2. FUNDAMENTALS OF ELECTROCHEMISTRY ELECTROCHEMISTRY IS THE BRANCH OF CHEMISTRY THAT DEALS WITH THE RELATIONSHIP BETWEEN ELECTRICITY AND CHEMICAL REACTIONS.  PRODUCTION OF ELECTRICAL CURRENT BY CHEMICAL REACTIONS (Batteries, Fuel cells)  CHEMICAL CHANGES PRODUCED BY ELECTRIC CURRENT (Electrolysis, Electroplating and refining of metals)  ELECTRIC CURRENT = Transfer of charge per unit time  BIOELECTROCHEMISTRY – study of electron transfer in biological regulations of organisms REDOX REACTIONS

3. BASIC CONCEPTS  A redox reaction involves transfer of electrons from one species to another.  Oxidation: loss electrons - Reducing agent  Reduction : gain electrons - Oxidizing agent

4.  Electric charge (q) = n x F = Coulombs F – Faraday constant = 96 485.3415 C/mol of e -  Electric current (I) – is the quantity of charge flowing each second through the circuit. Unit – Amperes (A) Eg. Calculate the mass of aluminum produced in 1 hour by electrolysis of molten AlCl 3 if the electrical current is 20 A. (Answer: 33 g)

5.  Potential difference (E) between 2 points is the work needed when moving an electric charge from 1 point to another, unit is Volt  Work = E. q OR J = CV  1 Joule is the energy gained or lost when 1 coulomb of charge moves between points whose potentials differ by 1V  The greater the potential difference between 2 points the stronger will be the “push” on a charged particle travelling between those points. A 12V battery pushes e - 8X harder than a 1.5V dry cell

6.  The free energy of change, ∆G, represents the maximum electrical work that can be done by the reaction on its surroundings.  Work done on surroundings = ∆G = - work = -E.q  ∆G = - nFE (-ve ∆G = spontaneous rxn)  Ohm’s Law, I = E/R  Unit of resistance is ohms or Greek symbol Ω (omega)  Power, P = work/time = E.q = E. q = E. I = I 2. R s s

7. BALANCING REDOX REACTIONS  In an acidic medium  In a basic medium Split the reaction into 2 components or half reactions by identifying which species are oxidised and which ones are reduced. Add electrons appropriately to match the change in oxidation state of each element  Introduce H 2 O to balance the oxygen atoms  Introduce H + to balance the charges as well as the hydrogen atoms  Multiply the half reactions so as to have an equal no. of e - and add the half reactions Whichever case, balance for the acid (H + ) and then for basic media add OH - to the side where there is H + to eliminate it as a H 2 O molecule (this is a 1:1 reaction)

8. STANDARD POTENTIALS The voltmeter tells how much work is done by e - flowing from one side to the other, +ve V means e - flow into negative terminal LHS – negative terminal Reference electrode Line notation for cell RHS connected to the positive terminal

9.  STANDARD REDUCTION POTENTIAL (E o ) for each half cell is measured or setup by the above experiment.  ‘Standard’ means the activities ( A ) of all species are unity.  The half reaction of interest is : 2Ag + + 2e - 2Ag(s)  A Ag+ = 1 by definition activity of Ag (s) = unity  Standard Hydrogen Electrode (SHE), consists of a catalytic Pt surface in contact with an acid solution H + (aq,1M), A H+ =1  By convention the LHS electrode (Pt) is attached to the negative terminal of the potentiometer  equilibrium rxn at SHE : 2H + (aq, A = 1) + 2e - H 2 (g, A = 1)  half reaction – always written as reduction reactions

10. BY INTERNATIONAL AGREEMENT THE SHE IS ARBITARILY ASSIGNED E o = 0.00V at 25 o C.  E 0 cell = + 0.799V  E 0 cell = E o red (cathode) – E o red (anode) or  E 0 cell = RHS electrode potential – LHS electrode potential E 0 cell = E 0 red (reduction process) – E 0 red (oxidation process)  A positive E 0 = spontaneous process  A negative E 0 = non spontaneous process  Sketch the cell construction : SHE ll Cd 2+ (aq, A=1) l Cd(s) The half reaction with a more positive E 0 is more reduced and that with a less positive E 0 is less reduced or more oxidised

