Applications of Aqueous Equilibria Chapter 18
What we learn from Chap 18 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.
CHAPTER OUTLINE 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. Acid.Base Titrations A. Strong-Acid.Strong-Base Titrations B. Acid-Base Titrations in Which One Component Is Weak and One Is Strong C. Indicators III. Solubility Equilibria A. Solubility, Precipitation, and Gravimetric Analysis B. To Precitpitate or Not Precipitate C. Acids, Bases, and Solubility IV. Complex-Ion Equilibria A. Introducing the Formation Constant B. Extending the Discussion to EDTA C. The Importance of the Conditional Formation Constant
18.1 Buffers 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.
Buffers Preparation : mix solution of weak acid and conjugate base Choice of system : K a and ratio of HB and B - Effect of adding strong acid or base : ph of a buffer does change slightly on moderate amounts of a strong acid or base Capacity
Preparation Prepared by adding both the weak acid HB and its conjugate base B to water. K a = [H+ ] [B - ] [HB] [H + ] = K a [HB] [B - ] = K a n HB n B - [HB] = n HB, [B ] = n B- B
Preparation example Calculate [H + ] in a solution prepared by adding 0.200 mol of acetic acid and 0.200 mol of sodium acetate to one liter. [H + ] = 1.8 10 5 = 1.8 10 5 M ph = 4.74 0.200 0.200
Choice of system Note that since HB and B must be present in roughly equal amounts, [H + ] of the buffer is roughly equal to K a of the weak acid. To establish a buffer of ph 7, choose a system such as H 2 PO 4, HPO 4 2, where K a = 6.2 10 8.
ph of a Buffer Solution A buffer solution is made from equal amounts of 0.15M proprionic acid (HPro) and 0.20M sodium proprionate (NaPro). The K a of proprionic acid is 1.3 10-5. What is the ph? HPro H + + Pro - K a = 1.3 10-5 Pro - + H 2 O HPro + OH - K b = 7.7 10-10 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 0.15 +x 0.20
ph of a Buffer Solution K a 5 = 1.3 10 = + - H Pro [ HPro] ( x)( 0.20) ( 0.15) x = 9.75 10 ph = 5.01-6
Dissociation of Hpro is suppressed by Pro - : Le Chatelier s Principle Ka H Pr o H + Pr o 0.15 M x x x K a + [ H ][Pr o ] = [ H Pr o] + [ H ] = x= K 0.15 = 1.4 10 a 3 M + ph=2.85 Kb Pr o + H2O H Pr o + OH 0.20 M x x [ OH ]= x = K 0.20 b 5 1.2 10 M = poh=4.91 ph=9.09
Henderson-Hasselbalch Equation A useful equation for solving buffer problems: a K a + HA H A K + [ H ][ A ] = [ HA] + + [ HA] [ H ] = Ka [ A ] [ HA] ph = pka log( ) [ A ] ph = pk a [ conjugate base] + log [ conjugate acid]
ph of a Buffer Solution A buffer solution is made from equal amounts of 0.15M proprionic acid (HPro) and 0.20M sodium proprionate (NaPro). The K a of proprionic acid is 1.3 10-5. What is the ph? ph = pk a [ conjugate base] + log [ conjugate acid] ( -5 ) ( 0.20) ( ) ph = -log 1.3 10 + log 0.15 ph = 5.01
Buffer Action 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)
Buffers
Buffers
Ex1) ph of 0.20M CH 3 COOH + 0.20M NaCH 3 COO a K a + HAc H Ac K + [ H ][ Ac ] = = 1.8 10 [ HAc] + [ HA] [ H ] = Ka [ A ] [ HA] ph = pka log( ) [ A ] = pk = 4.75 a + 5
Ex2) Prep. ph=10 ammoniaammonium buffer a + + 4 3 [ H ][ NH ] 1.0 10 = = = 5.6 10 [ ] 1.8 10 + 14 3 + 5 NH4 + + [ NH4 ] [ H ] = Ka [ NH ] 3 + 4 log( ) + 4 10 9.25 = log( ) 3 K a + NH H NH K ph = pk a [ NH ] [ NH ] 3 [ NH ] [ NH ] + [ NH4 ] = exp( 10 + 9.25) = 0.47 [ NH ] 3 10
Buffer Capacity Buffer Capacity : Moles of strong acid or base without a large change in ph of buffer (within 1.0 ph) 0.200M HCOOH+0.400M HCOONa:100mL + 1.00M HCl 50.0mL + HCOO ( aq) + H ( aq) HCOOH ( aq) K a = 5.6 10 Initial 0.0400 0.0200 Added 0.0500 change -0.0400-0.0400 +0.0400 Equil. 0 0.0100 0.0600 3 + 0.0100mol [ H ] = = 0.0667 M ph = 1.18 0.1500L ** Buffer capacity : 0.200mol strong base, 0.0400mol strong acid
Tris buffer
re Strong acids or Bases Buffers? Aspirin MgCO 3 (antacid: increase stomach ph) Tetracycline An antibiotic drug
