Chapter 8. Acids & Bases
8.1 Introduction to Acids & Bases
8.1.1
8.1.2
8.1.3
8.1.5
pOH = 3.328 (rounded to three decimal places)
pH greater than 7 indicates a basic solution.
Detailed explanation here.
PART A.
Rewritten equation: HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)
The rewritten equation includes water, showing that the H⁺ ion does not exist in isolation but forms H₃O⁺ by bonding with H₂O.
Detailed explanation here.
PART B.
H₃O⁺ is the better representation because it acknowledges the interaction between H⁺ and water molecules in the solution. In reality, protons do not exist freely in water; they are stabilized by forming the hydronium ion.
Detailed explanation here.
[H₃O⁺] = 2.35 × 10⁻⁷ M, [OH⁻] = 2.35 × 10⁻⁷ M
Detailed explanation here.
Neutrality does not always mean pH=7.0 because Kw changes with temperature.
Detailed explanation here.
8.3 Weak Acids and Base Equilibria
8.3.1
8.3.2
8.3.3
8.3.4
8.3.5
The pH of the 0.20 M nitrous acid solution is 2.02.
Detailed explanation here.
PART A.
the acids in order of increasing acid strength are: Phenol < Hypochlorous acid < Benzoic acid < Lactic acid < Formic acid.
Detailed explanation here.
PART B. Formic acid is the strongest acid (lowest pKa), so its conjugate base (formate ion) is the weakest.
Detailed explanation here.
PART C. Phenol has the strongest conjugate base.
Detailed explanation here.
The compounds ranked in order of increasing strength of their conjugate acid or base are: 5, 4, 2, 1, 6, 3.
Detailed explanation here.
Substitute pOH=2.67: pH = 11.33
Detailed explanation here.
Percent ionization is 1.89%
Detailed explanation here.
8.4 Acid-Base Reactions and Buffers
8.4.1
8.4.2
8.4.3
8.4.4
8.4.5
pH = 1.65
Detailed explanation here.
pH = 8.22
Detailed explanation here.
PART A. NH₃ + H⁺ → NH₄⁺
Detailed explanation here.
PART B. Since both NH₃ (a weak base) and NH₄⁺ (its conjugate acid) are present, a buffer solution is formed.
Detailed explanation here.
PART C. The pH of the buffer solution is 9.07
Detailed explanation here.
PART A. When H₂CO₃ (a weak acid) and NH₃ (a weak base) are mixed, they undergo an acid-base reaction: H₂CO₃ donates a proton (H⁺) to NH₃, forming HCO₃⁻ (bicarbonate ion) and NH₄⁺ (ammonium ion).
Detailed explanation here.
PART B.
Carbonic acid (H₂CO₃) and its conjugate base (HCO₃⁻).
Ammonia (NH₃) and its conjugate acid (NH₄⁺).
Detailed explanation here.
PART C.
The position of equilibrium depends on the relative strengths of the weak acid (H₂CO₃) and the weak base (NH₃). Because NH₃ is a stronger base than HCO₃⁻, the reaction slightly favors the products, but significant amounts of reactants remain.
Detailed explanation here.
Since Keq is a very small value (2.44 × 10⁻¹⁴), the equilibrium strongly favors the reactants (HCOOH and CH₃NH₂).
Detailed explanation here.
8.5 Acid-Base Titrations
8.5.1
8.5.2
8.5.3
8.5.4
PART A.
Detailed explanation here.
PART B.
Detailed explanation here.
PART C.
Detailed explanation here.
PART D.
Detailed explanation here.
PART E.
Detailed explanation here.
PART F.
Detailed explanation here.
37.5 mL of 0.100 M NaOH is required to reach the equivalence point.
Detailed explanation here.
The pKa of the weak acid is 5.22.
Detailed explanation here.
PART A.
The initial pH is determined by H₃PO₄, which acts as a weak acid.
The first buffering region appears near pH = 2.1.
The first equivalence point occurs when H₃PO₄ is completely converted to H₂PO₄⁻.
The second buffering region appears near pH = 7.2.
The second equivalence point occurs when H₂PO₄⁻ is completely converted to HPO₄²⁻.
The third buffering region appears near pH = 12.4.
The third equivalence point occurs when HPO₄²⁻ is completely converted to PO₄³⁻.
Detailed explanation here.
PART B.
Before Any NaOH Is Added:
Major species: H₃PO₄, H⁺
At V = 25.0 mL (Halfway to the First Equivalence Point):
Major species: H₃PO₄, H₂PO₄⁻
At the First Equivalence Point (V = 50.0 mL):
Major species: H₂PO₄⁻
At V = 75.0 mL (Halfway to the Second Equivalence Point):
Major species: H₂PO₄⁻, HPO₄²⁻
At the Second Equivalence Point (V = 100.0 mL):
Major species: HPO₄²⁻
At the Third Equivalence Point (V = 150.0 mL):
Major species: PO₄³⁻
Detailed explanation here.