12. NERNST EQUATION Le Chatelier`s principle tells us that conc. of reactants or products drives rxn to the right or left respectively. The net driving force is expressed by the NERNST EQUATION For the half reaction : a A + ne - ↔ b B Nernst Equation: for the half cell potential, E Q = Reaction quotient a A b B o A A nF RT EEln 

13. where E o = standard reduction potential ( A A = A B = 1) A i = activity of species i R = gas constant ( 8.314 J/K.mol) T = Temperature (K) N = number of electrons in the half reaction F = Faraday constant (9.6485 x 10 4 C/mol) [concentration in K c can be replaced by activities to account for ionic strength ] A c = [C] γ c γ = activity co efficient – measures the deviation from ideality, γ = 1 behavior is ideal, low ionic strength

14. Pure solids and liquids are omitted from Q, because their activities are (close to) unity. Concentration mol/L and pressure in bars When all activities are unity Q = 1, ln Q = 0 then E = E o Log form of the Nernst equation, at 25 o C T = 298K

15. Find the voltage for the cell if the right half cell contains 0.50M AgNO 3 (aq) and the left half cell 0.010M Cd(NO 3 ) 2 (aq). NERNST EQUATION FOR A COMPLETE CELL E = E + - E -

16. STEP 1 Right electrode : 2 Ag + + 2e - ↔ 2Ag(s) E o = 0.799V Left electrode : Cd 2+ + 2e - ↔ Cd(s) E o = - 0.402V STEP 2 Nernst equation for the right electrode = 0.781V STEP 3 Nernst equation for the left electrode = - 0.461V STEP 4 Cell Voltage E = E + - E - = 0.781 - (-0.461) = + 1.242V STEP 5 Net cell reaction: Eqn Right electrode – Eqn Left electrode Cd (s) + 2 Ag + ↔ Cd 2+ + 2 Ag (s)

17. E o and the Equilibrium Constant A galvanic cell produces electricity because the cell is not at equilibrium. Relating E to the reaction quotient Q : Consider: Right electrode: aA + ne - ↔ cC E o + Left electrode: dD + ne - ↔ bB E o - Note : Multiplying a half reaction by a number does not change E 0 nor E To figure half cell reactions look for the element in two different oxidations states

18. The Nernst equation will be: E = E + - E - = = E 0 Q = When the cell is at equilibrium E = 0 and Q = K E o can be calculated and used to find K for 2 half reactions!

19. Question 10 in Tutorial on Complexation rxns Use the following standard-state cell potentials to calculate the complex formation equilibrium constant for the Zn(NH 3 ) 4 2+ complex ion. Zn(NH 3 ) 4 2+ + 2e - ⇌ Zn + 4NH 3 E o red = -1.04 V Zn 2+ + 2e - ⇌ Zn E o red = - 0.7628 V SOLUTION: E 0 cell = E 0 red (reduction process) – E 0 red (oxidation process) (i) Zn + 4NH 3 ⇌ Zn(NH 3 ) 4 2+ + 2e - E 0 = + 1.04 V (ii) Zn 2+ + 2e - ⇌ Zn E 0 = - 0.7628 V (i) + (ii): Zn 2+ + 4NH 3 ⇌ Zn(NH 3 ) 4 2+ E 0 = + 0.28 V = 10 (2)(0.28)/0.05916 = 2.92 x 10 9

20. REDOX TITRATIONS  Theory of redox titrations and common titrants. A redox titration is based on an oxidation- reduction reaction between an analyte and titrant. Environmental and biological analytes can be measured by redox titrations. Other analytes include laser and superconductor materials. REDOX TITRATION CURVE Consider the potentiometric titration of Fe(II) and cerium (IV). Titration rxn: Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ (1) Each mole of Ce 4+ ion oxidizes 1 mole of ferrous ion, the titration creates a mixture of Ce 4+, Fe 2+, Ce 3+ and Fe 3+ ions.

