1. Chapter 15 Applying equilibrium

2. The Common Ion Effect l When the salt with the anion of a weak acid is added to that acid, l It reverses the dissociation of the acid. l Lowers the percent dissociation of the acid. l The same principle applies to salts with the cation of a weak base.. l The calculations are the same as last chapter.

3. Buffered solutions l A solution that resists a change in pH. l Either a weak acid and its salt or a weak base and its salt. l We can make a buffer of any pH by varying the concentrations of these solutions. l Same calculations as before. l Calculate the pH of a solution that is.50 M HAc and.25 M NaAc (Ka = 1.8 x 10 -5 )

4. Adding a strong acid or base l Do the stoichiometry first. l A strong base will grab protons from the weak acid reducing [HA] 0 l A strong acid will add its proton to the anion of the salt reducing [A - ] 0 l Then do the equilibrium problem. l What is the pH of 1.0 L of the previous solution when 0.010 mol of solid NaOH is added?

5. General equation l Ka = [H + ] [A - ] [HA] l so [H + ] = Ka [HA] [A - ] l The [H + ] depends on the ratio [HA]/[A - ] l taking the negative log of both sides l pH = -log(Ka [HA]/[A - ]) l pH = -log(Ka)-log([HA]/[A - ]) l pH = pKa + log([A - ]/[HA])

6. This is called the Henderson- Hasselbach equation l pH = pKa + log([A - ]/[HA]) l pH = pKa + log(base/acid) l Calculate the pH of the following mixtures l 0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) l 0.25 M NH 3 and 0.40 M NH 4 Cl l (Kb = 1.8 x 10 -5 )

7. Prove they’re buffers l What would the pH be if.020 mol of HCl is added to 1.0 L of both of the preceding solutions. l What would the pH be if 0.050 mol of solid NaOH is added to each of the proceeding.

8. Buffer capacity l The pH of a buffered solution is determined by the ratio [A - ]/[HA]. l As long as this doesn’t change much the pH won’t change much. l The more concentrated these two are the more H + and OH - the solution will be able to absorb. l Larger concentrations bigger buffer capacity.

9. Buffer Capacity l Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following: l 5.00 M HAc and 5.00 M NaAc l 0.050 M HAc and 0.050 M NaAc l Ka= 1.8x10 -5

10. Buffer capacity l The best buffers have a ratio [A - ]/[HA] = 1 l This is most resistant to change l True when [A - ] = [HA] l Make pH = pKa (since log1=0)

11. Titrations l Millimole (mmol) = 1/1000 mol l Molarity = mmol/mL = mol/L l Makes calculations easier because we will rarely add Liters of solution. l Adding a solution of known concentration until the substance being tested is consumed. l This is called the equivalence point. l Graph of pH vs. mL is a titration curve.

12. Strong acid with Strong Base l Do the stoichiometry. l There is no equilibrium. l They both dissociate completely. l The titration of 50.0 mL of 0.200 M HNO 3 with 0.100 M NaOH l Analyze the pH

13. Weak acid with Strong base l There is an equilibrium. l Do stoichiometry. l Then do equilibrium. l Titrate 50.0 mL of 0.10 M HF (Ka = 7.2 x 10 -4 ) with 0.10 M NaOH

14. Titration Curves

15. pH mL of Base added 7 l Strong acid with strong Base l Equivalence at pH 7

16. pH mL of Base added >7 l Weak acid with strong Base l Equivalence at pH >7

17. pH mL of Base added 7 l Strong base with strong acid l Equivalence at pH 7

18. pH mL of Base added <7 l Weak base with strong acid l Equivalence at pH <7

19. Summary l Strong acid and base just stoichiometry. l Determine Ka, use for 0 mL base l Weak acid before equivalence point –Stoichiometry first –Then Henderson-Hasselbach l Weak acid at equivalence point Kb l Weak base after equivalence - leftover strong base.

20. Summary l Determine Ka, use for 0 mL acid. l Weak base before equivalence point. –Stoichiometry first –Then Henderson-Hasselbach l Weak base at equivalence point Ka. l Weak base after equivalence - leftover strong acid.

21. Indicators l Weak acids that change color when they become bases. l weak acid written HIn l Weak base l HIn H + + In - clear red l Equilibrium is controlled by pH l End point - when the indicator changes color.

22. Indicators l Since it is an equilibrium the color change is gradual. l It is noticeable when the ratio of [In - ]/[HI] or [HI]/[In - ] is 1/10 l Since the Indicator is a weak acid, it has a Ka. l pH the indicator changes at is. l pH=pKa +log([In - ]/[HI]) = pKa +log(1/10) l pH=pKa - 1 on the way up

23. Indicators l pH=pKa + log([HI]/[In - ]) = pKa + log(10) l pH=pKa+1 on the way down l Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with base. l Choose the indicator with a pKa 1 greater than the pH at equivalence point if you are titrating with acid.

