1. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
after equivalence point
added 30.0 mL NaOH
5.0 x 10-4 mol NaOH xs
Tro, Chemistry: A Molecular Approach

2. Adding NaOH to HCHO2 added 10.0 mL NaOH 0.00150 mol HCHO2 pH = 3.56
mol NaOH xs
pH = 12.22
added 30.0 mL NaOH
mol NaOH xs
pH = 11.96
added 25.0 mL NaOH
equivalence point
mol CHO2−
[CHO2−]init = M
[OH−]eq = 1.7 x 10-6
pH = 8.23
added 5.0 mL NaOH
mol HCHO2
pH = 3.14
initial HCHO2 solution
mol HCHO2
pH = 2.37
added 12.5 mL NaOH
mol HCHO2
pH = 3.74 = pKa
half-neutralization
added 40.0 mL NaOH
mol NaOH xs
pH = 12.36
added 15.0 mL NaOH
mol HCHO2
pH = 3.92
added 50.0 mL NaOH
mol NaOH xs
pH = 12.52
added 20.0 mL NaOH
mol HCHO2
pH = 4.34
Tro, Chemistry: A Molecular Approach

3. Tro, Chemistry: A Molecular Approach

4. Titration of 25.0 mL of 0.100 M HCHO2 with 0.100 M NaOH
The 1st derivative of the curve is maximum at the equivalence point
pH at equivalence = 8.23
Since the solutions are equal concentration, the equivalence point is at equal volumes
Tro, Chemistry: A Molecular Approach

5. at pH 3.74, the [HCHO2] = [CHO2]; the acid is half neutralized
half-neutralization occurs when pH = pKa
Tro, Chemistry: A Molecular Approach

6. Titrating Weak Acid with a Strong Base
the initial pH is that of the weak acid solution
calculate like a weak acid equilibrium problem
e.g., 15.5 and 15.6
before the equivalence point, the solution becomes a buffer
calculate mol HAinit and mol A−init using reaction stoichiometry
calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−init
half-neutralization pH = pKa
Tro, Chemistry: A Molecular Approach

7. Titrating Weak Acid with a Strong Base
at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established
mol A− = original mole HA
calculate the volume of added base like Ex 4.8
[A−]init = mol A−/total liters
calculate like a weak base equilibrium problem
e.g., 15.14
beyond equivalence point, the OH is in excess
[OH−] = mol MOH xs/total liters
[H3O+][OH−]=1 x 10-14
Tro, Chemistry: A Molecular Approach

8. Ex 16. 7a – A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0
Ex 16.7a – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the volume of KOH at the equivalence point
Write an equation for the reaction for B with HA.
Use Stoichiometry to determine the volume of added B
HNO2 + KOH  NO2 + H2O
Tro, Chemistry: A Molecular Approach

9. HNO2 + KOH  NO2 + H2O HNO2 NO2- OH− mols Before mols added -
Ex 16.7b – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH after adding 5.00 mL KOH
Write an equation for the reaction for B with HA.
Determine the moles of HAbefore & moles of added B
Make a stoichiometry table and determine the moles of HA in excess and moles A made
HNO2 + KOH  NO2 + H2O
HNO2
NO2-
OH−
mols Before
≈ 0
mols added -
mols After
Tro, Chemistry: A Molecular Approach

10. HNO2 + H2O  NO2 + H3O+ HNO2 NO2- OH− mols Before mols added -
Ex 16.7b – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH after adding 5.00 mL KOH.
HNO2 + H2O  NO2 + H3O+
Write an equation for the reaction of HA with H2O
Determine Ka and pKa for HA
Use the Henderson-Hasselbalch Equation to determine the pH
Table Ka = 4.6 x 10-4
HNO2
NO2-
OH−
mols Before
≈ 0
mols added -
mols After
Tro, Chemistry: A Molecular Approach

11. HNO2 + KOH  NO2 + H2O HNO2 NO2- OH− mols Before mols added -
Ex 16.7b – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH at the half-equivalence point
Write an equation for the reaction for B with HA.
Determine the moles of HAbefore & moles of added B
Make a stoichiometry table and determine the moles of HA in excess and moles A made
HNO2 + KOH  NO2 + H2O
at half-equivalence, moles KOH = ½ mole HNO2
HNO2
NO2-
OH−
mols Before
≈ 0
mols added -
mols After
Tro, Chemistry: A Molecular Approach