18.2 Acid-Base Titrations 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.
Acid-Base Titrations Strong acid-strong base : at equivalence point, ph = 7 Weak acid-strong base : at equivalence point, ph > 7 Strong acid-weak base : at equivalence point, ph < 7 Choice of indicator
Strong Acid-Strong Base Titration
Strong Acid-Strong Base Titration 0.1000M HCl 50.00 ml + 0.2000M NaOH 5mL
Strong Acid-Strong Base Titration 0.1000M HCl 50.00 ml + 0.2000M NaOH 12.50mL
Strong Acid-Strong Base Titration 0.1000M HCl 50.00 ml + 0.2000M NaOH 24.00mL
Strong Acid-Strong Base Titration 0.1000M HCl 50.00 ml + 0.2000M NaOH 25.00mL
Strong Acid-Strong Base Titration 0.1000M HCl 50.00 ml + 0.2000M NaOH 40.00mL
Strong-Weak Titrations 0.1000M Hac 50.00mL + 0.2000M NaOH
Strong-Weak Titrations 0.1000M Hac 50.00mL + 0.2000M NaOH 5.00mL
Strong-Weak Titrations 0.1000M Hac 50.00mL + 0.2000M NaOH 12.500mL
Strong-Weak Titrations 0.1000M Hac 50.00mL + 0.2000M NaOH 24.00mL
Strong-Weak Titrations 0.1000M Hac 50.00mL + 0.2000M NaOH 25.00mL
Strong-Weak Titrations 0.1000M Hac 50.00mL + 0.2000M NaOH 40.00mL
K a vs Titration Curves
Titration Curves
Titration of HCl with NaOH
Titration of HC 2 H 3 O 2 with NaOH
Titration of NH 3 and HCl
Indicators 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.
Indicators Derived from weak acid HIn [HIn] = [In ] [H + ] K a of HIn Color depends upon the ratio of concentrations of HIn, In, which depends on [H + ] and K a of indicator. For bromthymol blue, K a = 10 7, HIn is yellow, In is blue ph < 6; [HIn] > 10[In ]; solution is yellow ph > 8; [HIn] < 0.1[In ]; solution is blue ph = 7; [HIn] = [In ]; solution is green
Acid-Base Indicators Color of HIn Color of In - K a ph at End Point Methyl Red Red Yellow 1 10-5 5 Bromothymol Blue Yellow Blue 1 10-7 7 Phenolphthale in Colorle ss Pink 1 10-9 9
Indicators
18.3 Solubility Equilibria 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 2-3 ]
K sp Values
Sample Problem 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 sp = [Pb 2+ ][I - ] 2 Let s be the amount that dissolves. PbI 2 Pb 2+ + 2I - I C -s +s +2s E s 2s
Sample Problem (cont) sp 2 2+ - = K Pb I 8 2 3 ( )( ) 1.4 10 = s 2s = 4s s = 3 s = 1.5 1.4 10 4 8-3 10 M
Sample Problem 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 sp ( )( ) ( -5 ) K = 1.3 10 K = 1.7 10 sp -10 2
Deviation from the simple calculation Side reactions that affect reactions of interest Hydrolysis of carbonate ions CO ( aq) + H O( l) HCO ( aq) + OH ( aq) K = 5.6 10 2 3 3 2 3 a 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) H2O( l) Atmospheric CO Molecular-Level Process Ion pair formation K sp 2+ 2 CaSO4( s) Ca ( aq) + SO4 ( aq) Ion activity Thermodynamic measurements
Solubility I C E K sp ( ) + ( ) AgBr ( ) 4 s Ag aq + Br aq K sp = 5.0 10-0 0 -s +s +s - +s +s -13 + K =[ Ag ][ Br ] sp 5.0 10 = s -13 2 s( molar solubility) = 7.1 10-7 M
Predicting Precipitation 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.
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 - ] = 1.2 10-3 M. Will a precipitate form? The K sp = 1.4 10-8 for PbI 2. PbI 2 Pb 2+ + 2I - K sp = [Pb 2+ ][I - ] 2 Q sp = (2.0 10-4 )(1.2 10-3 ) 2 Q sp = 2.9 10-10 Q sp < K sp so precipitate will not form
Acid, Bases & Solubility Sedimentation In waste water treatment + Al ( aq) + 3 OH ( aq) Al( OH ) ( s) K sp = 2 10 3-32 3 Coagulation Increasing the acidity of waterway: Increase metal concentrations
18.4 Complex-Ion Equilibria 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
K f 2+ 2+ Zn ( aq) + NH3( aq) Zn( NH3) ( aq) K f = 190 K f : formation constant or stability constant
Chelate Chelate : Ligand that form multiple bonds (coordinate covalent bonds) Monodentate : NH 3 Tetradentate : EDTA Hexadentate : EDTA
Metal-EDTA Complexes
Industrial Chelating Agents
Conditional formation constants ( ) + ( ) ( ) 2+ 4 2 Ca aq EDTA aq CaEDTA aq 2 + 2+ 3 CaEDTA aq H aq Ca aq HEDTA aq ' K = K fα α Ca EDTA ( ) + ( ) ( ) + ( ) 2 + 4 K f