1. Ch 17: Additional Aspects of Aqueous Equilibria Brown, LeMay Ch 17 AP Chemistry

2. 2 17.1: Common Ion Effect Addition of a “common ion”: solubility of solids decrease because of Le Châtelier’s principle. Ex: AgCl (s) ↔ Ag + (aq) + Cl - (aq) Addition of Cl - shifts equilibrium toward solid

3. 3 17.4: Solubility Equilibria Dissolving & precipitating of salts Solubility rules discussed earlier are generalized qualitative observations of quantitative experiments. Ex:PbCl 2 (s) ↔ Pb 2+ (aq) + 2 Cl - (aq) K sp = [Pb 2+ ][Cl - ] 2 = 1.6 x 10 -5 K sp = solubility-product constant (found in App. D) Recall that both aqueous ions and solid must be present in solution to achieve equilibrium Changes in pH will affect the solubility of salts composed of a weak acid or weak base ion.

4. 4 Molar solubility: moles of salt per liter of solution in a saturated solution Ex: What is the molar solubility of Ag 2 SO 4 at 25ºC? What mass of Ag 2 SO 4 does that represent per liter? K sp (Ag 2 SO 4 ) = 1.4 x 10 -5 ; MW Ag 2 SO 4 = 311.80 g Ag 2 SO 4 (s) ↔ 2 Ag + (aq) + SO 4 -2 (aq) K sp = [Ag + ] 2 [SO 4 -2 ] Ag 2 SO 4 (s)Ag + (aq)SO 4 2- (aq) Initial Change Equilibrium K sp = [Ag + ] 2 [SO 4 -2 ] = (2x) 2 (x) = 4x 3 x = (1.4 x 10 -5 /4) 1/3 = 0.015 M (-x) X + 2x M+ x M 2x Mx M X 0 M = [SO 4 2- ] = [Ag 2 SO 4 ]

5. 5 Ex: What mass of Ag 2 SO 4 does that represent per L? (MW Ag 2 SO 4 = 311.80 g, molar solubility, [Ag 2 SO 4 ] = 0.015 M) 4.7 g Ag 2 SO 4

6. Ex: 3.0 x 10 -5 g Pb 3 (AsO 4 ) 2 will dissolve in 1.0 L water to make a saturated solution. What is K sp of Pb 3 (AsO 4 ) 2 ? (MW of Pb 3 (AsO 4 ) 2 = 899.44 g/mol) Pb 3 (AsO 4 ) 2 (s) ↔ 3 Pb +2 (aq) + 2 AsO 4 -3 (aq) K sp = [Pb 2+ ] 3 [AsO 4 -3 ] 2 [Pb 3 (AsO 4 ) 2 ] = (3.0 x 10 -5 g)/(899.44 g/mol)/(1.0 L) = 3.3 x 10 -8 M Pb 3 (AsO 4 ) 2 = x K sp =[Pb 2+ ] 3 [AsO 4 -3 ] 2 = [3(3.3 x 10 -8 )] 3 [2(3.3 x 10 -8 )] 2 = = 4.5 x 10 -36 Pb 3 (AsO 4 ) 2 (s)Pb 2+ (aq)AsO 4 3- (aq) Initial Change Equilibrium (-x) X + 3x M+ 2x M 3x M2x M X 0 M

7. 7 Ex: What is the molar solubility of CaF 2 in a 1.0 L solution of 0.10 M Ca(NO 3 ) 2 ? K sp (CaF 2 ) = 3.9 x 10 -11 CaF 2 (s) ↔ Ca 2+ (aq) + 2 F - (aq) CaF 2 (s)Ca 2+ (aq)F - (aq) Initial Change Equilibrium (-x) X + x M+ 2x M (0.10 + x) M2x M Assume x is negligible: x = 9.9x10 -6 moles of CaF 2 are soluble per L. X 0.10 M0 M

8. 8  Predicting whether a precipitate (ppt) will form depends on concentration of ions present and K sp of all likely salts. Compare the ion-product, Q, with K sp Q has the same mathematical form as K sp Q must be calculated for each example If: Q > K sp precipitation occurs (continues until Q = K sp ) Q = K sp solution is at equilibrium Q < K sp solid dissolves (continues until Q = K sp )

9. 9 Ex: Will a precipitate form when 0.10 L of 3.0 x 10 -3 M Pb(NO 3 ) 2 is added to 0.400 L of 5.0 x 10 -3 M Na 2 SO 4 ? Possible new salts: PbSO 4 (K sp = 1.6 x 10 -8 ) NaNO 3 (K sp = 112.12) Q = [Pb 2+ ][SO 4 2- ] [Pb 2+ ] = (3.0 x 10 -3 M)(0.10 L) / (0.10 L + 0.400 L) = 3.0 x 10 -4 mol / 0.50 L = 6.0 x 10 -4 M Pb 2+ [SO 4 2- ] = (5.0 x 10 -3 M)(0.400 L) / (0.10 L + 0.400 L) = 2.0 x 10 -3 mol / 0.50 L = 4.0 x 10 -3 M SO 4 2- Q = [Pb 2+ ][SO 4 2- ] = (6.0x10 -4 )(4.0x10 -3 ) = 2.4 x 10 -6 ٭ Since Q > K sp, PbSO 4 will precipitate.

