Solution Manual For
SM Adsorption Theoretical Aspects and Environmental Applications, 1E Ronaldo Ferreira do
Nascimento
Chapters 1-9
Chapter 1
EASY LEVEL – SOLVED QUESTIONS
1. At pH 2.0, calculate the percentage of Zn²⁺ that exists as the free ion, assuming minimal
complexation under such acidic conditions.
Solution: At this low pH, acetic acid remains mostly undissociated, yielding negligible
CH₃COO⁻. Therefore, virtually no complexation occurs.
Answer: Approximately 100% of Zn²⁺ remains uncomplexed at pH 2.0.
2. .Determine the concentration of acetate ions (CH₃COO⁻) at pH 5.0, given acetic acid‘s
Ka = 1.8 × 10⁻⁵ and [CH₃COOH] = 5.0 × 10⁻³ M.
Solution: From ka = ([H⁺][CH₃COO⁻]/[CH₃COOH]), [H⁺] = 10⁻⁵ M at pH 5.
Substituting:
1.8×10⁻⁵ = (10⁻⁵ x)/(5×10⁻³ - x) ≈ (10⁻⁵ × x)/(5×10⁻³)
Solving for x: x ≈ 9.0 × 10⁻⁴ M
Answer: [CH₃COO⁻] ≈ 9.0 × 10⁻⁴ M
3. Explain why metal-acetate complexation increases with pH.
Solution: Higher pH values promote deprotonation of acetic acid, increasing CH₃COO⁻
availability for complex formation.
Answer: Because more acetate ions are available for binding as pH increases
4. At what approximate pH does Cd(CH₃COO)2 surpass free Cd²⁺ as the dominant species?
Solution: From the speciation diagram, Cd(CH₃COO) 2 becomes predominant around pH 6.
Answer: pH ≈ 6.0
5. If the total Cu²⁺ concentration is 1.0 × 10⁻³ M at pH 8.0 and Cu exists exclusively as
Cu(CH₃COO)₂, calculate the complex concentration.
Solution: All Cu²⁺ is converted to Cu(CH₃COO)₂.
Answer: [Cu(CH₃COO)₂] = 1.0 × 10⁻³ M
MEDIUM LEVEL – SOLVED QUESTIONS
6. Given that 75% of Cu²⁺ is complexed as Cu(CH₃COO)⁺ at pH 6.0 and [Cu]ₜₒₜₐₗ = 2.0 ×
10⁻³ M, determine [Cu²⁺].
,Solution: Free Cu²⁺ = 25% × 2.0 × 10⁻³ = 5.0 × 10⁻⁴ M
Answer: 5.0 × 10⁻⁴ M
7. Using K₁ = 10³ for Zn²⁺ + CH₃COO⁻ ⇌ Zn(CH₃COO)⁺, calculate [Zn(CH₃COO)⁺] if
[Zn²⁺] = 5.0×10⁻⁴ M and [CH₃COO⁻] = 1.0×10⁻³ M.
Solution:[Zn(CH₃COO)⁺] = K₁ × [Zn²⁺] × [CH₃COO⁻] = 10³ × 5.0×10⁻⁴ × 1.0×10⁻³ = 5.0 × 10⁻⁴
M
Answer:5.0 × 10⁻⁴ M
8. Provide a chemical explanation for why Zn(CH₃COO)₂ forms at lower pH than
Cu(CH₃COO)₂.
Solution: Zn²⁺ has a higher affinity for acetate than Cu²⁺, resulting in earlier complex formation.
Answer: Stronger interaction between Zn²⁺ and acetate allows complexation at lower pH.
9. Estimate the amount of acetate needed to fully complex 1.5×10⁻³ M Cu²⁺ into
Cu(CH₃COO)₂ at pH 10.
Solution: 1 mole of Cu(CH₃COO)₂ consumes 2 moles of acetate: Required acetate = 2 × 1.5×10⁻³
= 3.0 × 10⁻³ M
Answer:3.0 × 10⁻³ M
10. Describe the trend in metal-acetate complex formation from pH 3 to 9.
Solution: With increasing pH, acetate concentration increases, favoring formation of mono- and
di-acetate complexes.
Answer: Complexation increases in both quantity and coordination number as pH rises.
DIFFICULT LEVEL – SOLVED QUESTIONS
11. With K₁ = 10².⁸ and K₂ = 10².⁵ for Cd²⁺ complexation, calculate [Cd(CH₃COO)₂] from
1×10⁻³ M Cd²⁺ and 1×10⁻³ M CH₃COO⁻.
