1. Chapter 17 Additional Aspects of Aqueous Equilibria
Lecture Presentation
Chapter 17 Additional Aspects of Aqueous Equilibria
James F. Kirby
Quinnipiac University
Hamden, CT
© 2015 Pearson Education, Inc.

2. Some highlights from Chapter 16:
− pH of strong monoprotic acids:
pH = −log [H+] ≡
−log [HCl]0
HCl (aq) + H2O (l) H3O+ (aq) + Cl− (aq)
− pH of strong bases:
pOH = −log [OH−]
pH = – pH
Ca(OH)2(aq) Ca2+(aq) + 2OH-(aq)
− pH of weak acids:
ICE
pH = −log [H+]
HCN(aq) H+(aq) + CN–(aq)
− pH of weak bases:
ICE
pOH = −log [OH−]
pH = – pH
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

3. NaCl (s) Na+ (aq) + Cl− (aq)
Some highlights from Chapter 16:
- Salts that yield neutral solutions in water.
- Salts that yield basic solutions in water.
NaCl (s) Na+ (aq) + Cl− (aq)
H2O
CH3COONa (s) Na+ (aq) + CH3COO− (aq)
H2O
CH3COO− (aq) + H2O (l) CH3COOH (aq) + OH− (aq)

4. Some highlights from Chapter 16:
- Salts that yield acidic solutions in water.
- Salts containing small, highly charged metal cation (acidic solutions).
AlCl3, FeCl3, etc.
- Salts whose cation and anion can hydrolyze in water (acidic/neutral/basic solutions).
NH4Cl (s) NH4+ (aq) + Cl− (aq)
H2O
NH4+ (aq) NH3 (aq) + H+ (aq)

5. 15.1 Solutions of Acids or Bases Containing Common Ion
Some highlights from Chapter 16:
− pH of salt solutions:
NaF (s) Na+ (aq) + F− (aq)
H2O
F−(aq) + H2O(l) HF(aq) + OH−(aq)
ICE of the species that determines the pH.
If conjugate base, use Kb
pOH = −log [OH−]
pH = – pH
If conjugate acid, use Ka
pH = −log [H+]
If H2O
pH = 7.00

6. 17.1 The Common-Ion Effect Common ion effect
Consider separately a solution of acetic acid (weak acid) and a solution of sodium acetate (salt).
What happens when we add sodium acetate to the solution?
CH3COONa (s) Na+ (aq) + CH3COO− (aq)
salt of a weak acid
CH3COO−
is the common ion in both solutions
CH3COOH (aq) H+ (aq) + CH3COO− (aq)
weak acid
According to Le Chatelier’s principle, adding a salt solution to a weak acid solution, both containing a common ion, suppresses the ionization of the weak acid.

7. 17.2 Buffered Solutions A buffer solution is a solution of
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO− (aq) CH3COOH (aq)
Add strong base
OH− (aq) + CH3COOH (aq) CH3COO− (aq) + H2O (l)

8. Henderson-Hasselbalch equation
17.2 Buffered Solutions
pH of buffer solutions
Consider mixture of salt NaA and weak acid HA.
NaA (s) Na+ (aq) + A− (aq)
[H+][A−]
HA (aq) H+ (aq) + A− (aq)
Ka =
[HA] pH
[conjugate base]
[acid] =
pKa +
log
Henderson-Hasselbalch equation
pKa = −log Ka

9. EXERCISE #1
Which of the following solutions can be classified as a buffer?
a) HCl/NaCl b) NH3/NH4Br c) HCN/NaCN
d) CH3COOH e) LiNO2
Buffer: 1. weak acid (or base)
2. salt containing the conjugate base (or conjugate acid)

10. EXERCISE #2
Consider a buffer solution that is 0.45 M acetic acid (CH3COOH) and 0.85 M sodium acetate (CH3COONa). Ka(CH3COOH) = 1.8 x 10–5
pH = pKa + log
[conj. base]
[acid]
a) Calculate the pH of the buffer.
[CH3COO−]
[CH3COOH]
log
pH =
−log Ka +
log =
−log (1.8 x 10−5) +
= 5.02
b) Calculate the pH of this buffer after 0.02 mol of NaOH is added. Assume the volume of NaOH is negligible.
Assume 1.0 L of the buffer solution.
initial (mol)
0.45
0.02
0.85
Possible reaction:
CH3COOH OH− CH3COO− H2O
final (mol)
0.43
0.87
buffer
pH before = 2.54
log
pH =
−log (1.8 x 10−5) +
= 5.05
0.45 M acetic acid:
pH after = 3.41