10. 10 Ex: Will a precipitate form if 20.0 mL of 3.50 x 10 -3 M Hg 2 (NO 3 ) 2 solution are mixed with 40.0 mL of 2.00 x 10 -3 M NaCl solution? K sp (Hg 2 Cl 2 ) = 1.3 x 10 -18 [Hg 2 2+ ] = (3.50x10 -3 M)(0.0200 L)/(0.0200 L + 0.0400 L) = 0.00117 M [Cl - ] = (2.00 x 10 -3 M)(0.0400 L)/(0.0200 L + 0.0400 L) = 0.00133 M Q = [Hg 2 2+ ][Cl - ] 2 Q = (.00117) (.00133) 2 = 2.07 x 10 -9 ٭ Since Q > K sp, there will be a precipitate (Hg 2 Cl 2 )

11. 11 17.2: Buffers: ٭ Solutions that resist drastic changes in pH upon additions of small amounts of acid or base. Consist of a weak acid and its conjugate base (usually in salt form) Ex: acetic acid and sodium acetate: HC 2 H 3 O 2 + NaC 2 H 3 O 2 Or consist of a weak base and its conjugate acid (usually in salt form) Ex: ammonia and ammonium chloride: NH 3 + NH 4 Cl

12. 12 Consider a weak acid/salt buffer: HA (aq) ↔ H + (aq) + A - (aq) ٭ Thus, the pH of a buffer is determined by K a of weak acid and the ratio of the concentrations of an acid & its conjugate base.

13. 13 Henderson-Hasselbalch equation: Use for all buffer solutions! Lawrence Henderson (1878-1942) Karl Hasselbalch (1874-1962)

14. 14 Why/how do buffers work? ٭ A small addition of [H + ] will decrease the [A-] but increase the [HA]: H + (aq) + A - (aq) ↔ HA (aq) pH decreases, but  pH will be small ٭ A small addition of [OH - ] will decrease the [HA] but increase the [A-]: OH - (aq) + HA (aq) ↔ H 2 O (l) + A - (aq) pH increases, but  pH will be small

15. 15 ٭ In general, when choosing a buffer, select one in which the acid form has a pK a close to the desired pH. ٭ If [base] > [acid], then pH > pK a ٭ If [base] < [acid], then pH < pK a ٭ If [base] = [acid], then pH = pK a and [H + ] = K a

16. 16 Buffer capacity: ٭ Amount of acid or base a buffer can neutralize before the pH begins to change “significantly”. The greater the concentrations of acid & base in the buffer, the greater the buffer capacity. A “typical” buffer solution can hold the pH to ± 1.00 ٭ Buffer solutions can also be considered common- ion solutions.

17. 17 Ex: Calculate the pH of a buffer made by adding 0.300 mol acetic acid and 0.300 mol sodium acetate to enough water to make 1.00 L. (K a = 1.8 x 10 -5 ) CH 3 COOH (aq) ↔ CH 3 COO - (aq) + H + (aq) CH 3 COOH (aq) CH 3 COO - (aq)H + (aq) I 0.300 M 0 M C E -x M (0.300 – x) M + x M (0.300 + x) M x M

18. 18 Or, use Henderson-Hasselbalch: Assumes that x is negligible

19. 19 Ex: Calculate the pH of this buffer after the addition of 0.020 mol NaOH (assume volume remains constant.) 1. Neutralization: calculate the concentrations after the initial reaction of the strong base with the weak acid part of the buffer. CH 3 COOH (aq) + OH - (aq) → CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) OH - (aq) CH 3 COO - (aq) I 0.300 mol0.020 mol0.300 mol C F -0.020 mol 0.280 mol - 0.020 mol+ 0.020 mol 0 mol 0.320 mol Then, consider how these amounts behave in equilibrium!

