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

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion

Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] Also pK a = -log K a Henderson-Hasselbalch equation 16.2 pH = pK a + log [conjugate base] [acid]

What is the pH of a solution containing 0.30 M HCOOH (weak acid) and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x x x Common ion effect 0.30 – x  x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = log [0.52] [0.30] = Mixture of weak acid and conjugate base! HCOOH pK a = 3.77

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! mole of HCl is added to 1 L of pure water, the PH changes from its initial 7 to 2. (change of 5 PH units) mole of HCl is added to a solution containing both 0.1 M acetic acid (HC 2 H 3 O 2 ) and 0.1 M sodium acetate (NaC 2 H 3 O 2 ), the PH changes from initial value of 4.74 to (change of 0.08 PH units) Buffered solutions are important to living organisms whose cells can survive only in a very narrow PH range. Goldfish need buffering aqurium water at an appropriate PH. Human’s blood need to maintain the PH between 7.35 and 7.45.

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) Add strong base OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa

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 its 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) pH = log [0.25] [0.28] [NH 4 + ] = final volume = 80.0 mL mL = 100 mL [NH 3 ] = pK a = 9.25 end (moles)

Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4.7

Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l) 16.4

Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7): 16.4

Exactly 100 mL of 0.10 M HNO 2 are titrated with a 100 mL 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx [NO 2 - ] = = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x x = 2.2 x – x  0.05x  1.05 x = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): 16.4 H + (aq) + NH 3 (aq) NH 4 + (aq)

Acid-Base Indicators HIn (aq) H + (aq) + In - (aq)  10 [HIn] [In - ] Color of acid (HIn) predominates  0.1 [HIn] [In - ] Color of conjugate base (In - ) predominates 16.5

pH 16.5

Which indicator(s) would you use for a titration of HNO 2 with KOH ? Weak acid titrated with strong base. At equivalence point, pH > 7 Use cresol red or phenolphthalein 16.5

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 Q = [Ag + ] 0 [Cl - ] 0

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 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

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 = 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 Predicting precipitation reaction

The Common Ion Effect and Solubility The presence of a common ion decreases the solubility of the salt. What is the molar solubility of AgBr in (a) pure water and (b) M NaBr? AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x s 2 = K sp s = 8.8 x NaBr (s) Na + (aq) + Br - (aq) [Br - ] = M AgBr (s) Ag + (aq) + Br - (aq) [Ag + ] = s [Br - ] = s  K sp = x s s = 7.7 x

pH and Solubility The presence of a common ion decreases the solubility. Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutions Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 1.2 x K sp = (s)(2s) 2 = 4s 3 4s 3 = 1.2 x s = 1.4 x M [OH - ] = 2s = 2.8 x M pOH = 3.55 pH = At pH less than Lower [OH - ] OH - (aq) + H + (aq) H 2 O (l) Increase solubility of Mg(OH) 2 At pH greater than Raise [OH - ] Decrease solubility of Mg(OH)

Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq) + 4Cl - (aq) CoCl 4 (aq) 2- K f = [CoCl 4 ] [Co 2+ ][Cl - ] 4 2- The formation constant or stability constant (K f ) is the equilibrium constant for the complex ion formation. Co(H 2 O) 6 2+ CoCl KfKf stability of complex

16.10