Solution: First: [Cd(CH₃COO)⁺] = K₁ × [Cd²⁺] × [CH₃COO⁻] = 10².⁸ × 10⁻³ × 10⁻³ = 6.3×10⁻⁴ M
Then: [Cd(CH₃COO)₂] = K₂ × [Cd(CH₃COO)⁺] × [CH₃COO⁻]
= 10².⁵ × 6.3×10⁻⁴ × 10⁻³ ≈ 2.0 × 10⁻⁴ M
Answer:2.0 × 10⁻⁴ M
12. At pH 7.5, if 20% of Cd²⁺ remains uncomplexed from a 1×10⁻³ M solution, identify the
dominant complexed species.
Solution: Complexed fraction = 80% = 8×10⁻⁴ M Given the pH and graph, Cd(CH₃COO)₂
predominates.
Answer: Major species: Cd(CH₃COO)₂
13. Derive the total metal concentration expression (Mₜ) in terms of stepwise formation
constants β₁ and β₂ and [CH₃COO⁻].
Solution: Mₜ = [M²⁺] + β₁[L][M²⁺] + β₁β₂[L]²[M²⁺]
,Answer: Mₜ = [M²⁺](1 + β₁[L] + β₁β₂[L]²)
14. Explain mathematically how [M²⁺] varies with increasing pH, assuming constant acetate
concentration.
Solution: As pH rises, additional CH₃COO⁻ forms, increasing complexation.Thus, [M²⁺]
decreases due to equilibrium shift.
Answer: [M²⁺] ∝ 1 / (1 + β₁[L] + β₁β₂[L]²); it decreases with pH.
15. If total Cu²⁺ = 10⁻³ M and β₁ = 10³, β₂ = 10², calculate the concentrations of Cu²⁺,
Cu(CH₃COO)⁺, and Cu(CH₃COO)₂ at [CH₃COO⁻] = 1.0×10⁻³ M.
Solution: Let x = [Cu²⁺] Then:
Cu(CH₃COO)⁺ = β₁ × x × [L]
Cu(CH₃COO)₂ = β₁β₂ × x × [L]²
Total Cu = x + β₁x[L] + β₁β₂x[L]²
= x(1 + β₁[L] + β₁β₂[L]²)
Solving for x: x = 10⁻³ / (1 + 10³×10⁻³ + 10³×10²×10⁻⁶)
= 10⁻³ / (1 + 1 + 0.1) =4.76×10−4
LEARNING EXERCISES
EASY LEVEL
1. At 25 °C, determine the hydrogen ion concentration in a 0.01 mol/L acetic acid solution,
given Ka = 1.8 × 10⁻⁵.
Solution: CH₃COOH ⇌ CH₃COO⁻ + H⁺
Ka = [H⁺]² / [CH₃COOH] ≈ x² / 0.01
x² = 1.8 × 10⁻⁵ × 0.01 = 1.8 × 10⁻⁷
x = √(1.8 × 10⁻⁷) ≈ 1.34 × 10⁻⁴ mol/L
Answer: [H⁺] = 1.34 × 10⁻⁴ mol/L
2. If the total concentration of Zn²⁺ is 2.0 × 10⁻³ mol/L and no complexation occurs at pH 3,
determine the concentration of free Zn²⁺.
Solution: Since no complexation takes place, the total concentration is equal to the free ion
concentration.
Answer: [Zn²⁺] = 2.0 × 10⁻³ mol/L
INTERMEDIATE LEVEL
3. A solution contains 1.0 × 10⁻³ mol/L of Cu²⁺ at pH 7, and 80% is complexed as
Cu(CH₃COO)⁺. Calculate the concentration of free Cu²⁺.
, Solution: 20% remains free → 0.20 × 1.0 × 10⁻³ = 2.0 × 10⁻⁴ mol/L
Answer: [Cu²⁺] = 2.0 × 10⁻⁴ mol/L
4. Given Kf = 10⁴ for Zn²⁺ + CH₃COO⁻ ⇌ Zn(CH₃COO)⁺, calculate [Zn(CH₃COO)⁺] if [Zn²⁺] =
1.0 × 10⁻³ mol/L and [CH₃COO⁻] = 2.0 × 10⁻³ mol/L.
