Acid-Base Equilibria and Solubility Equilibria Chapter 16 Semester 1/2017 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Chapter 16 Semester 1 / 2016 16.1 Homogeneous versus Heterogeneous Solution Equilibria 16.3 Buffer Solution 16.6 Solubility Equilibria 16.8 The Common Ion Effect and Solubility

16.1 Homogeneous Versus Heterogeneous Solution Equilibria At equilibrium a weak acid solution contains nonionized acid as well as H+ ions and the conjugate base. All of these species are dissolved so the system is said to be Homogeneous. If we consider equilibrium of the dissolution and precipitation of slightly soluble substances, the system is said to be Heterogeneous.

16.3 Buffer Solutions A buffer solution is a solution of: A weak acid or a weak base and The salt of the weak acid or weak base (Both must be present!) Definition: A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Consider an equal molar mixture of CH3COOH and CH3COONa Add strong acid H+ (aq) + CH3COO- (aq) CH3COOH (aq) CH3COO- combines with H+ ions from strong acid to produce weak acid (weak dissociation) Add strong base OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l) CH3COOH combines with OH- ions from strong base to produce H2O(weak dissociation) 16.3

HCl + CH3COO- CH3COOH + Cl- HCl H+ + Cl- HCl + CH3COO- CH3COOH + Cl- 16.3

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3 (a) HF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is it conjugate acid buffer solution 16.3

NH4+ (aq) H+ (aq) + NH3 (aq) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? moles of NH3 =0.024= (0.30*80)/1000 NH4+ (aq) H+ (aq) + NH3 (aq) moles of NH4+ = 0.029=(0.36*80)/1000 pH = pKa + log [NH3] [NH4+] pH = 9.25 + log [0.30] [0.36] pKa = 9.25 = 9.17 moles of OH- =0.001=(0.05*20)/1000 start (moles) 0.029 0.001 0.024 NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq) end (moles) 0.028 0.0 0.025 final volume = 80.0 mL + 20.0 mL = 100 mL [NH4+] = 0.028 0.10 [NH3] = 0.025 0.10 pH = 9.25 + log [0.25] [0.28] = 9.20 16.3

Chemistry In Action: Maintaining the pH of Blood 16.3

16.6 Solubility Equilibria AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = [Ag+][Cl-] Ksp is the solubility product constant MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2 Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO32-] Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO33-]2 Dissolution of an ionic solid in aqueous solution: Q < Ksp Unsaturated solution No precipitate Q = Ksp Saturated solution No precipitate Q > Ksp Supersaturated solution Precipitate will form 16.6

16.6

Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution. Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution. 16.6

 What is the solubility of silver chloride in g/L ? AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = 1.6 x 10-10 Initial (M) Change (M) Equilibrium (M) 0.00 0.00 Ksp = [Ag+][Cl-] +s +s Ksp = s2 s = Ksp  s s s = 1.3 x 10-5 [Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M 1.3 x 10-5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x Solubility of AgCl = = 1.9 x 10-3 g/L 16.6

16.6

The ions present in solution are Na+, OH-, Ca2+, Cl-. If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form? The ions present in solution are Na+, OH-, Ca2+, Cl-. Only possible precipitate is Ca(OH)2 (solubility rules). Is Q > Ksp for Ca(OH)2? [Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M Q = [Ca2+]0[OH-]0 2 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8 Ksp = [Ca2+][OH-]2 = 8.0 x 10-6 Q < Ksp No precipitate will form 16.6

16.8 The Common Ion Effect and Solubility Consider a solution containing two dissolved substances that share a common ion (AgCl and AgNO3) AgCl (s) Ag+ (aq) + Cl- (aq) AgNO3 (s) Ag+ (aq) + NO3- (aq) Ag+ ions come from two sources: i.e AgCl & AgNO3 The total increase in [Ag+] ion concentration will make the ion product greater than the solubility product: Q = [Ag+]total[Cl-] > Ksp, To reestablish equilibrium, some AgCl will precipitate out of the solution, until the ion product is equal to Ksp  Adding a common ion decreases the solubility of the salt (AgCl) in solution.