Acid-Base Equilibria and Solubility Equilibria Chapter 16 Semester 2/2014 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 16 Semester 2 / 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: 1.A weak acid or a weak base and 2.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 Add strong acid H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) CH 3 COO- combines with H + ions from strong acid to produce weak acid (weak dissociation) Add strong base OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COOH combines with OH - ions from strong base to produce H 2 O(weak dissociation) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa

HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (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) CO 3 2- is a weak base and HCO 3 - is it conjugate acid buffer solution 16.3

= 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (moles) end (moles) pH = log [0.25] [0.28] [NH 4 + ] = final volume = 80.0 mL mL = 100 mL [NH 3 ] = moles of NH 4 + = 0.029=(0.36*80)/1000 moles of NH 3 =0.024= (0.30*80)/1000 moles of OH - =0.001=(0.05*20)/1000

Chemistry In Action: Maintaining the pH of Blood 16.3

16.6 Solubility Equilibria 16.6 AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO (aq) K sp = [Ag + ] 2 [CO ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO (aq) K sp = [Ca 2+ ] 3 [PO ] 2 Dissolution of an ionic solid in aqueous solution: Q = K sp Saturated solution Q < K sp Unsaturated solution No precipitate Q > K sp Supersaturated solution Precipitate will form No precipitate

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) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) s+s +s+s ss K sp = s 2 s = K sp  s = 1.3 x [Ag + ] = 1.3 x M [Cl - ] = 1.3 x M Solubility of AgCl = 1.3 x mol AgCl 1 L soln g AgCl 1 mol AgCl x = 1.9 x g/L K sp = 1.6 x

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

16.8 The Common Ion Effect and Solubility Consider a solution containing two dissolved substances that share a common ion (AgCl and AgNO 3 ) AgCl (s) Ag + (aq) + Cl - (aq) AgNO 3 (s) Ag + (aq) + NO 3 - (aq) Ag + ions come from two sources: i.e AgCl & AgNO 3 The total increase in [Ag+] ion concentration will make the ion product greater than the solubility product: Q = [Ag + ] total [Cl - ] > K sp, To reestablish equilibrium, some AgCl will precipitate out of the solution, until the ion product is equal to K sp  Adding a common ion decreases the solubility of the salt (AgCl) in solution.

Calculate the solubility of silver chloride (in g/L) in a 6.5 x M silver nitrate solution.

17 Let s be the molar solubility of AgCl in AgNO 3 solution. We summarize the changes in concentrations as follows: AgCl(s) Ag + (aq) +Cl - (aq) Initial (M):6.5 x Change (M):-s-s+s+s+s+s Equilibrium (M):(6.5 x s)s Step 3: K sp = [Ag + ][Cl - ] 1.6 x = (6.5 x s)(s)

18 Because AgCl is quite insoluble and the presence of Ag + ions from AgNO 3 further lowers the solubility of AgCl, s must be very small compared with 6.5 x Therefore, applying the approximation 6.5 x s ≈ 6.5 x 10 -3, we obtain 1.6 x = (6.5 x )s s = 2.5 x M Step 4: At equilibrium [Ag + ] = (6.5 x x ) M ≈ 6.5 x M [Cl + ] = 2.5 x M 16.12

19 and so our approximation was justified in step 3. Because all the Cl - ions must come from AgCl, the amount of AgCl dissolved in AgNO 3 solution also is 2.5 x M. Then, knowing the molar mass of AgCl (143.4 g), we can calculate the solubility of AgCl as follows: 16.12