Solution: [Zn(CH₃COO)⁺] = Kf × [Zn²⁺] × [CH₃COO⁻] = 10⁴ × 1.0 × 10⁻³ × 2.0 × 10⁻³ = 2.0 ×
10⁻² mol/L
Answer: [Zn(CH₃COO)⁺] = 2.0 × 10⁻² mol/L
ADVANCED LEVEL
5. . Consider the reactions: Cd²⁺ + CH₃COO⁻ ⇌ Cd(CH₃COO)⁺ K₁ = 10².⁶
Cd(CH₃COO)⁺ + CH₃COO⁻ ⇌ Cd(CH₃COO)₂ K₂ = 10².²
If [Cd²⁺] = 5.0 × 10⁻⁴ mol/L and [CH₃COO⁻] = 1.0 × 10⁻³ mol/L, calculate [Cd(CH₃COO)₂].
Solution:
[Cd(CH₃COO)⁺] = K₁ × [Cd²⁺] × [CH₃COO⁻]
= 10².⁶ × 5.0 × 10⁻⁴ × 1.0 × 10⁻³ ≈ 2.0 × 10⁻⁴ mol/L
[Cd(CH₃COO)₂] = K₂ × [Cd(CH₃COO)⁺] × [CH₃COO⁻]
= 10².² × 2.0 × 10⁻⁴ × 1.0 × 10⁻³ ≈ 3.2 × 10⁻⁵ mol/L
Answer: [Cd(CH₃COO)₂] ≈ 3.2 × 10⁻⁵ mol/L
6. Derive the expression for the total concentration of metal M considering the following
species: M²⁺, M(CH₃COO)⁺, M(CH₃COO)₂ and the cumulative formation constants β₁ and β₂.
Solution: Mtotal = [M²⁺] + β₁[L][M²⁺] + β₂[L]²[M²⁺] = [M²⁺](1 + β₁[L] + β₂[L]²)
Answer: Mtotal = [M²⁺](1 + β₁[L] + β₂[L]²)
Applied Questions
The point of zero charge (PZC) is a fundamental parameter in surface chemistry, representing the
pH at which the net surface charge of an adsorbent is zero. At this condition, the number of
positive and negative charges on the material's surface is balanced, resulting in no electrostatic
attraction or repulsion toward charged species in solution.
In environmental remediation, especially for the removal of ionic contaminants, the PZC serves
as a guiding reference for optimizing pH conditions. When the solution pH is lower than the
PZC, the adsorbent surface becomes positively charged, favoring the removal of anions.
Conversely, at pH values above the PZC, the surface acquires a negative charge, enhancing the
SM Adsorption Theoretical Aspects and Environmental Applications, 1E Ronaldo Ferreira do
Nascimento
Chapters 1-9
Chapter 1
EASY LEVEL – SOLVED QUESTIONS
1. At pH 2.0, calculate the percentage of Zn²⁺ that exists as the free ion, assuming minimal
complexation under such acidic conditions.
Solution: At this low pH, acetic acid remains mostly undissociated, yielding negligible
CH₃COO⁻. Therefore, virtually no complexation occurs.
Answer: Approximately 100% of Zn²⁺ remains uncomplexed at pH 2.0.
2. .Determine the concentration of acetate ions (CH₃COO⁻) at pH 5.0, given acetic acid‘s
Ka = 1.8 × 10⁻⁵ and [CH₃COOH] = 5.0 × 10⁻³ M.
Solution: From ka = ([H⁺][CH₃COO⁻]/[CH₃COOH]), [H⁺] = 10⁻⁵ M at pH 5.
Substituting:
1.8×10⁻⁵ = (10⁻⁵ x)/(5×10⁻³ - x) ≈ (10⁻⁵ × x)/(5×10⁻³)
Solving for x: x ≈ 9.0 × 10⁻⁴ M
Answer: [CH₃COO⁻] ≈ 9.0 × 10⁻⁴ M
3. Explain why metal-acetate complexation increases with pH.
Solution: Higher pH values promote deprotonation of acetic acid, increasing CH₃COO⁻
availability for complex formation.
Answer: Because more acetate ions are available for binding as pH increases
4. At what approximate pH does Cd(CH₃COO)2 surpass free Cd²⁺ as the dominant species?
Solution: From the speciation diagram, Cd(CH₃COO) 2 becomes predominant around pH 6.
Answer: pH ≈ 6.0
5. If the total Cu²⁺ concentration is 1.0 × 10⁻³ M at pH 8.0 and Cu exists exclusively as
Cu(CH₃COO)₂, calculate the complex concentration.
Solution: All Cu²⁺ is converted to Cu(CH₃COO)₂.
Answer: [Cu(CH₃COO)₂] = 1.0 × 10⁻³ M
MEDIUM LEVEL – SOLVED QUESTIONS
6. Given that 75% of Cu²⁺ is complexed as Cu(CH₃COO)⁺ at pH 6.0 and [Cu]ₜₒₜₐₗ = 2.0 ×
10⁻³ M, determine [Cu²⁺].
