1. The Equilibrium Constant We have the general reaction

2. Equilibrium Constant From this we can write an expression for our equilibrium Where we have the products over the reactants and each compound is represented by its concentration

3. A Word on Units What are the units on K? From the formula we might guess that they would depend on the specific equilibrium involved. This is not the case however. Let us visit the concept of standard state.

4. Standard States The terms and unit that we will use are dictated by convention Solutions M Gasbar The values used in our equilibrium expression will be based on some reference value

5. Standard States For solutions the standard state is 1.00M Gas 1.00 bar Pure solids or solvents the standard state is the substance. So if a solution is 2.31 x 10 -3 M then the value we use in the equilibrium expression is....

6. Standard State 2.31 x 10 -3 M / 1.00 M = 2.31 x 10 -3 with the units factoring out. Since gas values are usually expressed as pressure instead of seeing [X], you will see a gas expressed as P X Different standard states could be picked but convention is well established on these terms in Analytical Chemistry.

7. Equilibrium Constants The equilibrium expression will depend on how the expression is written. As a dissociation HA = H + + A -

8. Equilibrium But written as an association H + + A - = HA Which we can see that K 1 = 1/ K 2 So if we write an equation in reverse we must take the reciprocal of the constant

9. Equilibrium What about added equilibria. H 2 S = H + + HS - then HS - = H + + S 2-

10. Equilibrium If we add the two reactions we get. H 2 S = 2 H + + A 2- Giving us the product of the equilibria.

11. Thermodynamics How does this all relate to “thermo” Two contributions –Enthalpy  H, a measure of the heat of a reaction. + endothermic, - exothermic –Entropy  S, a measure of disorder in a system. + for more disorder, - for less disorder

12. Thermodynamics When HCl (which is a gas) is bubbled into water the following reaction occurs H 2 O + HCl(g) = H + (aq) + Cl - (aq) Which has a  H o of -75.15 kJ/mole @ 25 C Heat can be viewed as ending up in bonds. What bonds are in this reaction? The naught ( o ) means that the product and reactants are in the standard state.

13. Thermodynamics When we dissolve KCl, a salt similar to NaCl then we get an entropy effect. Since the ions in solution are more disordered than the ions in the solid. H 2 O + KCl(s) = K + (aq) + Cl - (aq)  S o is + 76 J/(K. mole) @ 25C and entropy favors this reaction

14. Thermodynamics But for the reaction of HCl with water the  S value is negative. Why would that be the case? What happens if the terms work against each other.

15. Thermodynamics This is where Gibbs free energy comes in.  G =  H - T  S For our reaction with HCl in water what do we get?  G = -75.15 kJ/mole - (298K)(-131.5J/K. mole) and calculating we get  G = -35.94 kJ/mole If  G is negative then the reaction is favored. Exergonic If  G is positive then the reaction is disfavored Endergonic

16. Thermodynamics If  G is zero then the reaction is at equilibrium. Remember that the values we have been looking at are in their standard state and these values will vary with concentration. What is the relationship between  G o and K

17. La Chatelier’s Principle If a change disturbs a equilibrium system that system will proceed back to equilibrium to partially offset the change. Use Q, same form as the equilibrium expression. If Q is less than K then the system must proceed to the right. If Q is more than K then the system must go to the left.

18. La Chatelier’s Principle Acetic acid in water HAc = H + + Ac - If you double the hydrogen ion concentration by adding a strong acid you will cause the amount of HAc to increase

19. Effect of heat on K The equilibrium constant of an endothermic reaction increases if temperature is raised. The equilibrium constant of an exothermic reaction decreases if temperature is raised. Look at it this way reactants = products + heat

20. Solubility A very important parameter in many areas of chemistry. –Likes dissolve likes –Polar compounds will dissolve in polar solvents sugar in water alcohols in water –Non polar compounds will dissolve in non polar solvents. oils in carbon tetrachloride flavor in fats

21. Salts in water Some salts are ‘soluble’ in water and some are not. –Soluble NaClKClNa 3 PO 4 –Not Soluble AgClLaF 3 Cu(OH) 2 Yet all lead to varying amounts of ions in solution that are very polar.

22. Solubility Products Let us look at an example AgCl(s) = Ag + + Cl - K sp = [Ag + ][Cl - ] If we look up the value for this K sp in appendix F we find that the value listed is 1.8 x 10 -10, what is [Ag + ] for a saturated solution? The K sp for NaCl would be 49. What does that mean?

23. K sp Problem What will be the concentration of La 3+ if 1.00 grams of the solid LaF 3 is added to 10 mL of water and allowed to come to equilibrium? K sp = 2 x 10 -19 Best way to solve. Set up a table!

24. Ksp solutions LaF 3 =La 3+ 3F - Initial Conc.solid 0 0 Final Conc.solid x 3x Where x is the concentration of La 3+ at equilibrium K sp = [La 3+ ][F - ] 3

25. Ksp Solutions Ksp = [La 3+ ][F - ] 3 From last slide [La 3+ ] = x and [F - ] = 3x so 2 x 10 -19 = x (3x) 3 = 27x 4 so x = 9 x 10 -6 M La 3+ and what value for F - ? ____________________

26. Common Ion Effect We can figure out how much AgCl would dissolve in pure water. How much would dissolve if the solution already had some of either ion already in it? Let’s look at the more complex example however.

