1. Chapter 17 Acids, Bases and Buffers

2. Overview strong acid : strong base strong acid : weak base weak acid : strong base weak acid : weak base common ion effect buffers Henderson-Hasselbalch Equation titration curves

3. strong acid : strong base at the equivalence point (OH - = H + ) the pH = 7 strong acid : weak base at the equivalence point, pH < 7 weak acid : strong base at the equivalence point, pH > 7 weak acid : weak base at the equivalence point, pH depends on the relative value of the K a and the K b of the acid and base

4. example: weak acid : strong base solution H 2 O + HCN H 3 O + + CN - K a H 3 O + + OH - 2H 2 O 1/K w HCN + OH - H 2 O + CN - K a /K w K a /K w is large so rxn. will be, essentially, complete WASB

5. example: strong acid : weak base solution H 2 O + CN - HCN + OH - K b H 3 O + + OH - 2H 2 O 1/K w CN - + H 3 O + H 2 O + HCN K b /K w K b /K w is large so rxn. will be, essentially, complete SAWB

6. example: weak acid : weak base solution H 2 O + HCN H 3 O + + CN - K a H 2 O + NH 3 OH - + NH 4 + K b pH depends upon relative magnitude of K a and K b

7. Common Ion Effect: HCN + H 2 O H 3 O + + CN - Initial addition of CN - (as NaCN) shifts equilibrium, decreasing H 3 O + thereby increasing the pH

8. How does this affect the pH quantitatively? HCN + H 2 O H 3 O + + CN - 0.20 M 0 0.10 M initail -x+ x + x change 0.20 - x x0.10 + x equil. 4.9 x 10 -10 = (x)(0.10 + x) 0.20 - x x = 9.8 x 10 -10 M = [H 3 O + ] = pH = 9.0 (vs. 9.9 x 10 - 6 M = pH = 5.0 with no added NaCN)

9. Buffers: Resist change in pH on addition of small amounts of strong acid or base Composed of : A weak acid and the salt of its conj. base or A weak base and the salt of its conj. acid Most effective when pH is  1 of the pK a

10. HX H + + X - [H + ] = K a [HX] [X - ] adding OH - causes inc. in X - dec. in HX OH - + HX H 2 O + X - adding H + causes dec. in X - inc. in HX H + + X - HX as long as amt. of OH - or H + is small compared to HX & X -, ratio changes little

11. [H + ] = K a [HX] [X - ] -log [H + ] = - log K a -log [HX] [X - ] pH = pK a - log [HX] [X - ] pH = pK a + log [X - ] [HX] Henderson-Hasselbalch Equation

12. Henderson-Hasselbalch Equation: Calculates pH of buffer solutions Can be used only for buffer solutions Can be used only when the equilibrium approximation can be used pH = pK a + log ([X - ] / [HX]) where HX is the weak acid and X - is its conjugate base

13. Titration Curves: strong acid : strong base Calculation of pH after addition of aliquots of base HCl + NaOH  H 2 O + NaCl pH 50 mL 0.10 0 mL 0.10 M 1.0 “ 10 mL 1.2 “ 20 mL 1.4 “ 49 mL 3.0 “ 50 mL 7.0 “ 55 mL11.7 “ 80 mL12.4 “ 100 mL12.5 Note: large increase in pH near equivalence pt.

14. 12 10 8 6 4 2 50 Volume of Base, mL pH Equivalence Pt. 50 mL 0.10 M HCl titrated with 0.10 M NaOH

15. Titration Curves: Weak acid : strong base Calculation of pH after addition of aliquots of base HC 2 H 3 O 2 + NaOH  H 2 O + NaC 2 H 3 O 2 pH 50 mL 0.10 0 mL 0.10 M 2.9 “ 10 mL 4.1 “ 20 mL 4.6 “ 25 mL half equivalence point [ pH = pK a ] 4.7 “ 49 mL 6.4 “ 50 mL equivalence point 8.7 “ 55 mL11.7 “ 80 mL12.4 “ 100 mL12.5 Note: large increase in pH near equivalence pt.

16. 12 10 8 6 4 2 50 Volume of Strong Base, mL pH Equivalence Pt. 50 mL 0.10 M HC 2 H 3 O 2 titrated with 0.10 M NaOH half equivalence point = pK a 25

17. Strong Base Titrated with Strong Acid

19. 12 10 8 6 4 2 50 Volume of Strong Acid, mL pH Equivalence Pt. 50 mL 0.10 M NH 3 titrated with 0.10 M HCl half equivalence point = pK b 25

20. 12 10 8 6 4 2 200 Volume of Base, mL pH second equivalence pt. Diprotic acid, H 2 C 2 O 4, titrated with 0.10 M NaOH first equivalence point = pK a 100

22. Indicators: Indicators are generally weak acids HInd + H 2 O H 3 O + + Ind - K a Ind = [H 3 O + ][Ind - ] [HInd] [H 3 O + ] = [HInd] this ratio controls color K a [Ind - ]

24. Solubility Equilibria Solubility and Solubility Product Precipitation of Insoluble Salts K sp and Q Common Ion Effect and Solubility Simultaneous Equilibria Solubility and pH Solubility and Complex Ions Separations and Qualitative Analysis