,Solution: Free Cu²⁺ = 25% × 2.0 × 10⁻³ = 5.0 × 10⁻⁴ M
Answer: 5.0 × 10⁻⁴ M
7. Using K₁ = 10³ for Zn²⁺ + CH₃COO⁻ ⇌ Zn(CH₃COO)⁺, calculate [Zn(CH₃COO)⁺] if
[Zn²⁺] = 5.0×10⁻⁴ M and [CH₃COO⁻] = 1.0×10⁻³ M.
Solution:[Zn(CH₃COO)⁺] = K₁ × [Zn²⁺] × [CH₃COO⁻] = 10³ × 5.0×10⁻⁴ × 1.0×10⁻³ = 5.0 × 10⁻⁴
M
Answer:5.0 × 10⁻⁴ M
8. Provide a chemical explanation for why Zn(CH₃COO)₂ forms at lower pH than
Cu(CH₃COO)₂.
Solution: Zn²⁺ has a higher affinity for acetate than Cu²⁺, resulting in earlier complex formation.
Answer: Stronger interaction between Zn²⁺ and acetate allows complexation at lower pH.
9. Estimate the amount of acetate needed to fully complex 1.5×10⁻³ M Cu²⁺ into
Cu(CH₃COO)₂ at pH 10.
Solution: 1 mole of Cu(CH₃COO)₂ consumes 2 moles of acetate: Required acetate = 2 × 1.5×10⁻³
= 3.0 × 10⁻³ M
Answer:3.0 × 10⁻³ M
10. Describe the trend in metal-acetate complex formation from pH 3 to 9.
Solution: With increasing pH, acetate concentration increases, favoring formation of mono- and
di-acetate complexes.
Answer: Complexation increases in both quantity and coordination number as pH rises.
DIFFICULT LEVEL – SOLVED QUESTIONS
11. With K₁ = 10².⁸ and K₂ = 10².⁵ for Cd²⁺ complexation, calculate [Cd(CH₃COO)₂] from
1×10⁻³ M Cd²⁺ and 1×10⁻³ M CH₃COO⁻.
Solution: First: [Cd(CH₃COO)⁺] = K₁ × [Cd²⁺] × [CH₃COO⁻] = 10².⁸ × 10⁻³ × 10⁻³ = 6.3×10⁻⁴ M
Then: [Cd(CH₃COO)₂] = K₂ × [Cd(CH₃COO)⁺] × [CH₃COO⁻]
= 10².⁵ × 6.3×10⁻⁴ × 10⁻³ ≈ 2.0 × 10⁻⁴ M
Answer:2.0 × 10⁻⁴ M
12. At pH 7.5, if 20% of Cd²⁺ remains uncomplexed from a 1×10⁻³ M solution, identify the
dominant complexed species.
Solution: Complexed fraction = 80% = 8×10⁻⁴ M Given the pH and graph, Cd(CH₃COO)₂
predominates.
Answer: Major species: Cd(CH₃COO)₂
13. Derive the total metal concentration expression (Mₜ) in terms of stepwise formation
constants β₁ and β₂ and [CH₃COO⁻].
Solution: Mₜ = [M²⁺] + β₁[L][M²⁺] + β₁β₂[L]²[M²⁺]
,Answer: Mₜ = [M²⁺](1 + β₁[L] + β₁β₂[L]²)
14. Explain mathematically how [M²⁺] varies with increasing pH, assuming constant acetate
concentration.
Solution: As pH rises, additional CH₃COO⁻ forms, increasing complexation.Thus, [M²⁺]
decreases due to equilibrium shift.
Answer: [M²⁺] ∝ 1 / (1 + β₁[L] + β₁β₂[L]²); it decreases with pH.
15. If total Cu²⁺ = 10⁻³ M and β₁ = 10³, β₂ = 10², calculate the concentrations of Cu²⁺,
Cu(CH₃COO)⁺, and Cu(CH₃COO)₂ at [CH₃COO⁻] = 1.0×10⁻³ M.
Solution: Let x = [Cu²⁺] Then:
Cu(CH₃COO)⁺ = β₁ × x × [L]
Cu(CH₃COO)₂ = β₁β₂ × x × [L]²
Total Cu = x + β₁x[L] + β₁β₂x[L]²
= x(1 + β₁[L] + β₁β₂[L]²)
Solving for x: x = 10⁻³ / (1 + 10³×10⁻³ + 10³×10²×10⁻⁶)
= 10⁻³ / (1 + 1 + 0.1) =4.76×10−4
LEARNING EXERCISES
EASY LEVEL
1. At 25 °C, determine the hydrogen ion concentration in a 0.01 mol/L acetic acid solution,
given Ka = 1.8 × 10⁻⁵.