11. Equivalence point – the point at which the reaction is complete
17.3 Acid-Base Titrations
Titrations
A solution of known concentration is used to determine the concentration of another solution of unknown concentration.
The solution of known concentration (titrant) is gradually added to the solution of unknown concentration (analyte) 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 (or near) the equivalence point
Base solution
Slowly add base
to unknown acid
UNTIL
known conc.
Acid solution
The indicator
changes color
(pink)
Equivalence point
unknown conc.
mol base = mol acid

12. 17.3 Acid-Base Titrations Strong Acid-Strong Base Titrations
Analyte: HNO3
Titrant: NaOH
Acid-base reaction:
HNO3 (aq) + NaOH (aq) H2O (l) + NaNO3 (aq)
At equivalence point:
H2O, Na+, NO3−
Possible rxns?
Rxns that really occur?
Reaction that controls the pH?
H2O H+ + OH−
pH = 7
Na+ + H2O NaOH + H+
NO3− + H2O HNO3 + OH−

13. 17.3 Acid-Base Titrations Weak Acid-Strong Base Titrations
Analyte: CH3COOOH
Titrant: NaOH
Acid-base reaction:
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)
At equivalence point:
H2O, CH3COO−, Na+
Possible rxns?
Rxns that really occur?
Reaction that controls the pH?
H2O H+ + OH−
Kw = 1.0 x 10−14
CH3COO−(aq) + H2O(l) CH3COOH(aq) + OH−(aq)
pH > 7
Na+ + H2O NaOH + H+
Kb(CH3COO−) = 5.6 x 10−10

14. 17.3 Acid-Base Titrations Strong Acid-Weak Base Titrations
Analyte: NH3
Titrant: HCl
Acid-base reaction:
HCl (aq) + NH3 (aq) NH4Cl (aq)
At equivalence point:
H2O, NH4+, Cl−
Possible rxns?
Rxns that really occur?
Reaction that controls the pH?
H2O(l) H+(aq) + OH−(aq)
Kw = 1.0 x 10−14
NH4+ (aq) NH3 (aq) + H+ (aq)
pH < 7
Ka(NH4+) = 5.6 x 10−10
Cl−(aq) + H2O(l) HCl(aq) + OH−(aq)
Copyright © Cengage Learning. All rights reserved

15. Solving Problems in Titrations
Consider the titration of 0.1 M CH3COOH with 0.1 M NaOH.
Before titration: only acid
pH is determined by the acid solution
 Weak acid problem (CH3COOH, H2O)
During titration (before equiv. point): acid and salt of acid
pH is determined by the buffer solution
 Buffer solution problem (CH3COOH, CH3COO−, Na+)
At equivalence point: only salt
pH is determined by the salt in the solution
 Salt hydrolysis (CH3COO−, Na+)
After equivalence point: salt and excess base
pH is determined by excess strong base
Analyte: CH3COOH
Titrant: NaOH
pH?

16. EXERCISE #3
Calculate the pH in the titration of 25.0 mL of M acetic acid by sodium hydroxide after the addition to the acid solution of 10.0 mL of M NaOH.
mol CH3COOH
= (0.100 M)(0.025 L)
= 2.5 x 10−3 mol
mol NaOH =
= (0.100 M)(0.010 L)
= 1.0 x 10−3 mol
Initial (mol)
Final (mol)
2.5 x 10−3
1.0 x 10−3
CH3COOH NaOH CH3COONa H2O
1.5 x 10−3
1.0 x 10−3
buffer
[CH3COO−]
[CH3COOH]
Ka(CH3COOH) = 1.8 x 10−5
pH = pKa + log
Vtot = L L = L
[CH3COOH]f =
1.5 x 10 −3 mol L
1.0 x 10 −3 mol L
= M
[CH3COO−]f =
= M
+ log
pH =
−log (1.8 x 10−5)
= (−0.176)
= 4.56

17. EXERCISE #4
Calculate the pH in the titration of 25.0 mL of M acetic acid by sodium hydroxide after the addition to the acid solution of 25.0 mL of M NaOH.
mol CH3COOH
= (0.100 M)(0.025 L)
= 2.5 x 10−3 mol
= mol NaOH
Initial (mol)
Final (mol)
2.5 x 10−3
2.5 x 10−3
CH3COOH NaOH CH3COONa H2O
2.5 x 10−3
Salt hydrolysis
= 2.5 x 10 −3 mol L
Vtot = L
[CH3COO−]f
= 0.05 M
Kb << [CH3COO−]f
 [CH3COO−]eq ≈ 0.05 M
Ka(CH3COOH) = 1.8 x 10−5
Kb = K w K a = 5.6 x 10−10
Kb = [OH − ][CH3COOH] [CH3COO − ]
CH3COO− + H2O CH3COOH + OH−
Kb ≈ x2 0.05
= 5.6 × 10–10
I (M)
C (M)
E (M)
0.05
x = 5.3 x 10−6 M
= [OH−] −x +x +x
pOH = −log [OH−]
= 5.28
0.05 – x x x
pH =
14 – 5.28
= 8.72