8.6 Molecular Structure of Acids and Bases
8.6.1
PART A.
Carbonic Acid (H₂CO₃): The acidic protons are the hydrogens bonded to the -OH groups. Carbonic acid is diprotic, meaning it can donate two protons.
Formic Acid (HCOOH): The acidic proton is the hydrogen atom bonded to the oxygen in the -COOH (carboxylic acid) group.
Methylammonium Ion (CH₃NH₃⁺): The acidic proton is one of the hydrogens bonded to the -NH₃⁺ group.
Detailed explanation here.
8.7 pH and pKa
8.7.1
8.7.2
PART B.
At pH = 6.21, the concentration of H₂PO₄⁻ is 10 times higher than that of HPO₄²⁻.
Detailed explanation here.
PART C.
At pH = 8.21, the concentration of HPO₄²⁻ is 10 times higher than that of H₂PO₄⁻.
Detailed explanation here.
PART D.
At pH = pKa: The concentrations of H₂PO₄⁻ and HPO₄²⁻ are equal.
At pH < pKa: The acid form (H₂PO₄⁻) predominates because the solution is more acidic than the pKa. The lower the pH below the pKa, the larger the ratio [H₂PO₄⁻]/[HPO₄²⁻].
At pH > pKa: The base form (HPO₄²⁻) predominates because the solution is more basic than the pKa. The higher the pH above the pKa, the larger the ratio [HPO₄²⁻]/[H₂PO₄⁻].
Detailed explanation here.
8.8 Properties of Buffers
8.8.1
PART A.
The H₂CO₃/HCO₃⁻ system is effective because H₂CO₃ (the conjugate acid) can react with added base (OH⁻), and HCO₃⁻ (the conjugate base) can react with added acid (H⁺). This dual capability helps stabilize the pH.
Detailed explanation here.
PART B.
If HCl is added, the HCO₃⁻ reacts with the added H⁺ to form H₂CO₃. This prevents a significant drop in pH.
Detailed explanation here.
PART C.
If NaOH is added, the H₂CO₃ reacts with the OH⁻ to form HCO₃⁻ and H₂O. This prevents a significant rise in pH.
Detailed explanation here.
PART D.
The buffer's ability to stabilize pH depends on the concentrations of H₂CO₃ and HCO₃⁻. If a large amount of HCl is added, it can exceed the capacity of HCO₃⁻ to neutralize H⁺. Once the HCO₃⁻ is depleted, the pH will drop significantly.
Detailed explanation here.
8.9 Henderson-Hasselbalch Equation
8.9.1
PART A.
The bicarbonate ion (HCO₃⁻), which is present in significant amounts in the buffer, reacts with the added H⁺ to form carbonic acid (H₂CO₃): HCO₃⁻ + H⁺ → H₂CO₃. This reaction removes the added H⁺ from the solution, preventing a significant drop in pH.
Detailed explanation here.
PART B.
The carbonic acid (H₂CO₃), which is present in significant amounts in the buffer, reacts with the added OH⁻ to form bicarbonate (HCO₃⁻) and water: H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O. This reaction removes the added OH⁻ from the solution, preventing a significant rise in pH.
Detailed explanation here.
8.10 Buffer Capacity
8.10.1
8.10.2
PART A.
The pH of Buffer X and Buffer Y is the same because the ratio of the buffer components is identical.
Detailed explanation here.
PART B.
Buffer Y has a greater capacity to neutralize added acid or base than Buffer X.
Detailed explanation here.
PART C.
Buffer Y, with higher concentrations (1.00 M), will experience a smaller change in pH compared to Buffer X (0.10 M).
Detailed explanation here.
PART A.
Buffer Y has more HCO₃⁻ than Buffer X (0.15 M vs. 0.05 M), so Buffer Y has a greater capacity to resist changes in pH when HCl is added.
Detailed explanation here.
PART B.
Buffer X has more H₂CO₃ than Buffer Y (0.15 M vs. 0.05 M), so Buffer X has a greater capacity to resist changes in pH when NaOH is added.
Detailed explanation here.
PART C.
The relative concentrations of H₂CO₃ (conjugate acid) and HCO₃⁻ (conjugate base) determine the buffer’s ability to neutralize added acids or bases:
A buffer with more conjugate base (HCO₃⁻) has a greater capacity to neutralize added acids (H⁺).
A buffer with more conjugate acid (H₂CO₃) has a greater capacity to neutralize added bases (OH⁻)
Detailed explanation here.