21. e - flow from the anode to the cathode, the circuit measures the potential for Fe 3+ /Ce 4+ reduction at the Pt surface by e - from the calomel electrode

22. Calomel reference electrode : 2Hg(l) + 2 Cl - Hg 2 Cl 2 (s) + 2e - Pt indicator electrode: 2 rxns coming to equilibrium: Fe 3+ + e- ↔ Fe 2+ E o = + 0.767V (2) Ce 4+ + e- ↔ Ce 3+ E o = + 1.70V (3) Cell rxns 2Fe 3+ + 2Hg(l) + 2Cl - ↔ Fe 2+ + Hg 2 Cl 2 (s) (4) 2Ce 4+ + 2Hg(l) + 2Cl - ↔ Ce 3+ + Hg 2 Cl 2 (s) (5) At equilibrium the potential driving rxns 1 and 2 must be the same. THE CELL RXNS ARE NOT THE SAME AS THE TITRATION RXN! The titration reaction goes to completion and is an oxidation of Fe 2+ and reduction of Ce 4+ Cell rxns proceed to negligible extent. The cell is used to measure activities, not to change them.

23. HOW CELL VOLTAGE CHANGES AS Fe 2+ IS TITRATED WITH Ce 4+ REGION 1 : BEFORE THE EQUIVALENCE POINT As each aliquot of Ce 4+ is added, it is consumed (eqn 1) and creates an equal number of moles of Ce 3+ and Fe 3+ Prior to the equivalence point excess unreacted Fe remains in solution Since the amounts of Fe 2+ and Fe 3+ are known, cell voltage can be calculated from eqn 2 rather than 3 E = E + - E - (calomel) When volume titrant is half the equivalence point ( V = 1/2V e ) the concentration of Fe 2+ and Fe 3+ are equal, thus E + = E o For an acid-base titration pH = pK a when V = 1/2Ve

24. SHAPES OF TITRATION CURVES E ~ E o (Ce 4+ I Ce 3+ ) - 0.241V = 1.46V 1:1 stoichiometry symmetric about equivalence point & same curve for diluted sample Not symmetric about the equivalence point – 2:1

25. REGION 2 : AT THE EQUIVALENCE POINT All cerium is in the form of Ce 3+ [Ce 3+ ] = [Fe 3+ ] Thus the equilibrium form of eqn 1 Ce 4+ + Fe 2+ ↔ Ce 3+ + Fe 3+ If a little Fe 3+ goes back to Fe 2+, an equal no. of moles Ce 4+ must be made and [Ce 4+ ] = [Fe 2+ ] Eqns 2 and 3 are in equilibrium at the Pt electrode, it is convenient to use both these eqns to calculate the cell voltage At equilibrium in a redox titration, Ecell = (E o 1 + E o 2 )/2

26. REGION 3 : AFTER THE EQUIVALENCE POINT Almost all the iron atoms are Fe 3+, the moles of Ce 3+ = moles Fe 3+, and there is a known excess of unreacted Ce 4+. We know [Ce 3+ ] and [Ce 4+ ] and so we can use eqn 3 to calc E.

27. FINDING THE ENDPOINT As in acid-base titration, indicators and electrodes are commonly used to find the endpoints of a redox titration. REDOX INDICATORS A redox indicator is a compound that changes colour when it goes from oxidized to a reduced state, eg. Ferroin changes from pale blue to red. By writing the Nernst equation we can predict the potential range over which the indicator will change : In (oxidised) + ne - ↔ In (reduced) (6)

28. As with acid base indicators, the colour of In (reduced) will be observed when: [In( reduced)] 10 [In(oxidised)] 1 And the colour of the In(oxidised) will be observed when: [In( reduced)] 1 [In(oxidised)] 10 Using these 2 quotients in eqn 6, tells us the colour range will occur over the range E o = 1.147V, we expect the colour change to occur ~ 1.088 – 1.206V wrt SHE

