1. 1 Titration Curve of a Weak Base with a Strong Acid

2. 2 Titration of a Polyprotic Acid  if K a1 >> K a2, there will be two equivalence points in the titration  the closer the K a ’s are to each other, the less distinguishable the equivalence points are  if K a1 >> K a2, there will be two equivalence points in the titration  the closer the K a ’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH

3. 3 Monitoring pH During a Titration  the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ]  using a probe that specifically measures just H 3 O +  the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve  if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator  the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ]  using a probe that specifically measures just H 3 O +  the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve  if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator

4. 4 Monitoring pH During a Titration

5. 5 Indicators  many dyes change color depending on the pH of the solution  these dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l) Ind - (aq) + H 3 O + (aq)  the color of the solution depends on the relative concentrations of Ind - :HInd ≈ 1, the color will be mix of the colors of Ind - and HInd  when Ind - :HInd > 10, the color will be mix of the colors of Ind -  when Ind - :HInd < 0.1, the color will be mix of the colors of HInd  many dyes change color depending on the pH of the solution  these dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l) Ind - (aq) + H 3 O + (aq)  the color of the solution depends on the relative concentrations of Ind - :HInd ≈ 1, the color will be mix of the colors of Ind - and HInd  when Ind - :HInd > 10, the color will be mix of the colors of Ind -  when Ind - :HInd < 0.1, the color will be mix of the colors of HInd

6. 6 Phenolphthalein

7. 7 Methyl Red

8. 8 Monitoring a Titration with an Indicator  for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point  an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH  pK a of HInd ≈ pH at equivalence point  for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point  an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH  pK a of HInd ≈ pH at equivalence point

9. 9 Acid-Base Indicators

10. 10 Solubility Equilibria  all ionic compounds dissolve in water to some degree  however, many compounds have such low solubility in water that we classify them as insoluble  we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water  all ionic compounds dissolve in water to some degree  however, many compounds have such low solubility in water that we classify them as insoluble  we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water

11. 11 Solubility Product  the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp  for an ionic solid M n X m, the dissociation reaction is: M n X m (s) nM m+ (aq) + mX n− (aq)  the solubility product would be K sp = [M m+ ] n [X n− ] m  for example, the dissociation reaction for PbCl 2 is PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq)  and its equilibrium constant is K sp = [Pb 2+ ][Cl − ] 2  the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp  for an ionic solid M n X m, the dissociation reaction is: M n X m (s) nM m+ (aq) + mX n− (aq)  the solubility product would be K sp = [M m+ ] n [X n− ] m  for example, the dissociation reaction for PbCl 2 is PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq)  and its equilibrium constant is K sp = [Pb 2+ ][Cl − ] 2

12. 12

13. 13 Molar Solubility  solubility is the amount of solute that will dissolve in a given amount of solution  at a particular temperature  the molar solubility is the number of moles of solute that will dissolve in a liter of solution  the molarity of the dissolved solute in a saturated solution  for the general reaction M n X m (s) nM m+ (aq) + mX n− (aq)  solubility is the amount of solute that will dissolve in a given amount of solution  at a particular temperature  the molar solubility is the number of moles of solute that will dissolve in a liter of solution  the molarity of the dissolved solute in a saturated solution  for the general reaction M n X m (s) nM m+ (aq) + mX n− (aq)

14. 14 Calculate the molar solubility of PbCl2 in pure water at 25°C Write the dissociation reaction and K sp expression Create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq) K sp = [Pb 2+ ][Cl − ] 2

15. 15 Calculate the molar solubility of PbCl2 in pure water at 25°C Substitute into the K sp expression Find the value of K sp from Table 16.2, plug into the equation and solve for S [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S K sp = [Pb 2+ ][Cl − ] 2 K sp = (S)(2S) 2

16. 16 Determine the K sp of PbBr 2 if its molar solubility in water at 25°C is 1.05 x 10 -2 M