12. Ex 16. 7b – A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0
Ex 16.7b – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH at the half-equivalence point.
HNO2 + H2O  NO2 + H3O+
Write an equation for the reaction of HA with H2O
Determine Ka and pKa for HA
Use the Henderson-Hasselbalch Equation to determine the pH
Table Ka = 4.6 x 10-4
HNO2
NO2-
OH−
mols Before
≈ 0
mols added -
mols After
Tro, Chemistry: A Molecular Approach

13. Titration Curve of a Weak Base with a Strong Acid
Tro, Chemistry: A Molecular Approach

14. Titration of a Polyprotic Acid
if Ka1 >> Ka2, there will be two equivalence points in the titration
the closer the Ka’s are to each other, the less distinguishable the equivalence points are
titration of 25.0 mL of M H2SO3 with M NaOH
Tro, Chemistry: A Molecular Approach

15. 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 [H3O+]
using a probe that specifically measures just H3O+
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
Tro, Chemistry: A Molecular Approach

16. Monitoring pH During a Titration
Tro, Chemistry: A Molecular Approach

17. HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq)
Indicators
many dyes change color depending on the pH of the solution
these dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution
HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq)
the color of the solution depends on the relative concentrations of Ind:HInd
when 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
Tro, Chemistry: A Molecular Approach

18. Phenolphthalein

19. Methyl Red
Tro, Chemistry: A Molecular Approach

20. 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
pKa of HInd ≈ pH at equivalence point
Tro, Chemistry: A Molecular Approach

21. Acid-Base Indicators

22. 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
Tro, Chemistry: A Molecular Approach

23. Solubility Product
the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp
for an ionic solid MnXm, the dissociation reaction is:
MnXm(s)  nMm+(aq) + mXn−(aq)
the solubility product would be
Ksp = [Mm+]n[Xn−]m
for example, the dissociation reaction for PbCl2 is
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
and its equilibrium constant is
Ksp = [Pb2+][Cl−]2
Tro, Chemistry: A Molecular Approach

24. Tro, Chemistry: A Molecular Approach

25. 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 MnXm(s)  nMm+(aq) + mXn−(aq)
Tro, Chemistry: A Molecular Approach

26. PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
[Pb2+]
[Cl−]
Initial
Change +S
+2S
Equilibrium S 2S
Tro, Chemistry: A Molecular Approach

27. Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Substitute into the Ksp expression
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
[Pb2+]
[Cl−]
Initial
Change +S
+2S
Equilibrium S 2S
Tro, Chemistry: A Molecular Approach

28. Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Tro, Chemistry: A Molecular Approach

29. PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2
[Pb2+]
[Br−]
Initial
Change
+(1.05 x 10-2)
+2(1.05 x 10-2)
Equilibrium
(1.05 x 10-2)
(2.10 x 10-2)
Tro, Chemistry: A Molecular Approach

30. Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Substitute into the Ksp expression
plug into the equation and solve
Ksp = [Pb2+][Br−]2
Ksp = (1.05 x 10-2)(2.10 x 10-2)2
[Pb2+]
[Br−]
Initial
Change
+(1.05 x 10-2)
+2(1.05 x 10-2)
Equilibrium
(1.05 x 10-2)
(2.10 x 10-2)
Tro, Chemistry: A Molecular Approach

31. Ksp and Relative Solubility
molar solubility is related to Ksp
but you cannot always compare solubilities of compounds by comparing their Ksps
in order to compare Ksps, the compounds must have the same dissociation stoichiometry
Tro, Chemistry: A Molecular Approach

32. 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 PbCl2 decreases the solubility of PbCl2
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
addition of Cl− shifts the equilibrium to the left
Tro, Chemistry: A Molecular Approach

33. CaF2(s)  Ca2+(aq) + 2 F−(aq)
Ex – Calculate the molar solubility of CaF2 in M NaF at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
CaF2(s)  Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
[Ca2+]
[F−]
Initial
0.100
Change +S
+2S
Equilibrium S S
Tro, Chemistry: A Molecular Approach

34. Ex 16. 10 – Calculate the molar solubility of CaF2 in 0
Ex – Calculate the molar solubility of CaF2 in M NaF at 25C
Substitute into the Ksp expression
assume S is small
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
Ksp = [Ca2+][F−]2
Ksp = (S)( S)2
Ksp = (S)(0.100)2
[Ca2+]
[F−]
Initial
0.100
Change +S
+2S
Equilibrium S S
Tro, Chemistry: A Molecular Approach

35. 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)  Mn+(aq) + nOH−(aq)
for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility
M2(CO3)n(s)  2 Mn+(aq) + nCO32−(aq)
H3O+(aq) + CO32− (aq)  HCO3− (aq) + H2O(l)
Tro, Chemistry: A Molecular Approach

