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386 Chapter 17 Aqueous Ionic Equilibrium 17.143 (a) Given: Au(OH)₃ in pure water Find: molar solubility (S) Other: = 5.5 X 10⁻⁴⁶ Conceptual Plan: Use equations derived in Problems 17.19 and 17.20 and solve for S. Then for ionic compound, = = check answer for validity. Solution: For Au(OH)₃, = 5.5 10⁻⁴⁶, A = Au³⁺, m = 1,X = and n = 3; so = = 5.5 X 10⁻⁴⁶ = Rearrange to solve for S.S 5.5 27 = 2.1 X 10⁻¹² M. This answer suggests that the = 3(2.1 X 10⁻¹² M) = 6.3 X 10⁻¹² M. This result is lower than what is found in pure water (1.0 X 10⁻⁷ M), so substitute this value for [OH⁻] and solve for S = [Au³⁺]. So Kₛₚ = [Au³⁺] = 5.5 X 10⁻⁴⁶ = S (1.0 10⁻⁷ M)³ and solving for S gives S = 5.5 X 10⁻²⁵ M. Check: The units (M) are correct. Because is so small, the autoionization of water must be considered and the solubility is smaller than what is normally anticipated. (b) Given: Au(OH)₃ in 1.0 M Find: molar solubility (S) Other: = 5.5 X 10⁻⁴⁶ Conceptual Plan: Because is a strong acid, it will neutralize the gold(III) hydroxide (through the reaction of with OH+ to form water (the reverse of the autoionization of water equilibrium). Write balanced equations for dissolving the solid and for the neutralization reaction. Add these two reactions to get the desired overall reaction. Using the rules from Chapter 15, multiply the individual reaction Ks to get the overall K for the sum of these reactions. Then M K S. ICE table Solution: Identify the solid as Au(OH)₃. Write the individual reactions and add them together. Au(OH)₃(s) + 3 + Au(OH)₃(s) + + Because the overall reaction is the sum of the dissolution reaction and three times the reverse of the autoionization of water reaction, the overall reaction K = 3 = (5.5 10⁻⁴⁶) 1 = 5.5 10⁻⁴ = [Au³⁺] [H⁺]³ then because is a strong acid, it will completely dissociate to and Set up an ICE table: Au(OH)₃(s) + Au³⁺(aq) + [H⁺] [Au³⁺] Initial 1.0 0.00 [Au³⁺] S K = = 5.5 X 10⁻⁴ = Change -3S +S (1.0 Equil 1.0 3S Assume that S is small (3S so (1.0 S = 5.5 X 10⁻⁴ M = (1.0)³ S = S. Confirm that the assumption is valid. 1.0 X 100% = 0.017% « 5%, so the assumption is valid. Check: The units (M) are correct. K is much larger than the original the solubility of Au(OH)₃ increases over that of pure water. 17.145 Given: 1.00 L of 0.100 M Find: volume of 0.100 M Na₂CO₃ to precipitate 99% of Mg²⁺ ions Other: = 6.82 10⁻⁶ Conceptual Plan: Because 99% of the Mg²⁺ ions are to be precipitated, 1% of the ions will be left in solution. Let x = required volume (in L). Calculate the amount of added and the amount of Mg²⁺ that does not precipitate and remains in solution. Use these to calculate the [Mg²⁺] and The solubility product constant (Kₛₚ) is the equilibrium expression for a chemical equation representing the dissolution of an ionic compound. The expression of the solubility product constant of is = Substitute these expressions in this equation to [Mg²⁺], x. for ionic compound, = Copyright © 2017 Pearson Education, Inc.