Chapter 17:
Additional Aspects of
Aqueous Equilibria
Section 1:
The Common Ion Effect
Objectives
• When you complete this presentation, you will be
able to
• describe the common ion effect.
Introduction
• Water is the most common and most important
solvent on Earth.
• We will be looking in some detail at the application
of equilibrium theory and practice to aqueous
solutions.
• Additional acid-base equilibria
• Buffers and acid-base titrations
• Solubility of compounds
• Formation of complex ions
The Common-Ion Effect
• We know that sodium salts are strong electrolytes
and dissociate completely in aqueous solution.
NaA(aq) → Na+(aq) + A−(aq)
• We also know that certain acids are weak
electrolytes and dissociate partially in solution.
HA(aq) ⇄ H+(aq) + A−(aq)
The Common-Ion Effect
• If we start with a solution of acetic acid, we will set
up the equilibrium for the weak acid.
HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)
• If we add sodium acetate to the solution, the
additional acetate will drive the equilibrium to the
left, decreasing the equilibrium [H+].
• The presence of the added acetate ion causes the
acetic acid to ionize less than it normally would.
The Common-Ion Effect
• Whenever a weak electrolyte and a strong electrolyte
contain a common ion, the weak electrolyte ionizes
less than it would if it were alone in the system.
• This is called the common-ion effect.
• We can calculate equilibrium concentrations of
systems with common ions.
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• Plan:
• Identify the major species in solution.
• Identify the major sources that affect the concentration of H+.
• Build “i-c-e” table.
• Use K expression to calculate [H+].
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• Major species:
•
HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• Major sources that affect the concentration of H+:
•
HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• “i-c-e” table:
•
HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)
initial
0.30 M
0.00 M
0.30 M
change
-x
+x
+x
(0.30 – x) M
xM
(0.30 + x) M
equilibrium
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• K expression:
+][CH COO−]
[H
3
Ka = 1.8 × 10−5 =
[HCH3COO]
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• K expression:
Ka = 1.8 ×
10−5
x(0.30 + x)
= 0.30 − x
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• K expression:
Ka = 1.8 ×
10−5
x(0.30 + x)
= 0.30 − x
assume x<<0.30
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• K expression:
Ka = 1.8 ×
10−5
=
x(0.30)
0.30
assume x<<0.30
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• K expression:
Ka = 1.8 ×
10−5
x(0.30)
=
0.30
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• K expression:
Ka = 1.8 ×
10−5
x(0.30)
=
=x
0.30
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• K expression:
Ka = 1.8 ×
10−5
x(0.30)
=
= x = [H+]
0.30
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• calculate pH:
pH = −log[H+]
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• calculate pH:
pH = −log[H+] = −log(1.8 × 10−5)
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• calculate pH:
pH = −log[H+] = −log(1.8 × 10−5) = 4.74
The Common-Ion Effect
• Sample Exercise 17.1 (pg. 720)
• What is the pH of a solution made by adding 0.30
mol of acetic acid and 0.30 mol of sodium acetate to
enough water to make a 1.0 L solution?
• calculate pH:
pH = −log[H+] = −log(1.8 × 10−5) = 4.74
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• Plan:
• Identify the major species in solution.
• Identify the major sources that affect the concentration of H+ & F−.
• Build “i-c-e” table.
• Use K expression to calculate [H+] & [F−].
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• Major species:
HF(aq) ⇄ H+(aq) + F−(aq)
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• Major sources that affect [H+]:
HF(aq) ⇄ H+(aq) + F−(aq)
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• “i-c-e” table:
HF(aq) ⇄ H+(aq) + F−(aq)
initial
0.20 M
0.10 M
0.00 M
change
-x
+x
+x
(0.20 – x)
M
(0.10 + x) M
xM
equilibrium
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• K expression:
+][F−]
[H
Ka = 6.8 × 10−4 =
[HF]
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• K expression:
Ka = 6.8×
10−4
(0.10 +x)x
=
0.20 − x
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• K expression:
Ka = 6.8×
10−4
(0.10 +x)x
=
0.20 − x
assume x<<0.30
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• K expression:
Ka = 6.8×
10−4
(0.10)x
=
0.20
assume x<<0.30
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• K expression:
Ka = 6.8×
10−4
(0.10)x
=
0.20
x = (6.8×
10−4)
0.20
0.10
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• K expression:
Ka = 6.8×
10−4
(0.10)x
=
0.20
x = (6.8×
10−4)
0.20
= [F−]
0.10
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• K expression:
Ka = 6.8×
10−4
(0.10)x
=
0.20
x = (6.8×
10−4)
0.20
= [F−] = 1.4 × 10−3 M
0.10
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• K expression:
Ka = 6.8×
10−4
(0.10)x
=
0.20
x = (6.8×
10−4)
0.20
= [F−] = 1.4 × 10−3 M
0.10
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• Find pH:
[H+] = 0.10 M − 1.4 × 10−3
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• Find pH:
[H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• Find pH:
[H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M
pH = −log[H+]
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• Find pH:
[H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M
pH = −log[H+] = −log(0.10)
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• Find pH:
[H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M
pH = −log[H+] = −log(0.10) = 1.00
The Common-Ion Effect
• Sample Exercise 17.2 (pg. 722)
• Calculate [F−] and pH of a solution that is 0.20 M in
HF and 0.10 M in HCl.
• Find pH:
[H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M
pH = −log[H+] = −log(0.10) = 1.00
The Common-Ion Effect
• Sample Exercises 17.1 and 17.2 both involve weak
acids.
• We can also use the same techniques with weak
bases.
• For example, adding NH4Cl to an aqueous solution
of NH3 will cause the NH3 to dissociate less and
lower the pH.
The Common-Ion Effect
• The techniques of Sample Exercises 17.1 and 17.2
may be used to solve homework problems 17.15 and
17.17.