20. 20 2. Equilibrium: CH 3 COOH (aq) ↔ CH 3 COO - (aq) + H + (aq) CH 3 COOH (aq) CH 3 COO - (aq) H + (aq) I 0.280 mol0.320 mol0 mol C E (0.280 – x) mol (0.320 + x) mol x mol -x mol+ x mol

21. 21 Or, use Henderson-Hasselbalch: The pH increased, but not by much.

22. 22 Ex: Calculate the pH of the original buffer after the addition of 0.020 mol HCl (assume volume remains constant.) 1. Neutralization: calculate the concentrations after the initial reaction of the strong acid with the weak conjugate base. CH 3 COO - (aq) + H + (aq) → CH 3 COOH (aq) CH 3 COO - (aq) H + (aq) CH 3 COOH (aq) I 0.300 mol0.020 mol0.300 mol C F -0.020 mol 0.280 mol - 0.020 mol+ 0.020 mol 0 mol 0.320 mol Then, consider how these amounts behave in equilibrium!

23. 23 2. Equilibrium: CH 3 COOH (aq) ↔ CH 3 COO - (aq) + H + (aq) CH 3 COOH (aq) CH 3 COO - (aq) H + (aq) I 0.320 mol0.280 mol0 mol C E (0.320 – x) mol (0.280 + x) mol x mol -x mol+ x mol

24. 24 Or, use Henderson-Hasselbalch: The pH decreased, but not by much.

25. 25 17.3: Acid-Base Titrations Titration curve: graph of pH vs. volume of titrant added ٭ Each type of titration has a uniquely shaped titration curve Every titration consists of four regions: 1. The initial pH 2. Between the initial pH and the equivalence point: pH is determined by amount of solution not yet neutralized 3. The equivalence point: moles of acid = moles of base, leaving a solution of the salt produced in the acid-base neutralization 4. After the equivalence point: pH is determined by amount of excess titrant

26. 26 Strong Acid – Strong Base Titration Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl HCl (aq) + NaOH (aq) → NaCl (aq) + H 2 O (l) 14 pH 7 0 0 50 Volume of NaOH added (mL) Region 1: Initial pH pH starts very low (here, pH = -log 0.100 = 1) because a strong acid is being titrated.

27. 27 Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl 14 pH 7 0 0 50 Volume of NaOH added (mL) Region 2: Between initial pH and equivalence point As base is added, pH increases slowly, then rapidly as it approaches the equivalence point. Before equivalence point, moles H + > moles OH - added; pH is determined by amount of solution not yet neutralized

28. Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl 14 pH 7 0 0 50 Volume of NaOH added (mL) moles H + from HCl = moles OH - added Region 3: Equivalence point Reached when moles H + = moles OH - added; here, at pH = 7 because only species present are H 2 O and the salt (Na +1 and Cl -1 ), which do not hydrolyze to produce an acidic or basic solution, and at V = 50 mL because M A V A = M B V B = (0.100)(.0500) = (0.100)V B Endpoint: point at which an indicator changes color; an approximation of the actual equivalence point

29. Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl 14 pH 7 0 0 50 Volume of NaOH added (mL) Region 4: After equivalence point  As base is added, pH rises rapidly, then continues to increase slowly.  Moles H + < moles OH - added  pH eventually approaches an upper limit as base is added; here at pH = 13 (-log [OH - ] = 1).

30. 30 pH Indicators: ٭ Chemicals with acid and base forms with significant color differences For a titration, choose an indicator with a range near the equivalence point * Methyl red: red (HMer, acid form) to yellow (Mer -, base form) at pH ≈ 4.2 – 6.3 Phenolphthalein: colorless (HPh: acid form above) to pink (Ph -, base form) at pH ≈ 8.5-9.5

31. 31 pH Indicators and Titrations ٭ Phenophthalein is often used as an indicator in the titration of a strong acid with a strong base since it turns pink around pH 9, only a small addition of titrant beyond the equivalence point at pH 7.

32. 32 Weak Acid – Strong Base Ex: 0.100 M NaOH (aq) is added to 50.0 mL of 0.100 M CH 3 COOH (aq) Equivalence point: M A V A = M B V B (0.100 M)(0.0500 L) = (0.100 M)(V B ) V B = 0.0500 L NaOH Equation 1: Neutralization: reaction goes to completion CH 3 COOH (aq) + OH - (aq) → H 2 O (l) + CH 3 COO - (aq) Equation 2: Equilibrium: extent of reaction based on K a CH 3 COOH (aq) ↔ H + (aq) + CH 3 COO - (aq)

33. 33 Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH Region 1: Initial pH pH starts below pH 7 because a weak acid is being titrated. 14 pH 7 0 0 50 Volume of NaOH added (mL)

