂, [Ba²⁺] = 0.10 M. BaF₂(s) + Initial 0.10 0.00 Change S 2S = = 2.45 X 10⁻⁵ = (0.10 + Equil 0.10 + S 2S Copyright © 2017 Pearson Education, Inc.