17. 17 Determine the K sp of PbBr 2 if its molar solubility in water at 25°C is 1.05 x 10 -2 M Write the dissociation reaction and K sp expression Create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Br − ] Initial00 Change+(1.05 x 10 -2 )+2(1.05 x 10 -2 ) Equilibrium(1.05 x 10 -2 )(2.10 x 10 -2 ) PbBr 2 (s) Pb 2+ (aq) + 2 Br − (aq) K sp = [Pb 2+ ][Br − ] 2

18. 18 Determine the K sp of PbBr 2 if its molar solubility in water at 25°C is 1.05 x 10 -2 M Substitute into the K sp expression plug into the equation and solve K sp = [Pb 2+ ][Br − ] 2 K sp = (1.05 x 10 -2 )(2.10 x 10 -2 ) 2 [Pb 2+ ][Br − ] Initial00 Change+(1.05 x 10 -2 )+2(1.05 x 10 -2 ) Equilibrium(1.05 x 10 -2 )(2.10 x 10 -2 )

19. 19 K sp and Relative Solubility  molar solubility is related to K sp  but you cannot always compare solubilities of compounds by comparing their K sp s  in order to compare K sp s, the compounds must have the same dissociation stoichiometry  molar solubility is related to K sp  but you cannot always compare solubilities of compounds by comparing their K sp s  in order to compare K sp s, the compounds must have the same dissociation stoichiometry

20. 20 The Effect of Common Ion on Solubility  addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt  for example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq)  addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt  for example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq) addition of Cl − shifts the equilibrium to the left

21. 21 Calculate the molar solubility of CaF2 in 0.100 M NaF at 25°C Write the dissociation reaction and K sp expression Create an ICE table defining the change in terms of the solubility of the solid [Ca 2+ ][F − ] Initial00.100 Change+S+2S EquilibriumS0.100 + 2S CaF 2 (s) Ca 2+ (aq) + 2 F − (aq) K sp = [Ca 2+ ][F − ] 2

22. 22 Calculate the molar solubility of CaF2 in 0.100 M NaF at 25°C Substitute into the K sp expression assume S is small Find the value of K sp from Table 16.2, plug into the equation and solve for S [Ca 2+ ][F − ] Initial00.100 Change+S+2S EquilibriumS0.100 + 2S K sp = [Ca 2+ ][F − ] 2 K sp = (S)(0.100 + 2S) 2 K sp = (S)(0.100) 2

23. 23 The Effect of pH on Solubility  for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide  and the lower the pH, the higher the solubility  higher pH = increased [OH − ] M(OH) n (s) M n+ (aq) + nOH − (aq)  for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M 2 (CO 3 ) n (s) 2 M n+ (aq) + nCO 3 2− (aq) H 3 O + (aq) + CO 3 2− (aq) HCO 3 − (aq) + H 2 O(l)  for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide  and the lower the pH, the higher the solubility  higher pH = increased [OH − ] M(OH) n (s) M n+ (aq) + nOH − (aq)  for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M 2 (CO 3 ) n (s) 2 M n+ (aq) + nCO 3 2− (aq) H 3 O + (aq) + CO 3 2− (aq) HCO 3 − (aq) + H 2 O(l)

24. 24 Precipitation  precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound  if we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur  Q = K sp, the solution is saturated, no precipitation  Q < K sp, the solution is unsaturated, no precipitation  Q > K sp, the solution would be above saturation, the salt above saturation will precipitate  some solutions with Q > K sp will not precipitate unless disturbed – these are called supersaturated solutions  precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound  if we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur  Q = K sp, the solution is saturated, no precipitation  Q < K sp, the solution is unsaturated, no precipitation  Q > K sp, the solution would be above saturation, the salt above saturation will precipitate  some solutions with Q > K sp will not precipitate unless disturbed – these are called supersaturated solutions