18. EXERCISE #5
Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point? Ka(HNO2) = 4.5 x 10-4
mol HNO2 =
= 0.01 mol
(0.10 M)(0.1 L)
= mol NaOH
Initial (mol)
Final (mol)
0.01
0.01
HNO2 + NaOH NaNO2 + H2O
0.01
Salt hydrolysis
Vtot = 200 mL = 0.2 L
Kb = K w K a = 2.22 x 10−11
Kb << [NO2−]f  [NO2−]eq ≈ 0.05 M
0.01 mol 0.2 L
[NO2−]f =
= 0.05 M
Kb = [OH − ][HNO2] [NO2 − ]
NO2− + H2O HNO2 + OH−
Kb ≈ x2 0.05
= 2.22 x 10–11
I (M)
C (M)
E (M)
0.05
x = 1.05 x 10−6 M
= [OH−] −x +x +x
pOH = −log [OH−]
= 5.98
0.05 – x x x
pH =
14 – 5.98
= 8.02

19. Soluble and Insoluble Compounds
Important Exceptions
Compounds containing NO3
none
Compounds containing C2H3O2
Compounds containing Cl
Salts of Ag+, Hg22+, Pb2+
Compounds containing Br
Compounds containing I
Compounds containing SO42
Salts of Ca2+, Sr2+, Ba2+, Hg22+, Pb2+
Insoluble compounds
 Compounds containing S2
Salts of ammonium, alkali metal cations and Ca2+, Sr2+, Ba2+
Compounds containing CO32
Salts of ammonium, alkali metal cations
Compounds containing PO43
Compounds containing OH

20. 17.4 Solubility Equilibria
Solubility Product, Ksp
AgCl (s) Ag+ (aq) + Cl− (aq) Kc =
[Ag+]
[Cl−]
= Ksp
MgF2 (s) Mg2+ (aq) + 2F− (aq)
Ksp = [Mg2+][F−]2
Ag2CO3 (s) Ag+ (aq) + CO32− (aq)
Ksp = [Ag+]2[CO32−]
Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43− (aq)
Ksp = [Ca2+]3[PO43−]2
Solubility product constant (Ksp) expresses the extent of dissociation of an ionic compound - similar to Ka or Kb.
Solubility product (Ksp) has only one value for a given solid at a given temperature. Solubility is an equilibrium position.
Molar solubility (s) is the number of moles of solute dissolved in 1 L of a saturated solution (mol/L) - similar to equilibrium concentrations in acid-base equilibria.

22. EXAMPLE #1
The molar solubility of silver chloride is 1.3 x 10−5 M. What is the solubility product of silver chloride?
AgCl(s) Ag+(aq) + Cl−(aq)
Ksp
= [Ag+][Cl−]
= s2
I (M)
C (M)
E (M)
Ksp = (1.3 x 10−5)2 +s +s
Ksp = 1.7 x 10−10 s s
EXAMPLE #2
Calculate the molar solubility (s) of silver chloride, the concentrations of silver ions and chloride ions, respectively, in the solution. Ksp(AgCl) = 1.6 x 10−10
Ksp =
[Ag+][Cl−] = s2
AgCl (s) Ag+ (aq) + Cl− (aq)
s = Ksp 
= x 10−10 
Initial (M)
Change (M) +s +s
s = 1.3 x 10−5 mol/L
Equilibrium (M) s s
[Ag+]
= s
= 1.3 x 10−5 M
[Cl−]
= s
= 1.3 x 10−5 M

23. Relationship between Ksp and molar solubility (s)
Compound
Ksp expression
Cation
Anion
Relation b/w Ksp and s
AgCl
[Ag+][Cl−] s s
Ksp = s2
s = (Ksp)1/2
Ksp = s2
s = (Ksp)1/2
BaSO4
[Ba2+][SO42−] s s
s = 𝐾 𝑠𝑝 4
1/3
Ag2CO3
[Ag+]2[CO32−] 2s s
Ksp = 4s3
s = 𝐾 𝑠𝑝 4
1/3
PbF2
[Pb2+][F−]2 s 2s
Ksp = 4s3
s = 𝐾 𝑠𝑝 27
1/4
Al(OH)3
[Al3+][OH−]3 s 3s
Ksp = 27s4
s = 𝐾 𝑠𝑝 108
1/5
Ca3(PO4)2
[Ca2+]3[PO43−]2 3s 2s
Ksp = 108s5