25. Solubility Product Constant: The equilibrium constant expression for the solution of a solid AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] All rules for equilibrium constants and expressions apply Solubility Product

26. Examples: CaF 2(s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ][F - ] 2 Ag 3 PO 4(s) 3Ag + (aq) + PO 4 3- (aq) K sp = [Ag + ] 3 [PO 4 3- ]

27. Ca 2+ F - Ca 2+ solid CaF 2(s) dissolved Ca 2+ (aq) & F - (aq) ions CaF 2(s) Ca 2+ (aq) + 2F - (aq) The quantity of CaF 2(s) dissolved is reflected by the quantity of Ca 2+ (aq) ions in solution The Molar Solubility of CaF 2(s) is equal to the [Ca 2+ ] at the eq. point = a saturated solution

28. Calculate the K sp from experimental data: Prob: [Ba 2+ ] = 7.5 x 10 -3 M in saturated BaF 2. Calculate K sp. BaF 2(s) Ba 2+ (aq) + 2F - (aq) x 2x 7.5 x 10 -3 2(7.5 x 10 -3 ) K sp = (7.5 x 10 -3 )(15.0 x 10 -3 ) 2 = 1.6 x 10 -6

29. Estimating Solubility: Much like doing equilibrium problems: Determine the molar solubility of CaCO 3 if the K sp is 3.8 x 10 -9 at 25  C CaCO 3(s) Ca 2+ (aq) + CO 3 2- (aq) initial 0 0 change +x+x equil. x x x 2 = 3.8 x 10 -9 x = 6.2 x 10 -5 M = [Ca 2+ ] = [CO 3 2- ]

30. Prob: Calculate the molar solubility of Mg(OH) 2 if the K sp is 1.5 x 10 -11 Mg(OH) 2(s) Mg 2+ (aq) + 2OH - (aq) initial0 0 change+x +2x equil. x 2x (x)(2x) 2 = 4x 3 = 1.5 x 10 -11 x = 1.6 x 10 -4 M [Mg 2+ ] = 1.6 x 10 -4 M [OH - ] = 3.2 x 10 -4 M K sp = [Mg 2+ ][OH - ] 2 (x) (2x)

31. Which is more soluble in water? AgClor Ag 2 CrO 4 AgCl (s) Ag + (aq) + Cl - (aq) K sp = 1.8 x 10 -10 Ag 2 CrO 4(s) 2Ag + (aq) + CrO 4 2- (aq) K sp = 9.0 x 10 -12 molar solubility = 1.3 x 10 -5 M molar solubility = 1.3 x 10 -4 M Direct comparisons of K sp ’s can only be used if the ion ratios are the same

32. Which is more soluble in water? AgClK sp = 1.8 x 10 -10 or AgCNK sp = 1.2 x 10 -16 Mg(OH) 2 K sp = 1.5 x 10 -11 or Ca(OH) 2 K sp = 7.9 x 10 -9

33. Q, Reaction Quotient Q = K sp system is at equilibrium Q > K sp system not at equilibrium solid forms (rxn. shifts ) Q < K sp system not at equilibrium solid dissolves (rxn. shifts ) Prob: PbI 2(s) (K sp = 8.7 x 10 -9 ) placed in solution where [Pb 2+ ] = 1.1 x 10 -3 M. Is the solution saturated? PbI 2(s) Pb 2+ (aq) + 2I - (aq) Q = 5.3 x 10 -9 < K sp No, more will dissolve

34. Concentrations required for precipitation: Prob: What is the minimum conc. of I - that can cause precipitation of PbI 2 from a 0.050 M solution of Pb(NO 3 ) 2 ? K sp (PbI 2 ) = 8.7 x 10 -9. K sp = [Pb 2+ ][I - ] 2 8.7 x 10 -9 = [I - ] 2 0.050 [I - ] = 4.2 x 10 -4 M How much Pb 2+ remains when [I - ] = 0.0015 M [Pb 2+ ] = 8.7 x 10 -9 = 3.9 x 10 -3 M (0.0015) 2

35. K sp and Precipitation: Prob: You have 100.0 mL of 0.0010 M AgNO 3. Does AgCl precipitate if you add 5.0 mL of 0.025 HCl? AgCl (s) Ag + (aq) + Cl - (aq) AgNO 3(aq) Ag + (aq) + NO 3 - (aq) HCl (aq) + H 2 O H 3 O + (aq) + Cl - (aq) 1.0 x 10 -4 mol 1.25 x 10 -4 mol 0.105 L 0.105 L 9.5 x 10 -4 M 1.2 x 10 -3 M Q = (9.5 x 10 -4 )(1.2 x 10 -3 ) Q = 1.1 x 10 -6 > K sp will precipitate

36. Solubility and Common Ion Effect: CaF 2(s) Ca 2+ (aq) + 2F - (aq) Adding extra Ca 2+ or F - shifts equilibrium causing a decrease in solubility of CaF 2(s) shift toward solid CaF 2