30. STARCH IODINE COMPLEX Many analytical procedures use redox titrations involving iodine. Starch is used as the indicator, since it forms an intense blue complex with iodine. Starch is not a redox indicator because it responds to I 2, not to a change in potential. Starch is readily biodegradable and must be freshly prepared. ADJUSTMENT OF ANALYTE OXIDATION STATES Before titration we adjust the oxidation state of the analyte, eg Mn 2+ can be pre-oxidised to MnO 4 - and then titrated with Fe 2+ Pre-adjustment must be quantitative and all excess reagent must be destroyed. Pre-oxidation : Persulphate S 2 O 8 2- is a powerful oxidant that requires Ag + as a catalyst: S 2 O 8 2- + Ag + SO 4 2- + SO 4 - + Ag 2+ Two powerful oxidants

31. Excess reagent is destroyed after by boiling the solution after oxidation is complete 2S 2 O 8 2- + 2 H 2 O boiling 4SO 4 2- + O 2 + 4H + H 2 O 2 is a good oxidant in basic solution and reductant in acidic solution. The excess spontaneously disproportionate in boiling water. H 2 O 2 H 2 O + O 2 PRE-REDUCTION Stannous & chromous chloride, SO 2, H 2 S are used to pre- reduce analytes to a lower oxidation state. An important pre-reduction technique uses a packed column to pre-reduce analyte to a lower oxidation state (analyte is drawn by suction).

32. Jones reductor, which contains Zn coated with Zn amalgam. Zn is a powerful reducing agent (E o = -0.764V) making the Jones reductor unselective, other species eg, Cr 3+ are reduced and may interfere with the titration analysis. OXIDATION WITH POTASSIUM PERMANGANATE KMnO 4 is a strong oxidant with an intense violet colour. In strongly acidic solutions (pH< 1) it is reduced to colourless Mn 2+ (manganous). In neutral or alkaline the product is a brown solid – MnO 2 In strongly alkaline solution, green manganate MnO 4 2- is produced.

33. KMnO 4 is not a primary standard, and can contain traces of MnO 2, thus it must be standardized with pure Fe wire or sodium oxalate (pink end-point) for greater accuracy, The KMnO 4 solution is unstable : 4MnO 4 - + 2H 2 O 4MnO 2 + 3O 2 + 4OH -

34. OXIDATION WITH Ce 4+ Reduction of Ce 4+ (yellow) to Ce 3+ (colorless) can be used in place of KMnO 4. Ce 4+ is used for the quantitative determination of malonic acid as well as alcohol, ketones and carboxylic acids. The primary standard is prepared by dissolving the salt in 1M H 2 SO 4 and is stable indefinitely. OXIDATION WITH K 2 Cr 2 O 7 Powerful oxidant – in acidic solution orange dichromate iron is reduced to green chromic ion. In 1M HCl the formal potential is 1.00V and 2M H 2 SO 4 it is 1.11V, thus less powerful than Ce 4+ and MnO 4 -

35. Stable primary standard that is employed to determine Fe 2+ Also used in environmental analysis of oxygen demand. COD or chemical oxygen demand is defined as the oxygen that is equivalent to the Cr 2 O 7 2- consumed by the oxidation of organics in water. METHODS INVOLVING IODINE

36. When a reducing analyte is titrated with iodine to produce I - the method is called Iodimetry (titration with I 3 - ). In the iodimetric determination of vitamin C-starch is added to give an intense blue end-point. Iodometry – oxidizing analyte added to I - to produce I 2 which is then titrated with thiosulfate standard, starch is added only before the endpoint. When we speak of using iodine as a titrant we mean a solution of I 2 plus excess I - I 2 (aq) + I - I 3 - K= 7x10 2 A 0.05M solution of I 3 - is prepared by dissolving0.12M KI plus 0.05M I 2 in water. Reducing agent + I 3 - 3I - Oxidizing agent + 3I - I 3 -

37. Precipitation Reactions Gravimetric Analysis: Solid product formed Relatively insoluble Easy to filter High purity Known Chemical composition Precipitation Conditions: Particle Size Small Particles: Clog & pass through filter paper Large Particles: Less surface area for attachment of foreign particles.

38. Crystallization 1. Nucleation 2. Particle Growth Molecules form small Aggregates randomly Addition of more molecules to a nucleus. Supersaturated Solution: More solute than should be present at equilibrium. Supersaturated Solution: Nucleation faster; Suspension (colloid) Formed. Less Supersaturated Solution: Nucleation slower, larger particles formed.

