1. Acid-Base Equilibria Chapter 15

2. 2 HF( aq ) + H 2 O( l ) H 3 O + ( aq ) + F - ( aq ) Addition of NaF will shift the equilibrium to the _______because of the addition of F -, which is already involved in the equilibrium reaction. A solution of HF and NaF is less acidic than a solution of HF alone. The presence of a common ion suppresses the ionization of a weak acid or a weak base. The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. left

3. A buffer solution is a solution of: 1.A weak acid and its salt or a weak base and its salt 2.A buffer must contain relatively large concentration of acid to react with added base (OH-) and must also contain similar concentration of base to react with added acid (H+). A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. 16.3

4. Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr (c) Na 2 CO 3 /NaHCO 3 (d) NaClO 4 /HClO 4 (e) NH 3 /NH 4 NO 3 Answer (a) HF is a weak acid and KF is its salt. Therefore, this is a buffer system. (b) HBr is a strong acid and hence this is not a buffer system. (c) NaHCO3 contains a weak acid (HCO3-) and Na2CO3 is a salt of weak acid. Therefore, this is a buffer system. (d) HClO4 is a strong acid and hence this is not a buffer system. (e) NH3 is a weak base and NH4NO3 is a salt of weak base, and therefore this is a buffer system.

5. 5 A buffered solution contains 0.50 M HC 2 H 3 O 2 (K a =1.8 x 10 -5 ) and 0.50 M NaC 2 H 3 O 2.Calculate the pH. List major species: HC 2 H 3 O 2, Na +, C 2 H 2 O 2 -, and H 2 O Adding more acid creates a shift left IF enough acetate ions are present CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) Consider an equal molar mixture of CH 3 COOH and NaCH 3 COO

6. 6 2. The acetic acid dissociation equilibrium will control the pH. HC 2 H 3 O 2 H + (aq) + C 2 H 3 O 2 - (aq) I.50.5 C-x+x+x E.5-xx.5+x Ka = 1.8x10 -5 = x(0.50 +x) = x(0.50) = 1.8x10 -5 0.50 0.50 pH = 4.74

7. 7 Calculate the change in pH when 0.010 mol solid NaOH is added to the 1.0 L of the buffered solution. 1.List major species: HC 2 H 3 O 2, Na +, C 2 H 2 O 2 -, OH -, and H 2 O (the strong base has an affinity for the p+, which will come from _____________) 2. OH - + HC 2 H 3 O 2 H 2 O + C 2 H 3 O 2 - the acetic acid Even though acetic acid is weak, NaOH is such a strong base the reaction will essentially go to completion.

8. 8 3.Stoichiometry first. OH- + HC 2 H 3 O 2 H 2 O + C 2 H 3 O 2 - before.010mol.5mol.5mol change.010mol-.010mol.5mol-.010mol.5mol+.010mol After0mol.49mol.51mol **note.010mol HC 2 H 3 O 2 is converted to C 2 H 3 O 2 - by the addition of NaOH

9. 9 Equilibrium problem 4.After reaction the major species HC 2 H 3 O 2, Na +, C 2 H 2 O 2 -, and H 2 O 5.HC 2 H 3 O 2 H + + C 2 H 3 O 2 -

10. 10 Ice Table HC 2 H 3 O 2 H + (aq) + C 2 H 3 O 2 - (aq) I.490.51 C-x+x+x E.49-xx.51+x

11. 11 ΔpH=.02 pH = 4.76-4.74=.02 )

12. 12 Solving Problems with Buffered Solutions

13. 13 Buffers

14. 14 Adding an Acid to a Buffer

15. 15 How Does a Buffer Work? 1.Contains large amounts of HA(weak acid) and A -. 2.When OH - added, weak acid is best source of p +. OH - + HA H 2 O + A - OH - are replaced by A - and the ratio?? Ka = [H + ][A-] [H+] = K a [HA] [HA] [A-] **pH is determined by the ratio [HA]/[A-] decreases

16. 16 Similar if you added H + ions. 1.Contains large amounts of HA(weak acid) and A -. 2.H + added, conjugate base A- is attracted to the p +. H + + A - HA H + are replaced by HA Ka = [H + ][A-] [H+] = K a [HA] [HA] [A-] **pH is determined by the ratio [HA]/[A-]

17. 17 Calculate the H + in a buffered solution of.10M HF. K a =7.2 x 10 -4 and.3M NaF is added. Take the –log of both sides [H+] = K a [HA] [A - ] [HF] [F - ] Another useful equation…. pH = pK a – log[HA] or, pH = pK a + log [A-] [A-] [HA]

18. 18 Henderson–Hasselbalch Equation For a particular buffering system (conjugate acid– base pair), all solutions that have the same ratio [A – ] / [HA] will have the same pH.

19. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (formic acid) HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x 16.2 Mixture of weak acid and conjugate base! K a for HCOOH = 1.8 x 10 -4 [H + ] [HCOO - ] K a = [HCOOH] x = 1.038 X 10 -4 pH = 3.98

20. OR…… Use the Henderson-Hasselbach equation Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] pK a = -log K a Henderson-Hasselbach equation 16.2 pH = pK a + log [conjugate base] [acid]

21. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = 3.77 + log [0.52] [0.30] = 4.01 16.2 Mixture of weak acid and conjugate base! K a for HCOOH = 1.8 x 10 -4

22. pH= 9.18 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Initial End 0.30 0.360 0.30 - x0.36 + xx 16.3 [NH 4 + ] [OH - ] [NH 3 ] K b = Change- x + x = 1.8 X 10 -5 (.36 + x)(x) (.30 – x) 1.8 X 10 -5 = 1.8 X 10 -5  0.36x 0.30 x = 1.5 X 10 -5 pOH = 4.82

23. pH = 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (M) end (M) 0.29 0.01 0.24 0.280.00.25 final volume = 80.0 mL + 20.0 mL = 100 mL 16.3 NH 4 + 0.36 M x 0.080 L = 0.029 mol /.1 L = 0.29 M OH - 0.050 x 0.020 L = 0.001 mol /.1 L = 0.01M NH 3 0.30 M x 0.080 = 0.024 mol /.1 L = 0.24M K a = = 5.6 X 10 -10 [H + ] [NH 3 ] [NH 4 + ] = 5.6 X 10 -10 [H + ] 0.25 0.28 [H + ] = 6.27 X 10 -10

24. = 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (M) end (M) 0.29 0.01 0.24 0.280.00.25 pH = 9.25 + log [0.25] [0.28] final volume = 80.0 mL + 20.0 mL = 100 mL 16.3

25. Chemistry In Action: Maintaining the pH of Blood 16.3

26. Titrations In a titration a solution of accurately known concentration is gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at the endpoint (hopefully close to the equivalence point) Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4.7

27. Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l) 16.4 100% ionization! No equilibrium

28. Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7): 16.4

29. Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): 16.4 H + (aq) + NH 3 (aq) NH 4 Cl (aq)

30. Acid-Base Indicators 16.5

31. pH 16.5

32. The titration curve of a strong acid with a strong base. 16.5

33. Which indicator(s) would you use for a titration of HNO 2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein 16.5

34. Finding the Equivalence Point (calculation method) Strong Acid vs. Strong Base –100 % ionized! pH = 7 No equilibrium! Weak Acid vs. Strong Base –Acid is neutralized; Need K b for conjugate base equilibrium Strong Acid vs. Weak Base –Base is neutralized; Need K a for conjugate acid equilibrium Weak Acid vs. Weak Base –Depends on the strength of both; could be conjugate acid, conjugate base, or pH 7

35. 35 Weak Acid - Strong Base Titration Step 1 -A stoichiometry problem - reaction is assumed to run to completion - then determine remaining species. Step 2 -An equilibrium problem - determine position of weak acid equilibrium and calculate pH.