37. Molar Solubility of CaF 2 (no added F - ): CaF 2(s) Ca 2+ (aq) + 2F - (aq) initial 0 0 change +x +2x equil. x 2x x = 2.1 x 10 -4 M = [Ca 2+ ] K sp = 3.9 x 10 -11 = (x)(2x) 2 molar solubility is 2.1 x 10 -4 moles CaF 2 / L

38. Molar Solubility of CaF 2 (with 0.010 M NaF): CaF 2(s) Ca 2+ (aq) + 2F - (aq) initial 0 0.010 change +x +2x equil. x 0.010 + 2x K sp = 3.9 x 10 -11 = (x)(0.010 + 2x) 2 x = 3.9 x 10 -7 M = [Ca 2+ ] molar solubility is 3.9 x 10 -7 moles CaF 2 / L with added NaF which suppresses solubility

39. What happens if acid is added to CaF 2 ? CaF 2(s) Ca 2+ (aq) + 2F - (aq) H 3 O + + F - HF + H 2 O SAWB The second rxn. has the effect of removing F - from the first equilibrium, affecting the solubility of CaF 2 H + addition shifts rxn, increasing solubility complete reaction

40. Which of the following would be more soluble in acid solution? PbCl 2 CaCO 3 Mg(OH) 2 the stronger the conj. base the more soluble if the anion is a hydrolyzing conjugate base, the stronger base it is, the more soluble the salt is in acid solution

41. Cl - + H 2 O HCl + OH - K very small CO 3 2- + H 2 O HCO 3 - + OH - K b 2.1 x 10 -4 OH - + H 2 O H 2 O + OH - K very large

42. For example: CaCO 3(s) Ca 2+ (aq) + CO 3 2- (aq) K sp 3.8 x 10 -9 CO 3 2- (aq) + H 2 O HCO 3 - + OH - K b 2.1 x 10 -4 OH - + H 3 O + 2H 2 O K w -1 1.0 x 10 14 CaCO 3(s) + H 3 O + Ca 2+ + HCO 3 - + H 2 O K = 79.8

43. Which of the following would be more soluble in acid solution? Ca 10 (PO 4 ) 6 (OH) 2 Ca 10 (PO 4 ) 6 F 2 hydroxy apatite - tooth enamel fluoro apatite - fluoridated tooth enamel

44. Formation of Complex Ions: Complex ions are large, polyatomic ions Most complex ions have large K values Formation of complex ions can affect solubility of some salts K f is the formation constant Most transition metals form stable complex ions- the transition metal is a Lewis Acid

45. Ag + + 2NH 3(aq) Ag(NH 3 ) 2 + K f 1.7 x 10 7 x M0.20 M (0.010 - x) M K = 1.7 x 10 7 = [Ag(NH 3 ) 2 + ] = 0.010 [Ag + ][NH 3 ] 2 (x)(0.20) 2 For example: Ex. 17.14 Calc. [Ag + ] present at eq. when conc. NH 3 added to 0.010 M AgNO 3 to give eq. [NH 3 ] = 0.20 M. Neglect vol. change. x = 1.5 x 10 -8 M = [Ag + ]

46. Selective Precipitation: Prob: 0.050 M Mg 2+ & 0.020 M Cu 2+. Which will ppt first as OH - is added? Mg(OH) 2 Mg 2+ + 2OH - K sp = 1.8 x 10 -11 Cu(OH) 2 Cu 2+ + 2OH - K sp = 2.2 x 10 -20 [OH - ] = (K sp / [M 2+ ]) 1/2 Cu(OH) 2

47. What concentration of OH - is necessary? [OH - ] = (K sp / [Cu 2+ ]) 1/2 = (2.2 x 10 -20 / 0.020) 1/2 = 1.0 x 10 -9 M [OH - ] = (K sp / [Mg 2+ ]) 1/2 = (1.8 x 10 -11 / 0.050) 1/2 = 1.9 x 10 -5 M takes less OH -

49. AgCl (s) Ag + (aq) + Cl - (aq) K sp 1.8 x 10 -10 Ag + + 2NH 3(aq) Ag(NH 3 ) 2 + K f 1.7 x 10 7 AgCl (s) + 2NH 3(aq) Ag(NH 3 ) 2 + + Cl - K = 3.1 x 10 -3 = [Ag(NH 3 ) 2 + ][Cl - ] [NH 3 ] 2 formation For example:

50. Prob: Does 100 mL of 4.0 M aqueous ammonia completely dissolve 0.010 mol of AgCl suspended in 1.0 L of solution? AgCl (s) + 2NH 3(aq) Ag(NH 3 ) 2 + + Cl - [NH 3 ] 2 = (1.0 x 10 -2 )(1.0 x 10 -2 ) = (0.032) 1/ 2 = 0.18 M (3.1 x 10 -3 ) we have 0.4 moles of NH 3 available which is plenty to provide the 0.020 mol necessary to form the complex plus 0.16 mol to achieve an equilibrium concentration of 0.18 M [NH 3 ] K = 3.1 x 10 -3 = [Ag(NH 3 ) 2 + ] [Cl - ] [NH 3 ] 2 0.010 moles0.020 moles0.010 moles