39. How to promote Crystal Growth 1. Raise the temperature Increase solubility Decrease supersaturation 2. Precipitant added slowly with vigorous stirring. 3. Keep low concentrations of precipitant and analyte (large solution volume).

40. Homogeneous Precipitation Precipitant generated slowly by a chemical reaction pH gradually increases Large particle size

41. Net +ve charge on colloidal particle because of adsorbed Ag +

42. Precipitation in the Presence of an Electrolyte Consider titration of Ag + with Cl - in the presence of 0.1 M HNO 3. Colloidal particles of ppt : Surface is + vely charged Adsorption of excess Ag + on surface (exposed Cl - ) Colloidal particles need enough kinetic energy to collide and coagulate. Addition of electrolyte (0.1 M HNO 3 ) causes neutralisation of the surface charges. Decrease in ionic atmosphere (less electrostatic repulsion)

43. Digestion and Purity Digestion: Period of standing in hot mother liquor. Promotion of recrystallisation Crystal particle size increases and expulsion of impurities. Purity: Adsorbed impurities: Surface-bound Absorbed impurities: Within the crystal Inclusions & Occlusions Inclusion: Impurity ions occupying crystal lattice sites. Occlusion: Pockets of impurities trapped within a growing crystal.

44. Coprecipitation:Adsorption, Inclusion and Occlusion Colloidal precipitates: Large surface area BaSO 4 ; Al(OH) 3 ; and Fe(OH) 3 How to Minimise Coprecipitation: 1. Wash mother liquor, redissolve, and reprecipitate. 2. Addition of a masking agent: Gravimetric analysis of Be 2+, Mg 2+, Ca 2+, or Ba 2+ with N-p-chlorophenylcinnamohydroxamic acid. Impurities are Ag +, Mn 2+, Zn 2+, Cd 2+, Hg 2+, Fe 2+, and Ga 2+. Add complexing KCN.

45. Ca 2+ + 2RH  CaR 2 (s) + 2H + Analyte Precipitate Mn 2+ + 6CN -  Mn(CN) 6 4- Impurity Masking agent Stays in solution Postprecipitation:Collection of impurities on ppt during Digestion: A supersaturated impurity e.g., MgC 2 O 4 on CaC 2 O 4. Peptization: Breaking up of charged solid particles when ppt is washed with water. AgCl is washed with volatile electrolyte (0.1 M HNO 3 ). Other electrolytes: HCl; NH 4 NO 3 ; and (NH 4 ) 2 CO 3.

46. Product Composition Hygroscopic substances: Difficult to weigh accurately Some ppts:Variable water quantity as water of Crystallisation. Drying Change final composition by ignition:

47. Thermogravimetric Analysis Heating a substance and measuring its mass as a function of temperature.

49. Example In the determination of magnesium in a sample, 0.352 g of this sample is dissolved and precipitated as Mg(NH 4 )PO 4. 6H 2 O. The precipitate is washed and filtered. The precipitate is then ignited at for 1 hour 1100 o C and weighed as Mg 2 P 2 O 7.The mass of Mg 2 P 2 O 7 is 0.2168 g. Calculate the percentage of magnesium in the sample.

50. Solution: The gravimetric factor is: Relative atomic mass Of Mg FM of Mg 2 P 2 O 7 Note: 2 mol Mg 2+ in 1 mol Mg 2 P 2 O 7.

51. Mass of Mg 2+ = 0.0471 g = 13.45 %

52. Combustion Analysis Determination of the carbon and Hydrogen content of organic compounds burned in excess oxygen. H 2 O absorption CO 2 Absorption Prevention of entrance of atmospheric O 2 and CO 2. Note: Mass increase in each tube.

