1. CHEMICAL EQUILIBRIUM

2. aA + bB cC + dD Equilibrium constant a,b,c,d – stoichiometry coefficients [A], [B], [C], [D] –concentrations of A, B, C, D in standard state: For solutes – 1M For gases – 1 atm For solids and pure liquids: [X] = 1 In these conditions K is dimensionless. K>1  Forward reaction is favoured the dynamic state in which rates of the forward and reverse reactions are identical. CHEMICAL EQUILIBRIUM

3. Equilibrium const. for a reverse reaction: Equilibrium const. for a reverse reaction:K 1 = 1/K Equilibrium const. for two reactions added: Equilibrium const. for two reactions added:K 3 = K 1 x K 2 H + + C CH + HA + C CH + + A - HA H + + A - cC + dD aA + bB

4. The equilibrium constant is derived from the thermodynamics of a chemical reaction. ENTHALPY ΔH - enthalpy change is the heat absorbed or released during reaction ΔH < 0 Heat liberated Exothermic reaction ΔH > 0 Heat absorbed Endothermic reaction THERMODYNAMICS

5. ENTROPY ΔS – a measure of ‘disorder’ of a substance ΔS > 0 Products more disordered than reactants ΔS < 0 Products less disordered than reactants Gas  Liquid  Solid Decrease in disorder Example: KCl(s) K + (aq) + Cl - (aq) ΔS 0 = +76 J/K mol at 25 0 C

6. FREE ENERGY Gibbs free energy: ΔG = ΔH - TΔS ΔG 1 ΔG > 0 The reaction is not favoured K<1 Free energy and equilibrium: OR

7. When a system at equilibrium is disturbed, the direction in which the system proceeds back to equilibrium is such that the change is partially offset. If reaction is at equilibrium and reactants are added (or products removed), the reaction goes to the right. If reaction is at equilibrium and products added ( or reactants are removed), the reaction goes to the left. LE CHATELIER’S PRINCIPLE A + B C + D

8. According to Le Châtelier: If Q > K  reaction will proceed in the reverse direction If Q < K  reaction will proceed in the forward direction Recall: Q = Reaction quotient has the same form as equilibrium constant (K), but the solution concentrations do not have to be equilibrium concentrations. Thus at equilibrium: Q = K

9. [A] = 0.002M [B] = 0.025M [C] = 5.0M [D] = 1.0M = 1x10 5 at 25 0 C To reach equilibrium: Q = K and the reaction must go to the right  spontaneous A + B C + D Say we add double reactant A, [A] = 0.004M = 0.5x10 5 at 25 0 C Q < K At equilibrium:

10. ΔG = ΔH - TΔS Independent of T For endothermic reactions (ΔH > 0): K increases if T increases. For exothermic reactions (ΔH < 0): K decreases if T increases. THE EFFECT OF TEMPERATURE ON K

11. K = K sp (solubility product) when the equilibrium reaction involves a solid salt dissolving to give its constituent ions in solution. Recall: [Solid] = 1 Saturated solution – in equilibrium with undissolved solid Thus if an aqueous solution is left in contact with excess solid, the solid will dissolve until K sp is satisfied. Thereafter the amount of undissolved solid remains constant. K AND SOLUBILITY

12. Example: Calculate the mass of PbCl 2 that dissolves in 100 ml water. (K sp = 1.7x10 -5 for PbCl 2 ) Initial: Final: (solid) m = 0.45 g PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq)

13. Now add 0.03M NaCl to the PbCl 2 solution  We added 0.03M Cl - THE COMMON ION EFFECT Initial: Final: (solid) PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) For this system to be at equilibrium when [Cl - ] is added, the [Pb 2+ ] decreases (reverse reaction). – this is an application of the Le Chatelier’s principle and is called THE COMMON ION EFFECT The salt will be less soluble if one of its constituent ions is already present in the solution.

14. The Common Ion Effect - experiment

15. THE NATURE OF WATER AND ITS IONS H + does not exist on its own in H 2 O  forms H 3 O + H3O+:H3O+:

16. In aqueous solution, H 3 O + is tightly associated with 3 molecules of H 2 O through exceptionally strong hydrogen bonds. One H 2 O is held by weaker ion-dipole attraction

17. Can also form H 5 O 2 + cation  H + shared by 2 water molecules H 3 O 2 - (OH -.H 2 O) has been observed in solids

18. AUTOPROTOLYSIS Water undergoes self-ionisation  autoprotolysis, since H 2 O acts as an acid and a base. H 2 O + H 2 O H 3 O + + OH - The extent of autoprotolysis is very small. For H 2 O: K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 (at 25 o C)

19. pH ≈ -log[H + ] Approximate definition of pH pH + pOH = -log(K w ) = 14.00 at 25 0 C pH

20. It is generally assumed that the pH range is 0-14. But we can get pH values outside this range. e.g. pH = -1  [H + ] = 10 M This is attainable in a strong concentrated acid.

