Acid-Base Equilibria and
Solubility Equilibria
Chapter 16
16.1-16.9
Acid-Base Equilibria and Solubility
Equilibria

We will continue to study acid-base
reactions
 Buffers
 Titrations

We will also look at properties of slightly
soluble compunds and their ions in
solution
Homogeneous vs. Heterogeneous

Homogeneous Solution Equilibria- a
solution that has the same composition
throughout after equilibrium has been
reached.

Heterogeneous Solution Equilibria- a
solution that after equilibrium has been
reached, results in components in more
than one phase.
The Common Ion Effect

Acid-Base Solutions
Common Ion
 The Common Ion Effect- the shift in

equilibrium caused by the addition of a
compound having an ion in common with
the dissolved substance.
 pH
The Common Ion Effect
The presence of a common ion suppresses
the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and
CH3COOH (weak acid).
CH3COONa (s)
CH3COOH (aq)
Na+ (aq) + CH3COO- (aq)
H+ (aq) + CH3COO- (aq)
common
ion
Henderson-Hasselbach Equation

Relationship between pKa and Ka.
pKa = -log Ka
Henderson-Hasselbalch
equation
[conjugate base]
pH = pKa + log
[acid]
pH Calculations
What is the pH of a solution containing 0.30 M
HCOOH and 0.52 M HCOOK? HCOOH pKa = 3.77
Mixture of weak acid and conjugate base!
Initial (M)
Change (M)
Equilibrium (M)
HCOOH (aq)
0.30
H+ (aq) + HCOO- (aq)
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
pH Calculations
[HCOO-]
pH = pKa + log
[HCOOH]
Common ion effect
0.30 – x  0.30
0.52 + x  0.52
[0.52]
pH = 3.77 + log
[0.30]
= 4.01
Buffer 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.
Buffer Solutions
IV solutions (~7.4)
 Blood (~7.4)
 Gastric Juice (~1.5)

Buffer Solutions
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq)
Add strong base
OH- (aq) + CH3COOH (aq)
CH3COOH (aq)
CH3COO- (aq) + H2O (l)
Buffer Solutions
Effectiveness of Buffer Solutions
Preparing a Buffer Solution with a
Specific pH
Work Backwards
 Choose a weak acid whose pKa is close to
the desired pH
 Substitute pKa and pH values into the
Henderson-Hasselbach Equation
 This will give a ratio of [conjugate
Base]/[Acid]
 Convert ratio to molar quantities

Acid-Base Titrations
In a titration a solution of accurately known
concentration is added gradually added to
another solution of unknown concentration until
the chemical reaction between the two solutions
is complete.
Types of Titrations
Those involving a strong acid and a strong base
Those involving a weak acid and a strong base
Those involving a strong acid and a weak base
Acid-Base Titrations

Indicator – substance that changes color at (or
near) the equivalence point
Equivalence point – the point at which the reaction is complete
Slowly add base
to unknown acid
UNTIL
The indicator
changes color
(pink)
Acid-Base Titrations
Strong Acid-Strong Base Titrations
Strong Acid-Strong Base Titrations
NaOH (aq) + HCl (aq)
OH- (aq) + H+ (aq)
H2O (l) + NaCl (aq)
H2O (l)
Strong Acid-Base Titrations
Strong Acid-Base Titrations

Calculate the pH after the addition of 10.0mL of 0.100M
NaOH to 25.0mL of 0.100M HCl.
# moles NaOH
10.0mL x (0.100mol NaOH/ 1L NaOH) x (1L/ 1000mL)
= 1.00 x 10-3 mol
#moles HCl
25.0mL x (0.100mol HCl/ 1 L HCl) x (1L/ 1000mL)
= 2.50 x 10-3 mol
Amount of HCl left after partial neutralization
2.50 x 10-3 mol - 1.00 x 10-3 mol = 1.5 x x 10-3 mol
Strong Acid-Base Titrations
Thus, [H+] = 1.5 x x 10-3 mol/ 0.035L
[H+] = 0.0429 M
pH = -log [H+]
pH = -log 0.0429
pH = 1.37
Strong Acid-Base Titrations