24. Solubility Equilibria Will it all dissolve, and if not, how much?

25. l All dissolving is an equilibrium. l If there is not much solid it will all dissolve. l As more solid is added the solution will become saturated. l Soliddissolved l The solid will precipitate as fast as it dissolves. l Equilibrium

26. General equation l M + stands for the cation (usually metal). l Nm - stands for the anion (a nonmetal). l M a Nm b (s) aM + (aq) + bNm - (aq) l K = [M + ] a [Nm - ] b /[M a Nm b ] l But the concentration of a solid doesn’t change. l K sp = [M + ] a [Nm - ] b l Called the solubility product for each compound.

27. Watch out l Solubility is not the same as solubility product. l Solubility product is an equilibrium constant. l it doesn’t change except with temperature. l Solubility is an equilibrium position for how much can dissolve. l A common ion can change this.

28. Calculating K sp l The solubility of iron(II) oxalate FeC 2 O 4 is 65.9 mg/L l The solubility of Li 2 CO 3 is 5.48 g/L

29. Calculating Solubility l The solubility is determined by equilibrium. l Its an equilibrium problem. l Calculate the solubility of SrSO 4, with a Ksp of 3.2 x 10 -7 in M and g/L. l Calculate the solubility of Ag 2 CrO 4, with a Ksp of 9.0 x 10 -12 in M and g/L.

30. Relative solubilities l Ksp will only allow us to compare the solubility of solids the at fall apart into the same number of ions. l The bigger the Ksp of those the more soluble. l If they fall apart into different number of pieces you have to do the math.

31. Common Ion Effect l If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve. l Calculate the solubility of SrSO 4, with a Ksp of 3.2 x 10 -7 in M and g/L in a solution of 0.010 M Na 2 SO 4. l Calculate the solubility of SrSO 4, with a Ksp of 3.2 x 10 -7 in M and g/L in a solution of 0.010 M SrNO 3.

32. pH and solubility l OH - can be a common ion. l More soluble in acid. l For other anions if they come from a weak acid they are more soluble in a acidic solution than in water. l CaC 2 O 4 Ca +2 + C 2 O 4 -2 l H + + C 2 O 4 -2 HC 2 O 4 - l Reduces C 2 O 4 -2 in acidic solution.

33. Precipitation l Ion Product, Q =[M + ] a [Nm - ] b l If Q>Ksp a precipitate forms. l If Q<Ksp No precipitate. l If Q = Ksp equilibrium. l A solution of 750.0 mL of 4.00 x 10 -3 M Ce(NO 3 ) 3 is added to 300.0 mL of 2.00 x 10 -2 M KIO 3. Will Ce(IO 3 ) 3 (Ksp= 1.9 x 10 -10 M)precipitate and if so, what is the concentration of the ions?

34. Selective Precipitations l Used to separate mixtures of metal ions in solutions. l Add anions that will only precipitate certain metals at a time. l Used to purify mixtures. l Often use H 2 S because in acidic solution Hg +2, Cd +2, Bi +3, Cu +2, Sn +4 will precipitate.

35. Selective Precipitation l In Basic adding OH - solution S -2 will increase so more soluble sulfides will precipitate. l Co +2, Zn +2, Mn +2, Ni +2, Fe +2, Cr(OH) 3, Al(OH) 3

36. Selective precipitation l Follow the steps first with insoluble chlorides (Ag, Pb, Ba) l Then sulfides in Acid. l Then sulfides in base. l Then insoluble carbonate (Ca, Ba, Mg) l Alkali metals and NH 4 + remain in solution.

37. Complex ion Equilibria l A charged ion surrounded by ligands. l Ligands are Lewis bases using their lone pair to stabilize the charged metal ions. l Common ligands are NH 3, H 2 O, Cl -,CN - l Coordination number is the number of attached ligands. l Cu(NH 3 ) 4 +2 has a coordination # of 4

38. The addition of each ligand has its own equilibrium l Usually the ligand is in large excess. l And the individual K’s will be large so we can treat them as if they go to equilibrium. l The complex ion will be the biggest ion in solution.

39. l Calculate the concentrations of Ag +, Ag(S 2 O 3 ) -, and Ag(S 2 O 3 ) -3 in a solution made by mixing 150.0 mL of AgNO3 with 200.0 mL of 5.00 M Na 2 S 2 O 3 l Ag + + S 2 O 3 -2 Ag(S 2 O 3 ) - K 1 =7.4 x 10 8 l Ag(S 2 O 3 ) - + S 2 O 3 -2 Ag(S 2 O 3 ) -3 K 2 =3.9 x 10 4