53. C, H, N, and S Analyser: Modern Technique Thermal Conductivity, IR,or Coulometry for Measuring products.

54. 2 mg sample in tin or silver capsule. Capsule melts and sample is oxidised in excess of O 2. ProductsHot WO 3 catalyst: Then, metallic Cu at 850 o C: Dynamic Flash combustion: Short burst of gaseous products

55. Oxygen Analysis: Pyrolysis or thermal decomposition in absence of oxygen. Gaseous products: Nickelised Carbon 1075 o C CO formed Halogen-containing compounds: CO 2, H 2 O, N 2, and HX products HX(aq) titration with Ag + coulometrically. Silicon Compounds (SiC, Si 3 N 4, & Silicates from rocks): Combustion with F 2 in nickel vessel Volatile SiF 4 & other fluorinated products Mass Spectrometry

56. Example 1: Write a balanced equation for the combustion of benzoic acid, C 6 H 5 CO 2 H, to give CO 2 and H 2 O. How many milligrams of CO 2 and H 2 O will be produced by the combustion of 4.635 mg of C 6 H 5 CO 2 H? Solution: C 6 H 5 CO 2 H + 15 / 2 O 2 FW = 122.123 7CO 2 + 3H 2 O 44.010 18.015 4.635 mg of C 6 H 5 CO 2 H = 1 mole C 6 H 5 CO 2 H yields 7 moles CO 2 and 3 moles H 2 O Mass CO 2 = 7 x 0.03795 mmol x 44.010 mg/mmol = 11.69 mg CO 2 Mass H 2 O = 3 x 0.03795 mmol x 18.015 mg/mmol = 11.69 mg H 2 O

57. Example 2: A 7.290 mg mixture of cyclohexane, C 6 H 12 (FW 84.159), and Oxirane, C 2 H 4 O (FW 44.053) was analysed by combustion, and 21.999 mg CO 2 (FW 44.010) were produced. Find the % weight of oxirane in the sample mixture. Solution: C 6 H 12 + C 2 H 4 O + 23 / 2 O 2 8CO 2 + 8H 2 O Let x = mg of C 6 H 12 and y = mg of C 2 H 4 O. X + y = 7.290 mg Also,CO 2 = 6(moles of C 6 H 12 ) + 2(moles of C 2 H 4 O)

58. X + y = 7.290 mg  x = 7.290 - y CO 2  y = mass of C 2 H 4 O = 0.767 mg Therefore, % Weight Oxirane = = 10.52 %

59. The Precipitation Titration Curve Reasons for calculation of titration curves: 1. Understand the chemistry occurring. 2. How to exert experimental control to influence the quality of analytical titration. In precipitation titrations: 1. Analyte concentration 2. Titrant concentration 3. K sp magnitude Influence the sharpness of the end point

60. Titration Curve A graph showing variation of concentration of one reactant with added titrant. Concentration varies over many orders of magnitude P function: pX = -log 10 [X] Consider the titration of 25.00 mL of 0.1000 M I - with 0.05000 M Ag +. I - + Ag +  AgI(s) There is small solubility of AgI: AgI(s)  I - + Ag + K sp = [Ag + ][I - ] = 8.3 x 10 -17

61. I - + Ag +  AgI(s) K =1/K sp = 1.2 x 10 16 V e = Volume of titrant at the equivalent point: Before the Equivalence Point: Addition of 20 mL of Ag + : This reaction: I - + Ag +  AgI(s) goes to completion.  V e = 0.05000 L = 50.00 mL (0.02500 L)(0.1000 mol I - /L)(V e )(0.05000 mol Ag + /L) = mol I - mol Ag +

62. [I - ] due to I - not precipitated by 20.00 mL of Ag +. Fraction of I - reacted: Fraction of I - remaining: Some AgI redissolves: AgI(s)  I - + Ag + Therefore, Fraction Remaining Original Conc. Dilution Factor Original volume of I - Total volume

63.  [I - ] = 3.33 x 10 -2 M   [Ag + ] = 2.49 x 10 -15 M pAg + = -log[Ag + ] = 14.60 The Equivalence Point: All AgI is precipitated AgI(s)  I - + Ag + Then,

64. K sp = [Ag + ][I - ] = 8.3 x 10 -17 And [Ag + ] = [I - ] = x K sp = (x)(x) = 8.3 x 10 -17  X = 9.1 x 10 -9 M  pAg + = -log x = 8.04 At equivalence point: pAg + value is independent of the original volumes or concentrations.