Calculate pH after the addition of 35.0mL of 0.100M
NaOH to 25.0mL of 0.100M HCl.
# moles NaOH
0.100 mol NaOH / 0.035 L NaOH
= 3.50 x 10-3 mol
#moles HCl
0.100 mol HCl / 0.025 L HCl
= 2.50 x 10-3 mol
Amount of NaOH left after full HCl neutralization
3.50 x 10-3 mol – 2.50 x 10-3 mol = 1.0 x x 10-3 mol
Strong Acid-Base Titrations
Thus, [NaOH] = 1.0 x x 10-3 mol/ 0.06L
[NaOH] = 0.0167 M
[OH-] = 0.0167 M
pOH = -log [H+]
pOH = -log 0.0167
pOH = 1.78
pH = 14.0-pOH
pH = 14.0-1.78
pH = 12.22
Weak Acid-Strong Base Titrations
Weak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq)
CH3COOH (aq) + OH- (aq)
CH3COONa (aq) + H2O (l)
CH3COO- (aq) + H2O (l)
At equivalence point (pH > 7):
CH3COO- (aq) + H2O (l)
OH- (aq) + CH3COOH (aq)
Weak Acid-Strong Base Titrations
Strong Acid-Weak Base Titrations
Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq)
NH4Cl (aq)
H+ (aq) + NH3 (aq)
NH4Cl (aq)
At equivalence point (pH < 7):
NH4+ (aq) + H2O (l)
NH3 (aq) + H+ (aq)
Strong Acid-Weak Base Titrations
Acid-Base Indicators
Equivalence point occurs when OH- = H+
originally present.
 Indicators
 End Point- Occurs when indicator
changes color
 End point ~ Equivalence point

Acid-Base Indicators
Acid-Base Indicators
Acid-Base Indicators
pH
Solubility Equilibria
Reactions that produce precipitates
 Importance

 Tooth
Enamel + Acid = tooth decay
 Barium Sulfate = used in x-rays
 Fudge!!!
Solubility Equilibria

Solubility Product Constant (Ksp)- the
product of the molar concentrations of the
constituent ions, each raised to the power of its
stoichiometric coefficient in the equilibrium
equation.
AgCl (s)
Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
Solubility Products

What are the correct solubility products of
the following equations?
MgF2 (s)
Ag2CO3 (s)
Ca3(PO4)2 (s)
Mg2+ (aq) + 2F- (aq)
Ksp = [Mg2+][F-]2
2Ag+ (aq) + CO32- (aq)
3Ca2+
(aq) +
2PO43-
Ksp = [Ag+]2[CO32-]
(aq)
Ksp = [Ca2+]3[PO33-]2
Solubility Product Constants
Solubility Constants
What does a large Ksp mean?
 What does a small value mean?

Molar Solubility and Solubility
Molar solubility (mol/L)- is the number of moles of
solute dissolved in 1 L of a saturated solution.
Solubility (g/L)- is the number of grams of solute dissolved in
1 L of a saturated solution.
Molar Solubility and Solubility
Molar Solubility and Solubility

The Ksp of silver bromide is 7.7 x 10-13. Calculate the
molar solubility.
Initial (M)
Change (M)
Equilibrium (M)
AgBr(s) ↔ Ag+(aq) + Br-(aq)
0
0
-s
+s
+s
s
s
Ksp= [Ag+][Br-]
7.7 x 10-13 = [Ag+][Br-]
7.7 x 10-13 = s2
S= 8.8 x 10-7 M
Ksp vs. Q
Dissolution of an ionic solid in aqueous
solution:
Q < Ksp
Unsaturated solution
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
No precipitate
Precipitate will form
Predicting Precipitation Reactions
Calculate Q for the reaction
 Is Q larger, smaller or equal to Ksp?

Separation of Ions by Fractional
Precipitation
Removal of ions from solution
 Useful in preparation of prescription
medications
 Ions can be removed by filtration

Fractional Precipitation
Ions + proper reagent
 Smallest → Largest Ksp


If AgNo3 is added to a solution containing Cl-,
Br- and I- ions, which compound will precipitate
out first?
Compound
AgCl
AgBr
AgI
Ksp
1.6 x 10 -10
1.7 x 10 -13
8.3 x 10 -17
The Common Ion Effect and
Solubility
What is the molar solubility of AgBr in (a) pure water
and (b) 0.0010 M NaBr?
AgBr (s)
Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
s2 = Ksp
s = 8.8 x 10-7
The Common Ion Effect and
Solubility
AgBr (s)
Ag+ (aq) + Br- (aq)
NaBr (s)
Na+ (aq) + Br- (aq)
[Br-] = 0.0010 M
[Ag+] = s
[Br-] = 0.0010 + s  0.0010
Ksp = 0.0010 x s
s = 7.7 x 10-10
pH and Solubility
The solubility of many substances depends
on the pH of the solution.
 ↑ pH
 ↓ pH
 Insoluble bases dissolve in acidic solutions
 Insoluble acids dissolve in basic solutions
