Aqueous Ionic Solutions and Equilibrium
Chapter 19
Common Ion Effect
Shift in equilibrium that occurs because of the
addition of an ion already involved in the
equilibrium reaction.
HF(aq)  H 2 O(l)
F (aq)  H 3O
-

(aq)
What happens to the equilibrium if 0.10 M NaF is added?
AgCl (s)
Ag

(aq)
 2 Cl

(aq)
What happens to the equilibrium if 0.10 M NaCl is added?
Buffers
Resist a change in pH when H+ or OH- is added
Components: conjugate acid-base pairs
acidic component reacts with OHbasic component reacts with H+
Example: 1.00 L of 0.500 M CH3COOH +
0.500 M CH3COONa
CH3COOH/CH3COO-
Reactions with H+ or OH-
Key Points on Buffers
1.Weak acids and bases containing common ion
2.Problems involve:
stoichiometry first
equilibrium second
Buffer Characteristics
-
Contain relatively large amounts of weak acid and
corresponding base.
-
Added H+ reacts to completion with weak base.
-
Added OH reacts to completion with weak acid.
-
pH is determined by ratio of concentrations of
weak acid and weak base.
Buffer Capacity
Amount of H+ or OH- it can absorb without a
significant change in pH.
For HA/A- system, buffer capacity depends on:
—
—
[HA] and [A-]
[HA] ratio
[A-]
(higher = higher)
(closer to 1 = higher)
Calculations with Ka
Calculate the pH of a buffer consisting of
0.50 M HF and 0.45 M F(a) before and
(b) after addition of 0.40 g NaOH to
1.0 L of the buffer.
Ka of HF = 6.8 x 10-4
Calculations
Find pH of a buffer
Buffer preparation
Find equilibrium concentrations
Helpful:
Hendeson-Hasselbach equation


-
-
[H 3O ][A ] [H ][A ]
For K a 

[HA]
[HA]
-
[A ]
pK a  pH  log
[HA]
Calculations with Ka
Calculate the pH of a buffer consisting of
0.50 M HF and 0.45 M F- using the
Henderson-Hasselbach equation.
-
[F ]
pK a  pH  log
[HF]
Preparing a buffer
1.Choose and acid-conjugate base pair
2.Calculate the ratio of the buffer component pairs
3.Determine the buffer concentration
How would you prepare a benzoic acid/benzoate
buffer with pH = 4.25, starting with 5.0 L of
0.050 M sodium benzoate (C6H5COONa)
solution and adding the acidic component?
Ka of benzoic acid (C6H5COOH) = 6.3 x 10-5
Titration (pH) Curves
Plot pH of solution vs. amount of titrant added
Equivalence point:
Enough titrant added to react exactly with
solution being analyzed.
Titration of Strong Acid with Strong Base
13.0
pH
15_327
Equivalence
point
7.0
1.0
0
50.0
100.0
Vol NaOH added (mL)
Weak Acid-Strong Base Titration
1.Stoichiometry
Reaction assumed to run to completion
2.Equilibrium
Use weak acid equilibrium to find pH
Weak Acid-Strong Base Titration
15_329
12.0
Equivalence
point
pH
9.0
3.0
25
50
Vol NaOH added (mL)
Differences
15_330
pH
Weak acid
Strong acid
Vol NaOH
Differences and Ka
15_331
12.0
Ka = 10– 10
10.0
Ka = 10– 8
8.0
pH
Ka = 10– 6
6.0
Ka = 10– 4
4.0
Ka = 10– 2
2.0
Strong acid
0
10
20
30
40
50
60
Vol 0.10 M NaOH added (mL)
Titration Calculations
1. Solution of HA
2. Solution of HA and added base
3. Equivalent amounts of HA and added base
4. Excess base
A chemist titrates 20.00 mL of 0.2000 M HBrO
(Ka = 2.3 x 10-9) with 0.1000 M NaOH. Find the pH:
(a) before any base is added
(b) when 30.00 mL of NaOH is added
(c) at the equivalence point
(d) when the moles of OH- added are twice the moles of
HBrO originally present?
Titration of Strong Base with Strong Acid
15_328
14.0
pH
Equivalence
point
7.0
50.0 mL
Vol 1.0 M HCl added
Weak Base-Strong Acid Titration
15_332
12
pH
10
Equivalence
point
8
6
4
2
0
10
20
30 40 50 60 70
Vol 0.10 M HCl (mL)
Acid-Base Indicator
Indicates endpoint of a titration
15_333
Endpoint is not necessarily the equivalence point
OH
HO
C
C
–
O
O
OH
C
O–
C
O
(Colorless acid form, HIn)
O–
O
(Pink base form, In– )
pH
15_334
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Crystal Violet
Cresol Red
Thymol Blue
Erythrosin B
2,4-Dinitrophenol
Bromphenol Blue
Methyl Orange
Bromcresol Green
Methyl Red
Eriochrome* Black T
Bromcresol Purple
Alizarin
Bromthymol Blue
Phenol Red
m - Nitrophenol
o-Cresolphthalein
Phenolphthalein
Thymolphthalein
Alizarin Yellow R
* Trademark
CIBA GEIGY CORP.
The pH ranges shown are approximate. Specific transition ranges depend on the indicator solvent chosen.
15_335AB
14
14
12
12
10
10
Phenolphthalein
pH
Equivalence
point
6
Methyl red
6
4
4
2
2
0
Equivalence
point
8
pH
8
Phenolphthalein
0
20 40 60 80 100 120
Vol 0.10 M NaOH added (mL)
0
Methyl red
0
20 40 60 80 100 120
Vol 0.10 M NaOH added (mL)
Polyprotic Acid, H2SO3
Solubility Constant, Ksp
For solids dissolving to form aqueous solutions.
For slightly soluble salts:
equilibrium between solid and component ions
PbCl 2(s)
Pb
2
(aq)
2
 2 Cl
 2
K sp  [ Pb ][Cl ]

