PowerPoint Lecture Presentation by J. David Robertson University of Missouri Acid-Base Equilibria and Solubility Equilibria Chapter 17 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) CH 3 COOH (aq) Na + (aq) + CH 3 COO - (aq) H + (aq) + CH 3 COO - (aq) common ion Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 17.2 Consider mixture of salt NaA and weak acid HA. NaA (s) HA (aq) Na + (aq) + A - (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] What is the ph of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! [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] pk a = -log K a Henderson-Hasselbalch equation ph = pk a + log [conjugate base] [acid] 17.2 Initial (M) Change (M) Equilibrium (M) Common ion effect 0.30 x! 0.30 0.52 + x! 0.52 HCOOH pk a = 3.77 HCOOH (aq) 0.30 0.00 -x +x 0.30 - x H + (aq) + HCOO - (aq) ph = pk a + log [HCOO- ] [HCOOH] x 0.52 +x 0.52 + x ph = 3.77 + log [0.52] [0.30] = 4.01 17.2
A buffer solution is a solution of: 1.! A weak acid or a weak base and HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl - 2.! The salt of the weak acid or weak base Both must be present! 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 CH 3 COOH and CH 3 COONa Add strong acid H + (aq) + CH 3 COO - (aq) Add strong base OH - (aq) + CH 3 COOH (aq) CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) KF 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 is a weak base and HCO 3 - is it conjugate acid buffer solution 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 0.050 M NaOH to 80.0 ml of the buffer solution? NH 4 + (aq) ph = pk a + log [NH 3 ] [NH 4+ ] start (moles) end (moles) H + (aq) + NH 3 (aq) pk a = 9.25 ph = 9.25 + log [0.30] [0.36] = 9.17 0.029 0.001 0.024 NH + 4 (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) 0.028 0.0 0.025 final volume = 80.0 ml + 20.0 ml = 100 ml [NH 4+ ] = 0.028 0.10 [NH 3 ] = 0.025 0.10 ph = 9.25 + log [0.25] [0.28] = 9.20
Chemistry In Action: Maintaining the ph of Blood 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) Strong Acid-Strong Base Titrations Weak Acid-Strong Base Titrations NaOH (aq) + HCl (aq) OH - (aq) + H + (aq) H 2 O (l) H 2 O (l) + NaCl (aq) 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) At equivalence point (ph > 7): CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) 17.4 17.4
Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) H + (aq) + NH 3 (aq) NH 4 Cl (aq) At equivalence point (ph < 7): NH + 4 (aq) + H 2 O (l) NH 3 (aq) + H + (aq) Exactly 100 ml of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the ph at the equivalence point? start (moles) end (moles) 0.01 0.01 HNO 2 (aq) + OH - (aq) NO - 2 (aq) + H 2 O (l) 0.0 0.0 0.01 Final volume = 200 ml NO 2 - (aq) + H 2 O (l) [NO 2 - ] = 0.01 0.200 = 0.05 M OH - (aq) + HNO 2 (aq) 17.4 Initial (M) Change (M) Equilibrium (M) K b = [OH- ][HNO 2 ] = [NO ] 0.05 0.00 -x +x 0.05 - x x 2 0.05-x = 2.2 x 10-11 0.05 x! 0.05 x! 1.05 x 10-6 = [OH - ] x poh = 5.98 0.00 +x x ph = 14 poh = 8.02 Acid-Base Indicators HIn (aq) [HIn] [In - ] [HIn] [In - ] " 10 # 10 H + (aq) + In - (aq) Color of acid (HIn) predominates Color of conjugate base (In - ) predominates 17.5 ph 17.5
The titration curve of a strong acid with a strong base. Which indicator(s) would you use for a titration of HNO 2 with KOH? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, ph > 7 Use cresol red or phenolphthalein 17.5 17.5 Solubility Equilibria 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 3 (aq) K sp = [Ag + ] 2 [CO 3 ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 3-4 (aq) K sp = [Ca 2+ ] 3 [PO 3-3 ] 2 Dissolution of an ionic solid in aqueous solution: Q < K sp Unsaturated solution No precipitate Q = K sp Saturated solution Q > K sp Supersaturated solution Precipitate will form
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. Initial (M) What is the solubility of silver chloride in g/l? Change (M) Equilibrium (M) AgCl (s) [Ag + ] = 1.3 x 10-5 M Ag + (aq) + Cl - (aq) 0.00 0.00 +s s [Cl - ] = 1.3 x 10-5 M +s s K sp = 1.6 x 10-10 K sp = [Ag + ][Cl - ] K sp = s 2 s = $ K sp s = 1.3 x 10-5 Solubility of AgCl = 1.3 x 10-5 mol AgCl 1 L soln x 143.35 g AgCl 1 mol AgCl = 1.9 x 10-3 g/l If 2.00 ml of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? 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 = 0.100 M [OH - ] 0 = 4.0 x 10-4 M Q = [Ca 2+ ] 0 [OH - 2 ] 0 = 0.10 x (4.0 x 10-4 ) 2 = 1.6 x 10-8 K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10-6 Q < K sp No precipitate will form
AgCl (s) What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br - and Cl - at a concentration of 0.02 M? AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x 10-13 [Ag + ] = [Ag + ] = Ag + (aq) + Cl - (aq) K sp = [Ag + ][Br - ] K sp 7.7 x 10 = -13 = 3.9 x 10 [Br - ] 0.020-11 M K sp = 1.6 x 10-10 K sp = [Ag + ][Cl - ] K sp 1.6 x 10 = -10 = 8.0 x 10 [Br - ] 0.020-9 M 3.9 x 10-11 M < [Ag + ] < 8.0 x 10-9 M 17.7 AgBr (s) 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) 0.0010 M NaBr? K sp = 7.7 x 10-13 s 2 = K sp s = 8.8 x 10-7 Ag + (aq) + Br - (aq) NaBr (s) [Br - ] = 0.0010 M AgBr (s) [Ag + ] = s Na + (aq) + Br - (aq) Ag + (aq) + Br - (aq) [Br - ] = 0.0010 + s! 0.0010 K sp = 0.0010 x s s = 7.7 x 10-10 17.8!!! 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) K sp = [Mg 2+ ][OH - ] 2 = 1.2 x 10-11 K sp = (s)(2s) 2 = 4s 3 4s 3 = 1.2 x 10-11 s = 1.4 x 10-4 M [OH - ] = 2s = 2.8 x 10-4 M poh = 3.55 ph = 10.45 remove add Mg 2+ (aq) + 2OH - (aq) At ph less than 10.45 Lower [OH - ] OH - (aq) + H + (aq) H 2 O (l) Increase solubility of Mg(OH) 2 At ph greater than 10.45 Raise [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) The formation constant or stability constant (K f ) is the equilibrium constant for the complex ion formation. 2+ Co(H 2 O) 6 CoCl 4 K f = [CoCl 4 ] [Co 2+ ][Cl - ] 4 K f stability of complex 17.9 Decrease solubility of Mg(OH) 2 17.10
17.10 17.11 Flame Test for Cations Qualitative Analysis of Cations lithium sodium potassium copper 17.11 17.11
Chemistry In Action: How an Eggshell is Formed Ca 2+ (aq) + CO 3 (aq) CaCO 3 (s) CO 2 (g) + H 2 O (l) carbonic anhydrase H 2 CO 3 (aq) H 2 CO 3 (aq) HCO 3 - (aq) H + (aq) + HCO 3 - (aq) H + (aq) + CO 3 (aq)