21. BUT in a real solution all charged ions are:  surrounded by ions with opposite charge – ionic atmosphere  hydrated - surrounded by tightly held water dipoles Equilibrium constant [A], [B], [C], [D] –concentrations of A, B, C, D ACTIVITY aA + bB cC + dD

22. Adding an “inert” salt to a sparingly soluble salt increases the solubility of the sparingly soluble salt. “inert” salt = a salt whose ions do not react with the compound of interest WHY? Consider: BaSO 4 (K sp = 1.1x10 -10 ) as the sparingly soluble salt and KNO 3 as the “inert” salt. In solution: The cation (Ba 2+ ) is surrounded by anions (SO 4 2-, NO 3 - )  net positive charge is reduced  The anion (SO 4 2- ) is surrounded by cations (Ba 2+, K + )  net negative charge is reduced  attraction between oppositely charged ions is decreased.

23. The net charge in the ionic atmosphere is less than the charge of the ion at the center. The greater the ionic strength of a solution, the higher the charge in the ionic atmosphere. Each ion-plus-atmosphere contains less charge and there is less attraction between any particular cation and anion. The ionic atmosphere decrease the attraction between ions.

24. Activity of the ion in a solution depends on its hydrated radius not the size of the bare ion.

25. A measure of the total concentration of ions in solution. The more highly charged an ion, the more it is counted. Where c i = concentration of the i th species z i = charge for all ions in solution IONIC STRENGTH, µ

26. Example: Find the ionic strength of 0.010 M Na 2 SO 4 solution.

27. Effect of ionic strength on solubility Explain all 4 cases

28. To account for the effect of ionic strength, concentrations are replaced by activities. Activity of C Activity coefficient And general form of equilibrium constant is: ACTIVITY COEFFICIENTS

29. Activity coefficient: Measure of deviation of behaviour from ideality (ideal   = 1) Allows for the effect of ionic strength Thus for the sparingly soluble salt BaSO 4, dissolving in the presence of the “inert” salt KNO 3 : K sp = a Ba a SO4 = [Ba 2+ ]  Ba [SO 4 2- ]  SO4  If more BaSO 4 dissolves in the presence of KNO 3, [Ba 2+ ] and [SO 4 2- ] increases and  Ba and  SO4 decreases At low ionic strength: activity coefficients  1 and K  concentration equilibrium

30. Extended Debye-Hűckel equation relates activity coefficients to ionic strength: at 25 0 C ACTIVITY COEFFICIENTS OF IONS  = effective hydrated radius of the ion

31. 3. The smaller the hydrated radius of the ion, the more important activity effects become. Effect of Ionic Strength, Ion charge and Ion Size on the Activity Coefficient Activity coefficients for differently charged ions with a constant hydrated radius of 500pm. (Over the range of ionic strength from 0 to 0.1M) 1.As ionic strength increases, the activity coefficient decreases.   1 as   0 2.As the charge of the ion increases, the departure of its activity coefficient from unity increases. Activity corrections are much more important for an ion with a charge of  3 than one with the charge  1.

32. Use interpolation to find values of  for ionic strengths not listed Obtain values for  from the table:

33. How to interpolate - SELF STUDY!! Linear interpolation:

34. At high ionic strengths: activity coefficients of most ions increase Concentrated salt solutions are not the same as dilute aqueous solutions  “different solvents” H + in NaClO 4 solution of varying ionic strengths

35. The real definition of pH is: pH ≈ -log[H + ] Approximate definition of pH NOTE: A pH electrode measures activity of H + and NOT concentration pH AND ACTIVITY COEFFICIENTS

36. The systematic procedure is to write as many independent algebraic equations as there are unknowns (species) in the problem. This includes all chemical equilibrium conditions + two balances: charge and mass balances. Chemical equilibrium provides a basis for most techniques in analytical chemistry and application of chemistry to other disciplines such as like biology, geology etc. The systematic treatment of equilibrium gives us the tool to deal with all types of complicated chemical equilibria. SYSTEMATIC TREATMENT OF EQUILIBRIUM

37. CHARGE BALANCE E.g. An aqueous solution of KH 2 PO 4 and KOH contains the following ionic species: H +, OH -, K +, H 2 PO 4 -, HPO 4 2-, PO 4 3- The charge balance is: The coefficient in front of each species = the magnitude of the charge on the ion [H + ] + [K + ] = [OH - ] + [H 2 PO 4 - ] + 2[HPO 4 2- ] + 3[PO 4 3- ] Charge neutrality The sum of positive charges in solution equals the sum of negative charges.