34. 34 Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH Region 2: Between initial pH and equivalence point Moles CH 3 COOH > moles OH - added pH is determined by considering extent of neutralization and equilibrium reactions 14 pH 7 0 0 50 Volume of NaOH added (mL)

35. 35 Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH Region 3: Equivalence point Equivalence point at pH > 7 because only species present are H 2 O and the salt (Na + and CH 3 COO - ); acetate ion is a weak base and hydrolyzes to produce a basic solution. 14 pH 7 0 0 50 Volume of NaOH added (mL) moles H + from CH 3 COOH = moles OH - added

36. 36 Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH Region 4: After equivalence point  Moles acid < moles base added  pH is determined by concentration of excess base in solution (as in strong acid/strong base); equilibrium can be ignored 14 pH 7 0 0 50 Volume of NaOH added (mL)

37. 37 Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH; calculate pH at regular intervals of titration (K a = 1.8 x 10 -5 ): (similar to Exercise 17.7, p. 637) After 30.0 mL of 0.100 M NaOH has been added: mol OH - = (0.100 M)(0.0300 L) = 3.00 x 10 -3 mol OH - mol CH 3 COOH = (0.100 M)(0.0500 L) = 5.00 x 10 -3 mol CH 3 COOH (aq) + OH - (aq ) → CH 3 COO - (aq) +H 2 O I 5.00 x 10 -3 mol3.00 x 10 -3 mol0 C F - 3.00 x 10 -3 mol 2.00 x 10 -3 mol - 3.00 x 10 -3 mol+ 3.00 x 10 -3 mol 0 3.00 x 10 -3 mol [CH 3 COOH] = (2.00 x 10 -3 mol)/(0.0800 L) = 0.0250 M [CH 3 COO - ] = (3.00 x 10 -3 mol)/(0.0800 L) = 0.0375 M

38. 38 After 30.0 mL of 0.100 M NaOH has been added: CH 3 COOH (aq) ↔ H + (aq) + CH 3 COO - (aq) I 0.0250 M00.0375 M C E - x 0.0250 - x + x 0.0375 + x Or use H-H since its a buffer solution:

39. 39 After 50.0 mL of 0.100 M NaOH has been added: ٭ Initial: mol OH - = (0.100 M)(0.0500 L) = 5.00 x 10 -3 mol OH - mol CH 3 COOH = (0.100 M)(0.0500 L) = 5.00 x 10 -3 mol CH 3 COOH (aq) + OH - (aq ) → CH 3 COO - (aq) +H 2 O I 5.00 x 10 -3 mol 0 C F [CH 3 COO - ] = (5.00 x 10 -3 mol)/(0.100 L) = 0.0500 M - 5.00 x 10 -3 mol 0 + 5.00 x 10 -3 mol 0 5.00 x 10 -3 mol

40. 40 After 50.0 mL of 0.100 M NaOH has been added: CH 3 COO - (aq) + H 2 O (l) ↔ OH - (aq) + CH 3 COOH (aq) I 0.0500 M00 C E - x 0.0500 - x + x x

41. 41 After 70.0 mL of 0.100 M NaOH is added: mol OH - = (0.100 M)(0.0700 L) = 7.00 x 10 -3 mol OH - mol CH 3 COOH = (0.100 M)(0.0500 L) = 5.00 x 10 -3 mol mol OH - in excess = (7.00 x 10 -3 – 5.00 x 10 -3 ) = 2.00 x 10 -3 mol OH - [OH - ] = (2.00 x 10 -3 mol)/(0.1200 L) = 0.0167 M OH - pOH = 1.778andpH = 12.22(pH + pOH = 14.00)

42. 42 Weak Base – Strong Acid Ex: 0.100 M HCl (aq) is added to 40.0 mL of 0.0750 M NH 3 (aq) Equivalence point: M A V A = M B V B (0.100 M)(V A ) = (0.0750 M)(0.0400 L) V A = 0.0300 L HCl Equation 1: Neutralization: reaction goes to completion NH 3 (aq) + H + (aq) → NH 4 + (aq) Equation 2: Equilibrium: extent of reaction based on K b NH 3 (aq) + H 2 O (l) ↔ OH - (aq) + NH 4 + (aq)

43. 43 Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH 3 (aq) Region 1: Initial pH pH starts above pH 7 because a weak base is being titrated. 14 pH 7 0 0 30 Volume of HCl added (mL)

44. 44 Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH 3 (aq) Region 2: Between initial pH and equivalence point Moles base > moles acid added pH is determined by considering extent of neutralization and equilibrium reactions. 14 pH 7 0 0 30 Volume of HCl added (mL)