24. 17.5 Factors Affecting Solubility Equilibria
Factors that affect solubility:
The common-ion effect pH
Complex-ion formation
Amphoteric nature of oxides and hydroxides

25. EXERCISE #1
Calculate the molar solubility of silver phosphate in water. Determine also the equilibrium concentrations of Ag+ and PO43−. Ksp(Ag3PO4) = 1.8 x 10–18
Ag3PO4(s) Ag+(aq) + PO43−(aq)
I (M)
C (M)
E (M)
+3s +s 3s s
Ksp =
[Ag+]3
[PO43−] =
(3s)3
(s) =
27s4 =
1.8 x 10−18 s =
1.6 x 10−5 mol/L
[Ag+]
= 1.6 x 10−5 mol/L
= 3s
= 4.8 x 10−5 mol/L
[PO43−]
= s

26. EXERCISE #2
How does the a) molar solubility, and b) Ksp of silver fluoride in water compare to that of silver fluoride in an acidic solution (made by adding nitric acid to the solution)? Explain.
Remove
AgF(s) Ag+(aq) + F−(aq)
F−(aq) + H+(aq) HF(aq)
HNO3(aq) H+(aq) + NO3−(aq)
a) Higher solubility in acid
b) Ksp remains the same

27. Ksp for AgCl is 1.6 × 10–10 Calculate the molar solubility of AgCl in:
EXERCISE #3
Ksp for AgCl is 1.6 × 10–10 Calculate the molar solubility of AgCl in:
a) 4.00 x 10-3 M calcium nitrate.
Ca(NO3)2 Ca2+(aq) + 2NO3−(aq)
AgCl(s) Ag+(aq) + Cl−(aq)
No common ion
Ksp =
[Ag+][Cl−]
= s2
= 1.6 x 10−10
s =
1.3 x 10−5 mol/L
b) 4.00 x 10-3 M calcium chloride.
Common ion
CaCl Ca2+(aq) + 2Cl−(aq)
AgCl(s) Ag+(aq) + Cl−(aq)
[Cl−]
= 2[CaCl2]
= (2)(4.0 x 10−3 M)
I (M)
C (M)
E (M)
0.008 +s +s
= 8.0 x 10− 3 M s s
Ksp << [Cl−]  [Cl−]eq ≈ M
Ksp
= [Ag+][Cl−] ≈
(0.008)
(s)
= 1.6 x 10−10
s =
2.0 x 10−8 mol/L

28. 17.6 Precipitation and Separation of Ions
Ion Product (Q)
AgCl (s) Ag+ (aq) + Cl− (aq)
Ksp = [Ag+][Cl−]
Solubility product constant
Q is the ion product; tells the current conditions of the reaction
Q = [Ag+]0[Cl−]0
Dissolution of an ionic solid in aqueous solution:
Q < Ksp
Unsaturated solution
No precipitation
Q = Ksp
Saturated solution
System is at equilibrium
Q > Ksp
Supersaturated solution
Precipitation until Q = Ksp

29. EXERCISE #4
Predicting precipitation reaction
If 2.00 mL of M NaOH are added to 1.00 L of M CaCl2, will a precipitate form? Ksp[Ca(OH)2] = 1.3 x 10−6
Is Q greater than, less than, or equal to Ksp?
2NaOH(aq) + CaCl2(aq) Ca(OH)2(s) + 2NaCl(aq)
Ca(OH)2 Ca2+ + 2OH−
Qsp =
[Ca2+]0
[OH−]02
mol OH−
= mol NaOH
= (0.002 L)(0.2 M)
= 4.0 x 10−4 mol
Vtot = L
mol Ca2+
= mol CaCl2
= (1.0 L)(0.1 M)
= 0.1 mol
4.0 x 10 −4 mol L
[Ca2+]0 =
0.1 mol L
[OH−]0 =
= x 10−4 M
= M
Qsp =
[Ca2+]0
[OH−]02 =
(0.0998)
(3.992 x 10−4)2 =
1.6 x 10−8
Qsp < Ksp
unsaturated
no precipitation

30. 17.6 Precipitation and Separation of Ions
One can use differences in solubilities of salts to separate ions in a mixture.
This has been used for qualitative analysis of the presence of ions in a solution.

31. END OF CHAPTER 17