36. 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 Ksp, we can determine if precipitation will occur
Q = Ksp, the solution is saturated, no precipitation
Q < Ksp, the solution is unsaturated, no precipitation
Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate
some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions
Tro, Chemistry: A Molecular Approach

37. a supersaturated solution will precipitate if a seed crystal is added
precipitation occurs if Q > Ksp
Tro, Chemistry: A Molecular Approach

38. 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 Ksp values are significantly different
Tro, Chemistry: A Molecular Approach

39. precipitating may just occur when Q = Ksp
Ex What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater?
precipitating may just occur when Q = Ksp
Tro, Chemistry: A Molecular Approach

40. precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Ex What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Tro, Chemistry: A Molecular Approach

41. precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Ex What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M
when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from M to 4.8 x M
Tro, Chemistry: A Molecular Approach

42. 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
Tro, Chemistry: A Molecular Approach

43. Qualitative Analysis
Tro, Chemistry: A Molecular Approach

45. Group 1 group one cations are Ag+, Pb2+, and Hg22+
all these cations form compounds with Cl− that are insoluble in water
as long as the concentration is large enough
PbCl2 may be borderline
molar solubility of PbCl2 = 1.43 x 10-2 M
precipitated by the addition of HCl
Tro, Chemistry: A Molecular Approach

46. Group 2
group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+
all these cations form compounds with HS− and S2− that are insoluble in water at low pH
precipitated by the addition of H2S in HCl

47. Group 3
group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides
all these cations form compounds with S2− that are insoluble in water at high pH
precipitated by the addition of H2S in NaOH

48. Group 4 group four cations are Mg2+, Ca2+, Ba2+
all these cations form compounds with PO43− that are insoluble in water at high pH
precipitated by the addition of (NH4)2HPO4
Tro, Chemistry: A Molecular Approach

49. Group 5 group five cations are Na+, K+, NH4+
all these cations form compounds that are soluble in water – they do not precipitate
identified by the color of their flame

50. Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
Complex Ion Formation
transition metals tend to be good Lewis acids
they often bond to one or more H2O molecules to form a hydrated ion
H2O is the Lewis base, donating electron pairs to form coordinate covalent bonds
Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
ions that form by combining a cation with several anions or neutral molecules are called complex ions
e.g., Ag(H2O)2+
the attached ions or molecules are called ligands
e.g., H2O
Tro, Chemistry: A Molecular Approach

51. 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(H2O)2+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) + 2 H2O(l)
generally H2O is not included, since its complex ion is always present in aqueous solution
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Tro, Chemistry: A Molecular Approach

52. Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Formation Constant
the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
the equilibrium constant for the formation reaction is called the formation constant, Kf
Tro, Chemistry: A Molecular Approach

53. Formation Constants
Tro, Chemistry: A Molecular Approach

54. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Ex – mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?
Write the formation reaction and Kf expression.
Look up Kf value
Determine the concentration of ions in the diluted solutions
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Tro, Chemistry: A Molecular Approach

55. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Ex – mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?
Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
[Cu2+]
[NH3]
[Cu(NH3)22+]
Initial
6.7E-4
0.11
Change
-≈6.7E-4
-4(6.7E-4)
+ 6.7E-4
Equilibrium x

56. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Ex – mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Substitute in and solve for x
confirm the “x is small” approximation
[Cu2+]
[NH3]
[Cu(NH3)22+]
Initial
6.7E-4
0.11
Change
-≈6.7E-4
-4(6.7E-4)
+ 6.7E-4
Equilibrium x
since 2.7 x << 6.7 x 10-4, the approximation is valid
Tro, Chemistry: A Molecular Approach

57. 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) Ksp = 1.77 x 10-10
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) Kf = 1.7 x 107
adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+
Tro, Chemistry: A Molecular Approach

59. 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 Al3+, Cr3+, Zn2+, Pb2+, and Sb2+
Tro, Chemistry: A Molecular Approach

60. Al3+ Al3+ is hydrated in water to form an acidic solution
Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+(aq) + H3O+(aq)
addition of OH− drives the equilibrium to the right and continues to remove H from the molecules
Al(H2O)5(OH)2+(aq) + OH−(aq)  Al(H2O)4(OH)2+(aq) + H2O (l)
Al(H2O)4(OH)2+(aq) + OH−(aq)  Al(H2O)3(OH)3(s) + H2O (l)
Tro, Chemistry: A Molecular Approach

61. Tro, Chemistry: A Molecular Approach