45. 45 Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH 3 (aq) Region 3: Equivalence point Equivalence point at pH < 7 because only species present are H 2 O and the salt (NH 4 + and Cl - ); ammonium ion is a weak acid and hydrolyzes to produce an acidic solution. 14 pH 7 0 0 30 Volume of HCl added (mL) moles NH 3 = moles H + added

46. 46 Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH 3 (aq) Region 4: After equivalence point  Moles base < moles acid added  pH is determined by concentration of excess acid in solution (as in strong acid/strong base); equilibrium can be ignored 14 pH 7 0 0 30 Volume of HCl added (mL)

47. 47 Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH 3 (aq); calculate pH at regular intervals of titration. (K b = 1.8 x 10 -5 ) After 15.0 mL of 0.100 M HCl has been added: mol H + = (0.100 M)(0.0150 L) = 1.50 x 10 -3 mol H + mol NH 3 (aq) = (0.0750 M)(0.0400 L) = 3.00 x 10 -3 mol NH 3 (aq) + H + (aq ) → NH 4 + (aq) I 3.00 x 10 -3 mol1.50 x 10 -3 mol0 C F - 1.50 x 10 -3 mol 1.50 x 10 -3 mol - 1.50 x 10 -3 mol+ 1.50 x 10 -3 mol 0 1.50 x 10 -3 mol [NH 3 ] = (1.50 x 10 -3 mol)/(0.0550 L) = 0.0273 M [NH 4 + ] = (1.50 x 10 -3 mol)/(0.0550 L) = 0.0273 M

48. 48 After 15.0 mL of 0.100 M HCl has been added: NH 3 (aq) + H 2 O (l) ↔ OH - (aq) + NH 4 + (aq) I 0.0273 M0 C E - x 0.0273 - x + x 0.0273 + x Or use H-H since its a buffer solution: Half-equivalence point: where half of base has been neutralized, producing ammonium; can be used to determine K b (see above)

49. 49 After 30.0 mL of 0.100 M HCl has been added: ٭ Initial: mol H + = (0.100 M)(0.0300 L) = 3.00 x 10 -3 mol H + mol NH 3 = (0.0750 M)(0.0400 L) = 3.00 x 10 -3 mol NH 3 (aq) + H + (aq ) → NH 4 + (aq) I 3.00 x 10 -3 mol 0 C F [NH 4 + ] = (3.00 x 10 -3 mol)/(0.0700 L) = 0.0429 M - 3.00 x 10 -3 mol 0 + 3.00 x 10 -3 mol 0 3.00 x 10 -3 mol

50. 50 After 30.0 mL of 0.100 M HCl has been added: NH 4 + (aq) ↔ H + (aq) + NH 3 (aq) I 0.0429 M00 C E - x 0.0429 - x + x x

51. 51 After 60.0 mL of 0.100 M HCl is added: mol H + = (0.100 M)(0.0600 L) = 6.00 x 10 -3 mol H + mol NH 3 = (0.0750 M)(0.0400 L) = 3.00 x 10 -3 mol mol H + in excess = (6.00 x 10 -3 – 3.00 x 10 -3 ) = 3.00 x 10 -3 mol H + [H + ] = (3.00 x 10 -3 mol)/(0.1000 L) = 0.0300 M H + pH = 1.523

52. 52 Titration of a Polyprotic Acid: Ex: 0.100 M NaOH (aq) is added to 50.0 mL of 0.100 M H 2 CO 3 (aq) Neutralization #1: H 2 CO 3 (aq) + OH - (aq) → HCO 3 - (aq) + H 2 O (l) Equivalence point #1: M A V A = M B V B (0.100 M H 2 CO 3 )(0.0500 L) = (0.100 M)(V B ) V B = 0.0500 L NaOH Neutralization #2: HCO 3 - (aq) + OH - (aq) → CO 3 2- (aq) + H 2 O (l) Equivalence point #2: M A V A = M B V B (0.100 M HCO 3 - )(0.0500 L) = (0.100 M)(V B ) V B = 0.0500 L NaOH (additionally)

53. 53 0.100 M NaOH (aq) is added to 50.0 mL of 0.100 M H 2 CO 3 (aq) 14 pH 7 0 0 50 100 Volume of NaOH added (mL) Eq pt #1: primarily HCO 3 - ; can be used to calculate orig. [H 2 CO 3 ] Half-eq. pt #1: mol H 2 CO 3 = mol HCO 3 - ; can be used to calculate K a1 Half-eq. pt #2: mol HCO 3 - = mol CO 3 2- ; can be used to calculate K a2 Eq pt #2: primarily CO 3 2- ; can be used to calculate orig. [H 2 CO 3 ]