Applications of Aqueous Equilibria Chapter 18 What we learn from Chap 18 18.2 This chapter is the third in the three-chapter sequence about equilibrium, this one building upon the core principles raised in the previous two chapters to expand and deepen the discussion to a broader range of applications. The chapter-wide case study concerns quality control, specifically the procedure for determining water hardness. We look at why each step is useful; first, EDTA, and second, an ammonia buffer. Next, we discuss acid-base titrations in general, then solubility equilibria, followed by complexometric equilibria. The result is a fairly full picture of how equilibria can be applied not only to an analytical procedure, but to many other chemical processes. The introduction considers quality control and some of the wide range of titrations we use to do chemical analysis. 1 2
CHAPTER OUTLINE 18.3 I. I. Buffers and the Common-Ion Effect A. The Impact of c on the Equilibria in the Buffer B. Buffer Preparation C. The Henderson-Hasselbalch Equation: We Proceed, but with Caution D. Buffer Capacity E. Food for Thought: Are Strong Acids and Bases Buffers? II. II. Acid.Base Titrations A. Strong-Acid.Strong-Base g Titrations B. Acid-Base Titrations in Which One Component Is Weak and One Is Strong C. Indicators III. III. Solubility Equilibria A. Solubility, Precipitation, and Gravimetric Analysis B. To Precitpitate or Not Precipitate C. Acids, Bases, and Solubility IV. IV. Complex-Ion Equilibria A. Introducing the Formation Constant B. Extending the Discussion to EDTA C. The Importance of the Conditional Formation Constant 1 3 18.1 Buffers 18.4 Buffers are composed of a weak acid and the salt of the acid (conjugate base) or weak base and the salt of the base (conjugate acid). Buffers resist changes in ph, if small amount of acid or base are added. 1 4
ph of a Buffer Solution 18.5 A buffer solution is made from equal amounts of 0.15M propionic acid (Hpro, CH 3 CH 2 CO 2 H) and 0.20M sodium propionate (NaPro). The K a of propionic acid is 1.3 10-5. What is the ph? HPro H + + Pro - K a = 1.3 10-5 Pro - +H - = 77 10-10 2 O HPro + OH K b 7.7 10 Since the K a >K b the buffer will be acidic. HPro H + + Pro - I 0.15 0 0.20 C -x +x +x E 0.15-x +x 0.20+x A 015 0.15 +x 020 0.20 1 5 ph of a Buffer Solution 18.6 1.3 10 K a + - H Pro HPro 5 x 0.20 0.15 x = 9.7510 ph = 5.01-6 1 6
Dissociation of Hpro is suppressed by Pro - : Le Chatelier s Principle K a H Pr o H Pr o 015 0.15 M x x x K a [ H ][Pr o ] [ H Pr o ] [ H ] x K 0.15 a 3 1.410 M ph=2.85 18.7 K b Pr o H2O H Pr o OH 020 0.20 M x x [ OH ]= x K 0.20 b poh=4.91 H 9 09 5 1.210 M ph=9.09 1 7 Henderson-Hasselbalch Equation 18.8 K a HA H A K [ H ][ A ] [ HA] a A useful equation for solving buffer problems: [ ] ph = pk a conjugate base log conjugate acid [ HA] [ H ] Ka [ A ] [ HA ] ph pk a log( ) [ A ] 1 8
ph of a Buffer Solution 18.9 A buffer solution o is made from 0.15M propionic o acid d( (HPro) and 0.20M sodium propionate (NaPro). The K a of propionic acid is 1.3 10-5. What is the ph? conjugate base ph = pk a log conjugate acid ph = -log1.3 10-5 log 0.20 015 0.15 ph = 5.01 1 9 Buffer Action 18.10 Buffer solutions have the ability to resist changes in ph upon the addition of small amounts of either acid or base. Consider an equal molar mixture of HPro and NaPro: HPro H + + Pro - K a = 1.3 10-5 Pro - + H 2 O HPro + OH - K b = 7.7 10-10 Adding strong acid: H + (aq) + Pro - (aq) HPro(aq) Adding strong base: OH - (aq) + HPro(aq) Pro - (aq) + H 2 O(l) 1 10