38. MASS BALANCE Quantity of all species in a solution containing a particular atom must equal the amount of that atom delivered to the solution. E.g. Mass balance for 0.02 M phosphoric acid in water: Conservation of matter. 0.02 M = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ]

39. SYSTEMATIC TREATMENT OF EQUILIBRIUM: Step 1. Write the pertinent reactions. Step 2. Write the charge balance equation. Step 3. Write the mass balance equations. Step 4. Write the equilibrium constant for each chemical reaction. Step 5. Count the equations and unknowns. Step 6. Solve for all the unknowns.

40. E.g.: The ionization of water H 2 O H + + OH - K w = 1.0x10 -14 at 25 0 C Find the concentrations of H + and OH - in pure water For pure water the ionic strength approaches 0 and we can write eq.3 as: Step 1. Pertinent reaction– only one above. Step 2. Charge balance: (1) Step 3. Mass balance: (2) Step 4. Equilibrium constants– the only one (3) Step 5. Count equations and unknowns – 2 eq. and 2 unknowns Step 6. Solve.

41. E.g.: The solubility of Hg 2 Cl 2 Step 6. Solve. Using eqn 2 we can write eqn 3 as: Find the concentration of Hg 2 2+ in a saturated solution of Hg 2 Cl 2 Step 2. Charge balance: (1) Step 3. Mass balance: (2) Step 4. Equilibrium constants: (3) (4) Step 5. Count equations and unknowns – 4 eqs. and 4 unknowns Step 1. Pertinent reactionsHg 2 Cl 2 Hg 2 2+ + 2Cl - K sp H 2 O H + + OH - K w

42. Coupled equilibria – the product of one reaction is reactant in the next reaction Problem: The mineral fluorite, CaF 2, has a cubic crystal structure and often cleaves to form nearly perfect octahedra. Find the solubility of CaF 2 in water. THE DEPENDENCE OF SOLUBILITY ON pH

43. Step 1. Pertinent reaction Step 2. Charge balance (1) Step 3. Mass balance Some fluoride ions react to give HF. (2) CaF 2 dissolves: CaF 2 (s) Ca 2+ + 2F - K sp = 3.9x10 -11 For every aqueous solution: H 2 O H + + OH - K w = 1x10 -14 The F - ions reacts with water to give HF: F - + H 2 O HF + OH - K b = 1.5x10 -11 Also

44. Step 4. Equilibrium constants (3) (4) (5) Step 5. Count equations and unknowns 5 eqs. and 5 unknowns: Step 6. Solve CaF 2 (s) Ca 2+ + 2F - K sp F - + H 2 O HF + OH - K b H 2 O H + + OH - K w

45. Using this expression in eqn 3: K sp = [Ca 2+ ][F - ] 2 = [Ca 2+ ](0.80[Ca 2+ ]) 2 To simplify the problem let us solve it for a fixed pH = 3 That means:[H + ] = and [OH - ] = Then from eqn 4: K b = [HF][OH - ]/[F - ] [HF]/[F - ] = K b /[OH - ] = 1.5x10 -11 /1.0x10 -11 = 1.5 Thus[HF] = 1.5[F - ] Substitute [HF] in the eqn 2: [F - ] + [HF] = 2[Ca 2+ ] And[F - ] = 0.80[Ca 2+ ] Thus[Ca 2+ ] = (K sp /0.80 2 ) 1/3 = 3.9x10 -4 M [F - ] + 1.5[F - ] = 2[Ca 2+ ] Now find: [F - ] = 3.1x10 -4 M [HF] = 4.7x10 -4 M

46. pH dependence of the conc. of Ca 2+, F - and HF in a saturated solution. NOTE: To fix the pH of a solution an ionic compound is added. Thus the charge balance equation as written not longer holds. Also [OH - ] = [H + ] + [HF] No longer holds

47. Found [Ca] in acid rain that has washed off marble stone (largely CaCO 3 ) increases as the [H + ] of acid rain increases. Applications of coupled equilibria in the modeling of environmental problems CaCO 3 (s) + 2H + (aq)  Ca 2+ (aq) + CO 2 (g) + H 2 O(l) SO 2 (g) + H 2 O(l)  H 2 SO 3 (aq) oxidation H 2 SO 4 (aq)

48. Total [Al] as a function of pH in 1000 Norwegian lakes. Al is usually “locked” into insoluble minerals e.g. kaolinite and bauxite. But due to acid rain, soluble forms of Al are introduced into the environment. (Similarly with other minerals containing Hg, Pb etc.)