36. 36 H + (aq) + OH - (aq) H 2 O(l) Millimole (mmol) since titrations are done in mL. A 1.0 M solution contains contains 1.0mmol of solute/mL solution. Number of mmol= volume (mL) x molarity

37. 37 Strong Acid-Strong Base Titration of 50.0mL of 0.200 M HNO 3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added. a. no NaOH added b. 10.0 mL NaOH c. 20.0 mL NaOH d. 50.0 mL NaOH e. 100.0 mL NaOH

38. 38 a)0 NaOH Strong acid goes to completion. 1)[H + ] =.200M and pH =.699

39. 39 Titration of 50.0mL of 0.200 M HNO 3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added. b) 10.0 mL NaOH Major species: H+, NO 3 -, Na+, OH-, and H 2 O H + + OH -  H 2 O Before 50.0 mL x 0.200M 10.0 mL x 0.100 M 10.0 mmol 1.00mmol After 9.0mmol 0

40. 40 c. 20.0 mL of NaOH H + + OH - H 2 O Before50.0 mL x 0.200M 20.0 mL x 0.100 M = 10.0 mmol =2.00mmol After=8.0mmol0 Titration of 50.0mL of 0.200 M HNO 3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added.

41. 41 d. 50.0 mL NaOH H + + OH -  H 2 O Before50.0 mL x 0.200M 50.0 mL x 0.100 M = 10.0 mmol =5.00mmol After=5.0mmol0 [H + ]= 5.0 mmol/100 mL =0.05M pH = 1.301 Titration of 50.0mL of 0.200 M HNO 3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added.

42. 42 e. 100 mL NaOH At this point the amount of NaOH that has been added is 100.0mL x 0.100M = 10.mmol Original amount of nitric acid was 50.0mL x 0.200 M = 10.0 mmol Look at what’s left in solution: Na +, NO 3 -, and H 2 O (none affect the pH) Equivalence point pH = 7.00 Titration of 50.0mL of 0.200 M HNO 3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added.

43. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 43

44. 44 d. 150.0 mL NaOH H + + OH -  H 2 O Before50.0 mL x 0.200M 150.0 mL x 0.100 M = 10.0 mmol =15.00mmol After=0.0mmol=5.0mmol [OH - ]= 5.0 mmol/200 mL =0.025M [H + ] [OH - ]= 1.0 x 10 -14 [H + ]= 1.0 x 10 -14 /.025 = 4.0 x 10-13M pH = 12.40 Titration of 50.0mL of 0.200 M HNO 3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added.

45. Exactly 100 mL of 0.10 M HNO 2 are titrated with 100 mL of a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) 0.01 0.0 0.01 NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0.050.00 -x-x+x+x 0.05 - x 0.00 +x+x xx [NO 2 - ] = 0.01 0.200 = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x2 0.05-x = 2.2 x 10 -11 0.05 – x  0.05x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

46. 46 Example: HCN K a =6.2 x 10 -10 If 50.0mL of.10M HCN is titrated with.100 M NaOH calculate the pH at the equivalence point and the half way point. Half way point, how many mmols of HCN? How many mmols of NaOH are needed?How many mL are needed? When does the pH = pK a ?

47. 47 At the halfway point: How many mmol of OH? How many mL of NaOH are needed? pH = pKa = -log(6.2 x 10 -10 ) = 9.21 Example: HCN K a =6.2 x 10 -10 If 50.0mL of.10M HCN is titrated w/.100 M NaOH calculate the pH at the equivalence point and the half way point. 50.0 MmL x 0.100M = 5.00 mmol HCN 2.5 mmol OH- is the halfway point Volume of NaOH = 25.0 mL

48. 48 At the equivalence point: How many mmol of OH? How many mL of NaOH are needed? Major species: CN -, Na +, H 2 O K b =K w /K a 1.6 x 10 -5 = [HCN][OH-] [CN-] CN - + H 2 O  HCN + OH - [OH-] = x = 8.9 x 10 -4 [H+] = 1.1 x 10 -11 pH = 10.96 Example: HCN K a =6.2 x 10 -10 If 50.0mL of.10M HCN is titrated w/.100 M NaOH calculate the pH at the equivalence point and the half way point. 5.00 mmol OH - which is 50.0 mL

49. 49 Weak acid and a strong base Reacts essentially to completion The equivalence point is always greater than 7. At the half way point [H+] = K a pH = pK a It is the amount of acid, not the strength that determines the equivalence point. pH values at equivalence point is effected by acid strength Acid strength effects the shape of the curve.

50. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 50 The weaker the Acid the greater The pH at the Equivalence point.

51. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 51 Acid-Base Indicator... marks the end point of a titration by changing color. The equivalence point is not necessarily the same as the end point.