Buffers 18.11 1 11 Buffers 18.12 1 12
Ex1) ph of 0.20M CH 3 COOH + 0.20M NaCH 3 COO 18.13 K a HAc H Ac K a [ H ][ Ac ] 1.810 [ HAc] [ HA] [ H ] Ka [ A ] ph pk a pk a 4.75 [ HA] log( ) [ A ] 5 1 13 Ex2) Prep. ph=10 ammonia-ammonium ammonium buffer 18.14 K a NH H NH K 4 3 [ H ][ NH ] 1.010 14 3 a 5 [ NH4 ] 1.810 [ NH ] [ ] 4 H Ka NH 3 [ ] ph pk a [ NH ] [ NH ] 4 log( ) 4 10 9.25 log( ) 3 3 [ NH ] [ NH ] [ NH 4 ] exp( 10 9.25) 0.47 [ NH ] 3 5.6 10 10 1 14
Buffer Capacity 18.15 Buffer Capacity : Moles of strong acid or base without a large change in ph of buffer (within ±1.0 0pH) 0.200M HCOOH+0.400M HCOONa:100mL + 1.00M HCl 10.0mL Initial conc. [HCOOH] initial [HCOONa] Final conc final HCOOH H + + HCOO - 1 15 Buffer Capacity 18.16 0.200M HCOOH+0.400M HCOONa:100mL + 1.00M HCl 50.0mL HCOO ( aq) H ( aq) HCOOH ( aq) K a 5.610 Initial 0.0400 0.0200 Added 0.0500 change -0.0400-0.0400 +0.0400 Equil. 0 0.0100 0.0600 0.0100mol [ H ] 0.0667 M ph 1.18 0.1500L 3 K a =1.8x10-4 ** Buffer capacity : 0.0200mol strong base, 0.0400mol strong acid 1 16
Buffer Capacity (cont.) 18.17 Useful ph range : pk a ±1 ph unit 1 17 Tris buffer 18.18 AH A - 1 18
Are Strong acids or Bases Buffers? 18.19 Aspirin MgCO 3 (antacid: increase stomach ph) Tetracycline An antibiotic drug 1 19 18.2 Acid-Base Titrations 18.20 A solution of known concentration (titrant) is added to a solution of unknown (sample, analyte)concentration until the reaction is complete. The equivalence point is the point at which the reaction is complete. 1 20
Strong Acid-Strong Base Titration 18.21 0.1000M HCl 50.00 ml + 0.2000M NaOH 0 ml [H + ]=0.100 M ph=1.0 1 21 Strong Acid-Strong Base Titration 18.22 0.1000M HCl 50.00 ml + 0.2000M NaOH 5mL [H + ]=0.0727 M ph=1.14 1 22
Strong Acid-Strong Base Titration 18.23 0.1000M HCl 50.00 ml + 0.2000M NaOH 12.50mL [H + ]=0.04 M ph=1.40 1 23 Strong Acid-Strong Base Titration 18.24 0.1000M HCl 50.00 ml + 0.2000M NaOH 24.00mL [H + ]=2.70x10-3 M ph=2.57 1 24
Strong Acid-Strong Base Titration 18.25 0.1000M HCl 50.00 ml + 0.2000M NaOH 25.00mL [H + ]=1.0x10-7 M ph=7.00 1 25 Strong Acid-Strong Base Titration 18.26 0.1000M HCl 50.00 ml + 0.2000M NaOH 40.00mL [OH - ]=0.0333 M ph=12.52 1 26
Strong-Weak Titrations 18.27 0.1000M HAc 50.00mL + 0.2000M NaOH [H + ]=1.30x10-3 M ph=2.87 1 27 Strong-Weak Titrations 18.28 0.1000M HAc 50.00mL + 0.2000M NaOH 5.00mL [H + ]=7.20x10-5 M ph=4.14 1 28
Strong-Weak Titrations 18.29 0.1000M HAc 50.00mL + 0.2000M NaOH 12.500mL [H + ]=1.8x10-5 M ph=4.74 1 29 Strong-Weak Titrations 18.30 0.1000M HAc 50.00mL + 0.2000M NaOH 24.00mL [H + ]=7.50x10-7 M ph=6.12 1 30
Strong-Weak Titrations 18.31 0.1000M HAc 50.00mL + 0.2000M NaOH 25.00mL [OH - ]=6.1x10-6 M ph=8.79 1 31 Strong-Weak Titrations 18.32 0.1000M HAc 50.00mL + 0.2000M NaOH 40.00mL [OH - ]=0.0333 M ph=12.52 1 32
K a vs Titration Curves 18.33 1 33 Titration Curves 18.34 1 34