27. Common Ion Solutions Let the initial Fluoride concentration be 0.001 M LaF 3 =La 3+ 3F - Initial Concsolid 0 0.001 Final Conc solid x 3x + 0.001 Where x is the concentration of La 3+ at equilibrium

28. Common Ion Ksp = [La 3+ ][F - ] 3 From last slide[La 3+ ] = x and [F - ] = 3x + 0.001 so 2 x 10 -19 = x (3x + 0.001) 3 = quadric equation to solve. Fun!!!!! What do we do??????

29. Common Ion We know that the amount of F - we are going to get will be small. We will make an approximation. We will assume that x is small in the F - term. Giving 2 x 10 -19 = x (0.001) 3 Which gives us x = 2 x 10 -10 M for La 3+ Now what is F - ? Does this check out?

30. Approximations Knowing when to make approximations will be very helpful to us. When approximations are made then we must check our answer to make sure that this approximation make some sense. Is the final F - concentration close to the 0.001 M as we assumed? If not then we will need to solve without making the approximation which can be difficult.

31. Separation by Pcpt In theory this will work great. Co-precipitation will lead to significant error.

32. Complex Formation Many different equilibria can be going on at the same time. The second type is the formation of complexes. The example in the book is the formation of iodides of lead(II).

33. Complexes With lead and iodide we can have solid PbI 2 formed. K sp = [Pb 2+ ][I - ] 2 With extra added iodide we have complex formed. As iodide is added we form the mono, di, tri and tetra iodide species. Each step has and equilibrium constant. These are formation constants.

34. Complexes Pb 2+ + I - = PbI - K 1 Pb 2+ +2I - = PbI 2  2 Pb 2+ +3I - = PbI 3 2-  3 Pb 2+ + 4I - = PbI 4 2-  4 Each step also has a stepwise function also. PbI 3 + I - = PbI 4 2- K 4

35. Complexes All equilibria must be satisfied at the same time. In this case here we can control the solubility of the PbI 2 by added iodide ion in increasing amounts giving a mixture of the species that were mentioned.

36. Protic Acid and Bases Bronsted and Lowry –Acids are proton donors –Bases are proton acceptors We increase or decrease the amount of hydronium ion (H 3 O + ) in solution with these acids and bases. We also have the Lewis acid base definition. Deals with electron pairs.

37. Salts When acids and bases are brought together they will neutralize each other. This produces a salt. Most salts added to water might not dissolve completely but will dissociate completely. NaClNa 2 SO 4 Na 3 PO 4 NH 4 Cl

38. Conjugates Products of acid/base reactions are also classified as acid/base. Acetic acid + Ammonia = Acetate Ion + Ammonium ion Acetic acidacid Acetate ionconjugate base Ammoniabase Ammonium ionconjugate acid

39. Water Water is our most common solvent. Its behavior and properties are important to world as we know it. Water can either gain a proton or lose a proton. We can use the hydronium ion H 3 O + in our reactions or we often get lazy and just use H +

40. Water Water can react with itself. One water could donate a proton and another water could lose that proton. This is called autoprotolysis. H 2 O + H 2 O = H + 3 O + OH - Other protic solvents can do this ie. Acetic acid 2CH 3 COOH = CH 3 COOH 2 + + CH 3 COO - There are solvents that are called aprotic. These solvents do not give up or accept the proton Diethyether, hexane, acetonitrile, toluene.

41. Water Autoprotolysis Constant H 2 O = H + + OH - has a special equilibrium constant. K w K w = [H + ][OH - ] K w = 1.01X 10 -14 @25C K w varies with temp. We will often see this expressed on the log scale. pK w = -log K w

42. Water What is pH? pH = -log[H + ] What is the pH of pure water. For each H 2 O that dissociates we get one H + and one OH -, so [H + ] = [OH - ] K w = [H+] 2 = 1 x 10 -14 H + = 1.00 x 10 -7 pH = 7.0 Water will usually have traces of CO 2 and other ions that change the pH

43. Acid and Base Strength Some acids and base are considered strong! This means that they fully dissociate in water. –AcidsHCl, H 2 SO 4, HNO 3, HClO 4 –BasesNaOH, KOH Many other acid are weak. This only means that they do not fully dissociate in water. Do not infer that they are less hazardous. –AcidsAcetic, H 3 PO 4, Benzoic etc. –BasesAmmonia, other amines.

44. Weak Acids The dissociation of weak acid have and equilibrium constant called K a. If more that one proton can dissociate, i.e. H 3 PO 4 each added loss will have its own equilibrium constant.

45. Acid/Conjugate Relationship Acetic acid HA HA = H + + A - K a = [H+][A - ]/[HA] The conjugate base. NaA Sodium Acetate NaA + H 2 O = HA + OH - + Na + K b = [HA][OH - ]/[A - ]

46. Acid/Conjugate Relationship What is K a *K b K a *K b = [H+][A - ]/[HA]*[HA][OH - ]/[A - ] = [H + ][OH - ] = K w So once you know the K a for the acid form you know the K b for the conjugate base. It must be the same conjugate pair. Goodbye bases, You will note that there are no K b values listed in the Appendix in the text. This the the way that biochemists look at the world.