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

53. 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

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

55. Look out Solubility is not the same as solubility product. Solubility product is an equilibrium constant. (amount nor size of particle matters) (it doesn’t change except with temperature.) Solubility is an equilibrium position for how much can dissolve. A common ion can change this.

56. Calculating Solubility The solubility is determined by equilibrium. Its an equilibrium problem. Watch the coefficients

57. 57 Copper(I) bromide has a measured solubility of 2.0 x 10 -4 mol/L at 25C°. Calculate the K sp. Write the equation. What is the K sp equation? What are the initial and final concentrations? CuBr(s)  Cu + (aq) + Br - (aq) K sp = [Cu + ][Br - ] Initial : [Cu+] 0 =0 [Br-] 0 =0

58. 58 [Cu+] = [Cu + ] 0 + change to reach equilibrium [Br-] = [Br-] 0 + change to reach equilibrium Solubility of CuBr(s) is 2.0 x 10 -4 mol/L. 0 + 2.0 x 10 -4 mol/L of each ion. K sp = [Cu + ][Br - ] K sp = (2.0 x 10 -4 m/L)(2.0 x 10 -4 mol/L) Units are usually omitted =(4.0 x 10 -8 ) CuBr(s)  Cu + (aq) + Br - (aq)

59. 59 Solubility Product Calculate the K sp value for bismuth sulfide (Bi 2 S 3 ), which has a solubility of 1.0 x 10 -15 mol/L at 25C. Hint**coefficients** Ksp = [Bi 3+ ] 2 [S 2- ] 3 1.0 x 10 -15  [0 + 2(1.0 x 10 -15 )] + [0 + 3(1.0 x 10 -15 )] Ksp = [2.0 X 10 -15 ] 2 [3.0 X 10 -15 ] 3 1.1 x 10 -73

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

61. Common Ion Effect If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve.

62. 62 Common Ion Calculate the solubility of solid CaF 2 (K sp =4.0 x 10 -11 ) in a 0.025 M NaF solution. CaF 2  Ca 2+ (aq) + 2F - (aq) K sp =4.0 x 10 -11 = [Ca 2+ ] [F - ] 2 [Ca 2+ ] 0 = 0 [Ca 2+ ] = x [F - ] 2 0 = 0.025M [F - ] 2 =.025 + 2x K sp =4.0 x 10 -11 = [Ca 2+ ] [F - ] 2 (x)(.025 + 2x) 2 =(x)(.025) 2 x = 6.4 x 10 -8 Approximation is ok.

63. 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 an acidic solution than in water because they produce strong conjugate bases.. 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.

64. General Rule for pH and solubility If the anion X - is an effective base (HX is a weak acid) the salt MX will show increased solubility in an acidic solution. OH -, S 2-, CO 3 2-, C 2 O 4 2-, and CrO 4 2-

65. Precipitation  Ion Product, Q =[M + ] a [Nm - ] b  Q = [Ca 2+ ] 0 [F - ] 0 2  If Q>K sp a precipitate forms.  If Q<K sp No precipitate.  If Q = K sp equilibrium.

66. 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 ) precipitate and if so, what is the concentration of the ions? [Ce 3+ ] 0 = (750.0 mL)(4.00 x 10 -3 M) 1050mL [IO 3 - ] 0 3 = (300.0 mL)(2.00 x 10 -2 M) 1050mL Q = [Ce 3+ ] 0 [IO 3 - ] 0 3 = 2.86 x 10 -3 M 5.71 x 10 -3 M (2.86 x 10 -3 M)(5.71 x 10 -3 M) 3 (5.32 x 10 -10 M) Q>Ksp (precip)

67. Selective Precipitations Used to separate mixtures of metal ions in solutions…. Adding anions that will only precipitate certain metals at a time. Used to purify mixtures. Solution containing Ba 2+ and Ag +. If NaCl is added, AgCl precips, but Ba 2+ does not.