25. 25 precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added

26. 26 Selective Precipitation  a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others  a successful reagent can precipitate with more than one of the cations, as long as their K sp values are significantly different  a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others  a successful reagent can precipitate with more than one of the cations, as long as their K sp values are significantly different

27. 27 What is the minimum [OH − ] necessary to just begin to precipitate Mg 2+ (with [0.059]) from seawater assuming K sp =2.06x10 -13 )? precipitating may just occur when Q = K sp

28. 28 What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH − ] = 1.9 x 10 -6 M

29. 29 What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH − ] = 1.9 x 10 -6 M precipitating Ca 2+ begins when [OH − ] = 2.06 x 10 -2 M when Ca 2+ just begins to precipitate out, the [Mg 2+ ] has dropped from 0.059 M to 4.8 x 10 -10 M

30. 30 Qualitative Analysis  an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme  wet chemistry  a sample containing several ions is subjected to the addition of several precipitating agents  addition of each reagent causes one of the ions present to precipitate out  an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme  wet chemistry  a sample containing several ions is subjected to the addition of several precipitating agents  addition of each reagent causes one of the ions present to precipitate out

31. 31 Qualitative Analysis

32. 32

33. 33 Group 1  group one cations are Ag +, Pb 2+, and Hg 2 2+  all these cations form compounds with Cl − that are insoluble in water  as long as the concentration is large enough  PbCl 2 may be borderline  molar solubility of PbCl 2 = 1.43 x 10 -2 M  precipitated by the addition of HCl  group one cations are Ag +, Pb 2+, and Hg 2 2+  all these cations form compounds with Cl − that are insoluble in water  as long as the concentration is large enough  PbCl 2 may be borderline  molar solubility of PbCl 2 = 1.43 x 10 -2 M  precipitated by the addition of HCl

34. 34 Group 2  group two cations are Cd 2+, Cu 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, and Hg 2+  all these cations form compounds with HS − and S 2− that are insoluble in water at low pH  precipitated by the addition of H 2 S in HCl  group two cations are Cd 2+, Cu 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, and Hg 2+  all these cations form compounds with HS − and S 2− that are insoluble in water at low pH  precipitated by the addition of H 2 S in HCl

35. 35 Group 3  group three cations are Fe 2+, Co 2+, Zn 2+, Mn 2+, Ni 2+ precipitated as sulfides; as well as Cr 3+, Fe 3+, and Al 3+ precipitated as hydroxides  all these cations form compounds with S 2− that are insoluble in water at high pH  precipitated by the addition of H 2 S in NaOH  group three cations are Fe 2+, Co 2+, Zn 2+, Mn 2+, Ni 2+ precipitated as sulfides; as well as Cr 3+, Fe 3+, and Al 3+ precipitated as hydroxides  all these cations form compounds with S 2− that are insoluble in water at high pH  precipitated by the addition of H 2 S in NaOH

36. 36 Group 4  group four cations are Mg 2+, Ca 2+, Ba 2+  all these cations form compounds with PO 4 3− that are insoluble in water at high pH  precipitated by the addition of (NH 4 ) 2 HPO 4  group four cations are Mg 2+, Ca 2+, Ba 2+  all these cations form compounds with PO 4 3− that are insoluble in water at high pH  precipitated by the addition of (NH 4 ) 2 HPO 4

37. 37 Group 5  group five cations are Na +, K +, NH 4 +  all these cations form compounds that are soluble in water – they do not precipitate  identified by the color of their flame  group five cations are Na +, K +, NH 4 +  all these cations form compounds that are soluble in water – they do not precipitate  identified by the color of their flame

