2. Buffered Solutions Buffered solutions contain either:Buffered solutions contain either: –A weak acid and its salt –A weak base and its salt A solution that resists a change in pH when either hydroxide ions or protons are added. A solution that resists a change in pH when either hydroxide ions or protons are added.

3. Acid/Salt Buffering Pairs Weak Acid Formula of the acid Example of a salt of the weak acid Hydrofluoric HF KF – Potassium fluoride Formic HCOOH KHCOO – Potassium formate Benzoic C 6 H 5 COOH NaC 6 H 5 COO – Sodium benzoate Acetic CH 3 COOH NaH 3 COO – Sodium acetate Carbonic H 2 CO 3 NaHCO 3 - Sodium bicarbonate Propanoic HC 3 H 5 O 2 NaC 3 H 5 O 2 - Sodium propanoate Hydrocyanic HCN KCN - potassium cyanide The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

4. Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO 3 ) Base Formula of the base Example of a salt of the weak acid Ammonia NH 3 NH 4 Cl - ammonium chloride Methylamine CH 3 NH 2 CH 3 NH 2 Cl – methylammonium chloride Ethylamine C 2 H 5 NH 2 C 2 H 5 NH 3 NO 3 - ethylammonium nitrate Aniline C 6 H 5 NH 2 C 6 H 5 NH 3 Cl – aniline hydrochloride Pyridine C 5 H 5 N C 5 H 5 NHCl – pyridine hydrochloride

5. Titration of an Unbuffered Solution A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH

6. Titration of a Buffered Solution A solution that is 0.10 M CH 3 COOH and 0.10 M NaCH 3 COO is titrated with 0.10 M NaOH

7. Comparing Results Unbuffered Buffered  In what ways are the graphs different?  In what ways are the graphs similar?

8. Comparing Results Buffered Unbuffered

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

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

11. This is called the Henderson-Hasselbalch Equation

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

13. Prove they’re buffers What would the pH be if.020 mol of HCl is added to 1.0 L of both of the following solutions. –0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) –0.25 M NH 3 and 0.40 M NH 4 Cl (Kb = 1.8 x 10 -5 ) What would the pH be if 0.050 mol of solid NaOH is added to each of the proceeding.

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

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

17. Weak Acid/Strong Base Titration A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH Endpoint is above pH 7

18. Strong Acid/Strong Base Titration A solution that is 0.10 M HCl is titrated with 0.10 M NaOH Endpoint is at pH 7

19. Strong Base/Strong Acid Titration A solution that is 0.10 M NaOH is titrated with 0.10 M HCl Endpoint is at pH 7 It is important to recognize that titration curves are not always increasing from left to right.

20. Weak Base/Strong Acid Titration

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

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

23. Selection of Indicators

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

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

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

27. K sp Values for Some Salts at 25  C NameFormulaK sp Barium carbonate BaCO 3 2.6 x 10 -9 Barium chromate BaCrO 4 1.2 x 10 -10 Barium sulfate BaSO 4 1.1 x 10 -10 Calcium carbonate CaCO 3 5.0 x 10 -9 Calcium oxalate CaC 2 O 4 2.3 x 10 -9 Calcium sulfate CaSO 4 7.1 x 10 -5 Copper(I) iodide Cu I 1.3 x 10 -12 Copper(II) iodate Cu( I O 3 ) 2 6.9 x 10 -8 Copper(II) sulfide CuS 6.0 x 10 -37 Iron(II) hydroxide Fe(OH) 2 4.9 x 10 -17 Iron(II) sulfide FeS 6.0 x 10 -19 Iron(III) hydroxide Fe(OH) 3 2.6 x 10 -39 Lead(II) bromide PbBr 2 6.6 x 10 -6 Lead(II) chloride PbCl 2 1.2 x 10 -5 Lead(II) iodate Pb( I O 3 ) 2 3.7 x 10 -13 Lead(II) iodide Pb I 2 8.5 x 10 -9 Lead(II) sulfate PbSO 4 1.8 x 10 -8 NameFormulaK sp Lead(II) bromide PbBr 2 6.6 x 10 -6 Lead(II) chloride PbCl 2 1.2 x 10 -5 Lead(II) iodate Pb( I O 3 ) 2 3.7 x 10 -13 Lead(II) iodide Pb I 2 8.5 x 10 -9 Lead(II) sulfate PbSO 4 1.8 x 10 -8 Magnesium carbonate MgCO 3 6.8 x 10 -6 Magnesium hydroxide Mg(OH) 2 5.6 x 10 -12 Silver bromate AgBrO 3 5.3 x 10 -5 Silver bromide AgBr 5.4 x 10 -13 Silver carbonate Ag 2 CO 3 8.5 x 10 -12 Silver chloride AgCl 1.8 x 10 -10 Silver chromate Ag 2 CrO 4 1.1 x 10 -12 Silver iodate Ag I O 3 3.2 x 10 -8 Silver iodide Ag I 8.5 x 10 -17 Strontium carbonate SrCO 3 5.6 x 10 -10 Strontium fluoride SrF 2 4.3 x 10 -9 Strontium sulfate SrSO 4 3.4 x 10 -7 Zinc sulfide ZnS 2.0 x 10 -25

28. Solving Solubility Problems For the salt AgI at 25  C, K sp = 1.5 x 10 -16 AgI(s)  Ag + (aq) + I - (aq) I C E O O +x x x 1.5 x 10 -16 = x 2 x = solubility of AgI in mol/L = 1.2 x 10 -8 M

29. Solving Solubility Problems For the salt PbCl 2 at 25  C, K sp = 1.6 x 10 -5 PbCl 2 (s)  Pb 2+ (aq) + 2Cl - (aq) I C E O O +x +2x x 2x 1.6 x 10 -5 = (x)(2x) 2 = 4x 3 x = solubility of PbCl 2 in mol/L = 1.6 x 10 -2 M

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

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

32. Solving Solubility with a Common Ion For the salt AgI at 25  C, K sp = 1.5 x 10 -16 What is its solubility in 0.05 M NaI? AgI(s)  Ag + (aq) + I - (aq) I C E 0.05 O +x 0.05+x x 1.5 x 10 -16 = (x)(0.05+x)  (x)(0.05) x = solubility of AgI in mol/L = 3.0 x 10 -15 M

33. Precipitation and Qualitative Analysis

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

35. Precipitation Ion Product, Q =[M + ] a [Nm - ] b If Q>Ksp a precipitate forms. If Q<Ksp No precipitate. If Q = Ksp equilibrium. 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?

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

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

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

39. Complex Ions A Complex ion is a charged species composed of: 1. A metallic cation 2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation

40. NH 3, CN -, and H 2 O are Common Ligands

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

42. Coordination Number  Coordination number refers to the number of ligands attached to the cation  2, 4, and 6 are the most common coordination numbers Coordination number Example(s) 2Ag(NH 3 ) 2 + 4CoCl 4 2- Cu(NH 3 ) 4 2+ 6Co(H 2 O) 6 2+ Ni(NH 3 ) 6 2+

43. Complex Ions and Solubility AgCl(s)  Ag + + Cl - K sp = 1.6 x 10 -10 Ag + + NH 3  Ag(NH 3 ) + K 1 = 1.6 x 10 -10 Ag(NH 3 ) + NH 3  Ag(NH 3 ) 2 + K 2 = 1.6 x 10 -10 AgCl + 2NH 3  Ag(NH 3 ) 2 + + Cl - K = K sp  K 1  K 2