65. After the Equivalence Point After the Equivalence Point: [Ag + ] is in excess after the equivalence point. Note: V e = 50.00 mL Suppose that 52.00 mL is added: Therefore, 2.00 mL excess Ag+ Original Ag + Concentration Dilution Factor Volume of excess Ag + Total volume of solution

66. [Ag + ] = 1.30 x 10 -3 M pAg + = -log[Ag + ] = 2.89 Shape of the Titration Curve: Steepest slope:has maximum value Equivalence point: point of maximum slope Inflection point:

67. Titration Curves: Effect of Diluting the reactants 1.0.1000 M I - vs 0.05000 M Ag+ 2. 0.01000 M I - vs 0.005000 M Ag+ 3. 0.001000 M I - vs 0.0005000 M Ag+

68. Titrations involving 1:1 stoichiometry of reactants Equiv. Point: Steepest point in titration curve Other stoichiometric ratios: 2Ag + + CrO 4 2-  Ag 2 CrO 4 (s) 1. Curve not symmetric near equiv. point 2. Equiv. Point: Not at the centre of the steepest section of titration curve 3. Equiv. Point: not an inflection point In practice: Conditions chosen such that curves are steep enough for the steepest point to be a good estimate of the equiv. point

69. Effect of K sp on the Titration Curve AgI is least soluble Sharpest change at equiv. point Least sharp, but steep enough for Equiv. point location K = 1/K sp largest 

70. Titration of a Mixture Less soluble precipitate forms first. Titration of KI & KCl solutions with AgNO 3 K sp (AgI) << K sp (AgCl) First precipitation of AgI nearly complete before the second (AgCl) commences. When AgI pption is almost complete, [Ag+] abruptly increeases and AgCl begins to precipitate. Finally, when Cl - is almost completely consumed, another abrupt change in [Ag+] occurs.

71. Titration Curve for 40.00 mL of 0.05000 M KI and 0.05000 M KCl with 0.084 M AgNO 3.

72. I - end point: Intersection of the steep and nearly horizontal curves. Note: Precipitation of AgI not quite complete when AgCl begins to precipitate. End of steep portion better approximation of the equivalence point. AgCl End Point: Midpoint of the second steep section.

73. The AgI end point is always slightly high for I - /Cl - mixture than for pure I -. 1. Random experimental error: both +tive and –tive. 2. Coprecipitation: +ve error High nitrate concentration to minimise coprecipitation. Example: Some Cl - attached to AgBr ppt and carries down an equivalent amount of Ag +. NO 3 - competes with Cl - for binding sites. Coprecipitation error lowers the calculated concentration of the second precipitated halide.

74. Separation of Cations by Precipitation Consider a solution of Pb 2+ and Hg 2 2+ : Each is 0.01 M PbI 2 (s) ⇌ Pb 2+ + 2I - Hg 2 I 2 (s) ⇌ Hg 2 2+ + 2I - K sp = 7.9 x 10 -9 K sp = 1.1 x 10 -28 Smaller K sp Considerably Less soluble Is separation of Hg 2 2+ from Pb 2+ “complete”? Is selective precipitation of Hg 2 2+ with I - feasible?

75. Can we lower [Hg 2 2+ ] to 0.010 % of its original value without precipitating Pb 2+ ? From 0.010 M to 1.0 x 10 –6 M? Add enough I - to precipitate 99.990 % Hg 2 2+. Hg 2 I 2 (s) ⇌ Hg 2 2+ + 2I - Initial Concentration: 0 0.010 0 Final Concentration: solid 1.0 x 10 -6 x  (1.0 x 10 -6 )(x) 2 = 1.1 x 10 -28 X = [I - ] = 1.0 x 10 –11 M  X = [I - ] = 1.0 x 10 –11 M

76. [I - ] = 1.0 x 10 –11 M Will this [I - ] = 1.0 x 10 –11 M precipitate 0.010 M Pb 2+ ? Q = 1.0 x 10 -24 << 7.9 x 10 –9 = K sp for PbI 2 Therefore, Pb 2+ will not precipitate. Prediction: All Hg 2 2+ will virtually precipitate before any Pb 2+ precipitates on adding I -.