chapter 17: additional aspects of aqueous equilibria
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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
equilibrium
(0.30 – x) M x M (0.30 + x) M
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 =
[H+][CH3COO−][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
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 = [H+]
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?
• 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
equilibrium
(0.20 – x) M
(0.10 + x) M
x 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.
• K expression:
Ka = 6.8 × 10−4 =[H+][F−]
[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)x0.20 − 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)x0.20 − 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 =
x = (6.8× 10−4)
(0.10)x0.20
0.200.10
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 =
x = (6.8× 10−4) = [F−]
(0.10)x0.20
0.200.10
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 =
x = (6.8× 10−4) = [F−] = 1.4 × 10−3 M
(0.10)x0.20
0.200.10
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 =
x = (6.8× 10−4) = [F−] = 1.4 × 10−3 M
(0.10)x0.20
0.200.10
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.
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