(aq)
Ksp
Write the expression for Ksp for
(a) CaSO4
(b) Cr2CO3
(c) Mg(OH)2
(d) As2S3
Solubility Product
Solubility s:
Amount of PbCl2 that dissolves
For:
PbCl 2(s)
Pb
2
(aq)
 2 Cl

(aq)
[PbCl2]dissolved = [Pb2+] = ½[Cl-]
s: varies, especially if common ion is present
Calculations
1.5 x 10-4 g of CaF2 dissolves in 10.00 mL solution
at 18°C.
Write the expression for Ksp.
Find the molar solubility of CaF2.
Find the [Ca2+] and [F-].
Calculate Ksp.
What is the molar solubility of Mg(OH)2 if the
value of Ksp is 6.3 x 10-10?
Complex Ions
Complex Ion:
Charged species - metal ion surrounded by ligands
(Lewis bases).
Coordination Number:
No. of ligands attached to a metal ion.
(Common: 6 and 4.)
Formation (Stability) Constants, Kf:
Equilibrium constants for stepwise addition of
ligands to metal ions.
Overall Formation Constant
Al(H 2 O)6
3
 6NH 3
Kf 
Al(NH 3 ) 6
3
 6H 2 O
3
[Al(NH 3 ) 6 ]
3
6
[Al(H 2 O)6 ][NH 3 ]
K f  K f1  K f2  K f3  K f4  K f5  K f6
Complex Ions
Write the stepwise formation constants for
Cr(NH3)63+, starting from Cr(H2O)63+ and NH3(aq)
What is the coordination number of Cr3+?
Applications
Selective precipitation
Exploit differences in Ksp
Qualitative analysis
1.
2.
3.
4.
5.
Insoluble chlorides (Ag+, Hg22+, Pb2+)
Acid-insoluble sulfides (Cu2+, Cd2+, Hg2+, As3+, Sb3+, Bi3+,
Sn2+, Sn4+, Pb2+)
Base-insoluble sulfides and hydroxides (Zn2+,Mn2+, Ni2+,
Fe2+, Co2+ as sulfides; Al3+, Cr3+ as hydroxides)
Insoluble phosphates (Mg2+, Ca2+, Ba2+)
Alkali metal and ammonium ions