38. 38 Complex Ion Formation  transition metals tend to be good Lewis acids  they often bond to one or more H 2 O molecules to form a hydrated ion  H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag + (aq) + 2 H 2 O(l) Ag(H 2 O) 2 + (aq)  ions that form by combining a cation with several anions or neutral molecules are called complex ions  e.g., Ag(H 2 O) 2 +  the attached ions or molecules are called ligands  e.g., H 2 O  transition metals tend to be good Lewis acids  they often bond to one or more H 2 O molecules to form a hydrated ion  H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag + (aq) + 2 H 2 O(l) Ag(H 2 O) 2 + (aq)  ions that form by combining a cation with several anions or neutral molecules are called complex ions  e.g., Ag(H 2 O) 2 +  the attached ions or molecules are called ligands  e.g., H 2 O

39. 39 Complex Ion Equilibria  if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H 2 O) 2 + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) + 2 H 2 O (l)  generally H 2 O is not included, since its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq)  if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H 2 O) 2 + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) + 2 H 2 O (l)  generally H 2 O is not included, since its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq)

40. 40 Formation Constant  the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq)  the equilibrium constant for the formation reaction is called the formation constant, K f  the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq)  the equilibrium constant for the formation reaction is called the formation constant, K f

41. 41 Formation Constants

42. 42 200.0 mL of 1.5 x 10 -3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Write the formation reaction and K f expression. Look up K f value Determine the concentration of ions in the diluted solutions Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 4 2+ (aq)

43. 43 200.0 mL of 1.5 x 10 -3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Create an ICE table. Since K f is large, assume all the Cu 2+ is converted into complex ion, then the system returns to equilibrium [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] Initial6.7E-40.110 Change-≈6.7E-4-4(6.7E-4)+ 6.7E-4 Equilibriumx0.116.7E-4 Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 4 2+ (aq)

44. 44 200.0 mL of 1.5 x 10 -3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 2 2+ (aq) Substitute in and solve for x confirm the “x is small” approximation [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] Initial6.7E-40.110 Change-≈6.7E-4-4(6.7E-4)+ 6.7E-4 Equilibriumx0.116.7E-4 since 2.7 x 10 -13 << 6.7 x 10 -4, the approximation is valid

45. 45 The Effect of Complex Ion Formation on Solubility  the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands AgCl (s) Ag + (aq) + Cl − (aq) K sp = 1.77 x 10 -10 Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7  adding NH 3 to a solution in equilibrium with AgCl (s) increases the solubility of Ag +  the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands AgCl (s) Ag + (aq) + Cl − (aq) K sp = 1.77 x 10 -10 Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7  adding NH 3 to a solution in equilibrium with AgCl (s) increases the solubility of Ag +

46. 46

47. 47 Solubility of Amphoteric Metal Hydroxides  many metal hydroxides are insoluble  all metal hydroxides become more soluble in acidic solution  shifting the equilibrium to the right by removing OH −  some metal hydroxides also become more soluble in basic solution  acting as a Lewis base forming a complex ion  substances that behave as both an acid and base are said to be amphoteric  some cations that form amphoteric hydroxides include Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sb 2+  many metal hydroxides are insoluble  all metal hydroxides become more soluble in acidic solution  shifting the equilibrium to the right by removing OH −  some metal hydroxides also become more soluble in basic solution  acting as a Lewis base forming a complex ion  substances that behave as both an acid and base are said to be amphoteric  some cations that form amphoteric hydroxides include Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sb 2+

48. 48 Al 3+  Al 3+ is hydrated in water to form an acidic solution Al(H 2 O) 6 3+ (aq) + H 2 O (l) Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq)  addition of OH − drives the equilibrium to the right and continues to remove H from the molecules Al(H 2 O) 5 (OH) 2+ (aq) + OH − (aq) Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH − (aq) Al(H 2 O) 3 (OH) 3(s) + H 2 O (l)  Al 3+ is hydrated in water to form an acidic solution Al(H 2 O) 6 3+ (aq) + H 2 O (l) Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq)  addition of OH − drives the equilibrium to the right and continues to remove H from the molecules Al(H 2 O) 5 (OH) 2+ (aq) + OH − (aq) Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH − (aq) Al(H 2 O) 3 (OH) 3(s) + H 2 O (l)

49. 49