68. A solution contains 1.0 x 10 -4 M Cu + and 2.0 x 10 -3 M Pb 2+. If a source of I - is added gradually to this solution, will PbI 2 (Ksp = 1.4 x 10 -8 ) o CuI (Ksp = 5.3 x 10 -12 ) precipitate first? Specify the concentrations of I- necessary to begin precipitation for each salt. PbI 2 Ksp = 1.4 x 10 -8 =[Pb 2+ ][I - ] 2 = (2.0 x 10 -3 ) (I - ) 2 (I - ) = 2.6 x 10 -3 M CuI Ksp = 5.3 x 10 -12 = [Cu+][I-] =(1.0 x 10 -4 ) (I - ) = 5.3 x 10 -8 M

69. Selective Precipitation Often use H 2 S because in acidic solution Hg +2, Cd +2, Bi +3, Cu +2, Sn +4 will precipitate. Then add 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

71. Selective precipitation Follow the steps 1.With dilute HCl insoluble chlorides (Ag, Pb, Hg, Ba) 2.Add H 2 S most insoluble sulfides (Hg, Cd, Bi, Cu, Sn) in acid. 3.Then sulfides are in basic solution. (Co, Zn, Mn, Ni, Fe, Cr, Al) 4.Then insoluble carbonate (Ca, Ba, Mg) 5.Alkali metals and NH 4 + remain in solution.

74. We have a solution that contains the following cations: Ag +, Cu 2+, and Mg 2+ We wish to isolate each ion by causing one at a time to precipitate out of solution. Once a solid precipitate forms, we can filter the solid precipitate out, leaving the other ions still in solution. We are given the following solutions to use: Na 2 S, NaCl, and NaOH.

75. To help us analyze our situation, we'll prepare a chart, with the cations we need to separate along one axis and anions along the other axis.

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

77. 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 completion. The complex ion will be the biggest ion in solution.

78. 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 0.010 M AgNO 3 with 200.0 mL of 5.00 M Na 2 S 2 O 3 Ag + + S 2 O 3 -2 Ag(S 2 O 3 ) - K 1 =7.4 x 10 8 Ag(S 2 O 3 ) - + S 2 O 3 -2 Ag(S 2 O 3 ) 2 -3 K 2 =3.9 x 10 4

79. Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq) + 4Cl - (aq) CoCl 4 (aq) 2- Co(H 2 O) 6 2+ CoCl 4 2- 16.10

80. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 80 Equilibria Involving Complex Ions Complex Ion: A charged species consisting of a metal ion surrounded by ligands (Lewis bases). Coordination Number: Number of ligands attached to a metal ion. (Most common are 6 and 4.) Formation (Stability) Constants: The equilibrium constants characterizing the stepwise addition of ligands to metal ions.

81. 16.10

82. Complex Ion Formation These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions). As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH 3 ) 4 2+ (ammonia is used as a test for Cu 2+ ions), and Ag(NH 3 ) 2 +. Memorize the common ligands.

83. Common Ligands LigandsNames used in the ion H2OH2Oaqua NH 3 ammine OH-hydroxy Cl-chloro Br-bromo CN-cyano SCN-thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)

84. Names Names: ligand first, then cation Examples: –tetraamminecopper(II) ion: Cu(NH 3 ) 4 2+ –diamminesilver(I) ion: Ag(NH 3 ) 2 +. –tetrahydroxyzinc(II) ion: Zn(OH) 4 2- The charge is the sum of the parts (2+) + 4(-1)= -2.

85. When Complexes Form Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H 2 O) 6 3+ Transitional metals, such as Iron, Zinc and Chromium, can form complex ions. The odd complex ion, FeSCN 2+, shows up once in a while Acid-base reactions may change NH 3 into NH 4 + (or vice versa) which will alter its ability to act as a ligand. Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu 2+ + a little NH 4 OH will form the light blue precipitate, Cu(OH) 2. With excess ammonia, the complex, Cu(NH 3 ) 4 2+, forms. Keywords such as "excess" and "concentrated" of any solution may indicate complex ions. AgNO 3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl 2 -, forms and the solution clears.

86. Coordination Number Total number of bonds from the ligands to the metal atom. Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.

87. Some Coordination Complexes molecular formula Lewis base/ligand Lewis aciddonor atom coordination number Ag(NH 3 ) 2 + NH 3 Ag + N2 [Zn(CN) 4 ] 2- CN-Zn 2+ C4 [Ni(CN) 4 ] 2- CN-Ni 2+ C4 [PtCl 6 ] 2- Cl-Pt 4+ Cl6 [Ni(NH 3 ) 6 ] 2+ NH 3 Ni 2+ N6