Acid-Base Titration 18.35 1 35 Indicators 18.36 Indicators are conjugate acid pairs of organic compounds that change color in acid or base. In a titration, an indicator is chosen with a pk a close to the ph at the equivalence point. (EP pk a ±1.0) The visual color change indicates the endpoint of the titration. 1 36
Indicators 18.37 1 37 18.3 Solubility Equilibria 18.38 An equilibrium between a sparingly soluble compound and its ions. CaCO 3 (s) Ca 2+ (aq) + CO 2-3 (aq) The solubility product constant is: K sp = [Ca 2+ ][CO 3 2- ] 1 38
18.39 1 39 Sample Problem 18.40 The K sp of PbI 2 is 1.4 10-8. What is the molar solubility of this compound? PbI 2 Pb 2+ + 2I - K =[Pb 2+ - 2 sp ][I ] Let s be the amount that dissolves. PbI 2 Pb 2+ + 2I - I C -s +s +2s E s 2s 1 40
Sample Problem (cont) 18.41 2+ - Ksp Pb I 2 1.4 10 s 2s 4s 8 3 s = 1.4 10 3 4 s = 1.5 2 8-3 10 M 1 41 Sample Problem 18.42 Calculate the solubility product constant (K sp ) of AgCl which has a molar solubility of 1.3 10-5 M. AgCl(s) Ag + (aq) + Cl - (aq) s s + - K sp Ag Cl K s s sp K = 1.310 sp K = 1.710 sp -5-10 2 1 42
Deviation from the simple calculation 18.43 Side reactions that affect reactions of interest Hydrolysis of carbonate ions K b1 =K w /K a2 =1.78x10-4 CO ( aq) H O( l) HCO ( aq) OH ( aq) K a 5.610 2 3 3 2 3 K K 2 2 2 3 3 CO aq H O l H CO aq H aq HCO aq a1 ( ) ( ) ( ) ( ) ( ) 1/ Kw H ( aq ) OH ( aq ) HOl ( ) Atmospheric p CO 2 Molecular-Level Process K sp 2 2 2 CaSO4( s) Ca ( aq) SO4 ( aq) Ion pair formation : Ca 2+ & SO 2-4 can exist as ion pair Ion activity: active ion concentration Thermodynamic properties: enthalpy, entropy changes 1 43 Solubility 18.44 K sp AB AgBr4 ( s) Ag ( aq ) Br ( aq ) K sp 5.0 10 I - 0 0 C -s +s +s E - +s +s -13 K =[ Ag ][ Br ] sp 5.010 s -13 2 s ( molar solubility ) 7.110-7 M If we add little sodium bromide, what will happen? 1 44
Predicting Precipitation 18.45 Consider: AgCl(s) Ag + (aq) + Cl - (aq) Let Q sp = [Ag + ] o [Cl - ] o If Q sp > K sp then a precipitate will form. If Q sp < K sp then a precipitate will not form. 1 45 Sample Problem A solution is mixed from one containing lead ions and another containing iodide ions. The solution contains [Pb 2+ ] = 2.0 10-4 M and [I - ] =12 10 1.2 10-3 M. Will a precipitate form? The K sp = 1.4 10-8 for PbI 2. PbI 2+ + - 2 Pb 2I K sp = [Pb 2+ ][I - ] 2 18.46 Q sp = (2.0 10-4 )(1.2 10-3 ) 2 Q sp = 2.9 10-10 p Q sp < K sp so precipitate will not form 1 46
Acid, Bases & Solubility 18.47 Sedimentation In waste water treatment Al ( aq) 3 OH ( aq) Al( OH ) ( s) K sp 210 3-32 3 Coagulation Increasing the acidity of waterway: Increase metal concentrations 1 47 18.4 Complex-Ion Equilibria 18.48 Complex ions are formed by a metal ion and ligands. Ligands are Lewis bases such as Cl -, CN -, NH 3, H 2 O. Complexes may be cations or anions: Ag(NH 3 ) + 2 Fe(CN) 3-6 1 48
18.49 K f 2 2 Zn ( aq) NH3( aq) Zn( NH3) ( aq) K f 190 K f : formation constant or stability constant 1 49 18.50 1 50
Chelate 18.51 Chelate : Ligand that form multiple bonds (coordinate covalent bonds) Monodentate : NH 3 Tetradentate : EDTA Hexadentate : EDTA 1 51 Metal-EDTA Complexes 18.52 1 52
Industrial Chelating Agents 18.53 1 53 18.54 1 54
Conditional formation constants 18.55 ( ) ( ) ( ) 2 4 2 Ca aq EDTA aq CaEDTA aq 2 4 K f 2 2 3 CaEDTA aq H aq Ca aq HEDTA aq ' K K f Ca EDTA ( ) ( ) ( ) ( ) 1 55 Problems 18.56 8, 18,26,36,44,58,70,81,82 1 56