Solution: CH₃COOH ⇌ CH₃COO⁻ + H⁺
Ka = [H⁺]² / [CH₃COOH] ≈ x² / 0.01
x² = 1.8 × 10⁻⁵ × 0.01 = 1.8 × 10⁻⁷
x = √(1.8 × 10⁻⁷) ≈ 1.34 × 10⁻⁴ mol/L
Answer: [H⁺] = 1.34 × 10⁻⁴ mol/L
2. If the total concentration of Zn²⁺ is 2.0 × 10⁻³ mol/L and no complexation occurs at pH 3,
determine the concentration of free Zn²⁺.
Solution: Since no complexation takes place, the total concentration is equal to the free ion
concentration.
Answer: [Zn²⁺] = 2.0 × 10⁻³ mol/L
INTERMEDIATE LEVEL
3. A solution contains 1.0 × 10⁻³ mol/L of Cu²⁺ at pH 7, and 80% is complexed as
Cu(CH₃COO)⁺. Calculate the concentration of free Cu²⁺.
, Solution: 20% remains free → 0.20 × 1.0 × 10⁻³ = 2.0 × 10⁻⁴ mol/L
Answer: [Cu²⁺] = 2.0 × 10⁻⁴ mol/L
4. Given Kf = 10⁴ for Zn²⁺ + CH₃COO⁻ ⇌ Zn(CH₃COO)⁺, calculate [Zn(CH₃COO)⁺] if [Zn²⁺] =
1.0 × 10⁻³ mol/L and [CH₃COO⁻] = 2.0 × 10⁻³ mol/L.
Solution: [Zn(CH₃COO)⁺] = Kf × [Zn²⁺] × [CH₃COO⁻] = 10⁴ × 1.0 × 10⁻³ × 2.0 × 10⁻³ = 2.0 ×
10⁻² mol/L
Answer: [Zn(CH₃COO)⁺] = 2.0 × 10⁻² mol/L
ADVANCED LEVEL
5. . Consider the reactions: Cd²⁺ + CH₃COO⁻ ⇌ Cd(CH₃COO)⁺ K₁ = 10².⁶
Cd(CH₃COO)⁺ + CH₃COO⁻ ⇌ Cd(CH₃COO)₂ K₂ = 10².²
If [Cd²⁺] = 5.0 × 10⁻⁴ mol/L and [CH₃COO⁻] = 1.0 × 10⁻³ mol/L, calculate [Cd(CH₃COO)₂].
Solution:
[Cd(CH₃COO)⁺] = K₁ × [Cd²⁺] × [CH₃COO⁻]
= 10².⁶ × 5.0 × 10⁻⁴ × 1.0 × 10⁻³ ≈ 2.0 × 10⁻⁴ mol/L
[Cd(CH₃COO)₂] = K₂ × [Cd(CH₃COO)⁺] × [CH₃COO⁻]
= 10².² × 2.0 × 10⁻⁴ × 1.0 × 10⁻³ ≈ 3.2 × 10⁻⁵ mol/L
Answer: [Cd(CH₃COO)₂] ≈ 3.2 × 10⁻⁵ mol/L
6. Derive the expression for the total concentration of metal M considering the following
species: M²⁺, M(CH₃COO)⁺, M(CH₃COO)₂ and the cumulative formation constants β₁ and β₂.
Solution: Mtotal = [M²⁺] + β₁[L][M²⁺] + β₂[L]²[M²⁺] = [M²⁺](1 + β₁[L] + β₂[L]²)
Answer: Mtotal = [M²⁺](1 + β₁[L] + β₂[L]²)
Applied Questions
The point of zero charge (PZC) is a fundamental parameter in surface chemistry, representing the
pH at which the net surface charge of an adsorbent is zero. At this condition, the number of
positive and negative charges on the material's surface is balanced, resulting in no electrostatic
attraction or repulsion toward charged species in solution.
In environmental remediation, especially for the removal of ionic contaminants, the PZC serves
as a guiding reference for optimizing pH conditions. When the solution pH is lower than the
PZC, the adsorbent surface becomes positively charged, favoring the removal of anions.
Conversely, at pH values above the PZC, the surface acquires a negative charge, enhancing the
