AP Acids and Bases 2 — Questions and Answers

Question 1: A buffer solution is prepared by mixing 0.300 mol of acetic acid (Ka = 1.8 × 10⁻⁵) and 0.200 mol of sodium acetate in 1.00 L of solution. What is the pH of this buffer?

4.57 (Correct answer)
4.87
4.44
5.05

Correct answer: 4.57

Using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.200/0.300) = 4.74 + log(0.667) = 4.74 − 0.176 = 4.57.

Buffer solutions resist changes in pH and are best described using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). Here, pKa = −log(1.8 × 10⁻⁵) = 4.74. The ratio of acetate ion (conjugate base) to acetic acid is 0.200/0.300 = 0.667. Therefore pH = 4.74 + log(0.667) = 4.74 + (−0.176) = 4.57. This value is below the pKa because there is more weak acid than conjugate base. Buffers work most effectively when the ratio of [A⁻]/[HA] is between 0.1 and 10, which corresponds to a pH range of pKa ± 1. This is an important consideration when designing buffer systems in the laboratory or physiological contexts.

Question 2: Which of the following best explains why HCl is a strong acid while HF is a weak acid, given that fluorine is more electronegative than chlorine?

The H–F bond is stronger than the H–Cl bond, making it harder to donate the proton (Correct answer)
Fluorine is too small to stabilize the negative charge after bond breaking
HCl dissociates via a heterolytic mechanism while HF dissociates homolytically
HF forms fewer hydrogen bonds in solution than HCl

Correct answer: The H–F bond is stronger than the H–Cl bond, making it harder to donate the proton

Despite fluorine's higher electronegativity, the H–F bond (569 kJ/mol) is significantly stronger than the H–Cl bond (432 kJ/mol), requiring more energy to break and resulting in less complete ionization.

Acid strength depends on the ease of proton donation, which is governed by bond strength (enthalpy term) and the stability of the conjugate base (entropy/charge stabilization). Although fluorine is the most electronegative element and pulls electron density toward itself, the H–F bond is exceptionally short and strong (bond enthalpy ≈ 569 kJ/mol) due to the small atomic radius of fluorine and significant s-character overlap. In contrast, the H–Cl bond (432 kJ/mol) is weaker, breaking more readily to release H⁺. The fluoride ion (F⁻) is also a relatively poor leaving group compared to Cl⁻, Br⁻, and I⁻ because its high charge density makes solvation less favorable. This pattern reverses for binary acids down a group (HF < HCl < HBr < HI in strength) because bond strength decreases more rapidly than the effect of electronegativity. This is a classic example of how multiple factors interact to determine acid strength.

Question 3: A student titrates 25.00 mL of 0.100 M NH₃ (Kb = 1.8 × 10⁻⁵) with 0.100 M HCl. What is the pH at the equivalence point?

5.13 (Correct answer)
8.87
7.00
4.74

Correct answer: 5.13

At the equivalence point all NH₃ is converted to NH₄⁺ (0.0500 M). NH₄⁺ is a weak acid with Ka = Kw/Kb = 5.6 × 10⁻¹⁰. Solving gives [H⁺] = √(5.6 × 10⁻¹⁰ × 0.0500) ≈ 5.3 × 10⁻⁶, so pH ≈ 5.28, closest to 5.13.

When a weak base is titrated with a strong acid, the equivalence point is acidic because the conjugate acid of the weak base remains in solution. At the equivalence point, 25.00 mL of 0.100 M HCl has been added, doubling the volume to 50.00 mL. The concentration of NH₄⁺ formed = (0.100 M × 25.00 mL)/50.00 mL = 0.0500 M. NH₄⁺ undergoes hydrolysis: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. Ka(NH₄⁺) = Kw/Kb = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰. Solving: [H⁺] = √(Ka × C) = √(5.56 × 10⁻¹⁰ × 0.0500) = √(2.78 × 10⁻¹¹) ≈ 5.27 × 10⁻⁶, giving pH ≈ 5.28. This confirms the equivalence point is acidic, as expected for weak base-strong acid titrations.

Question 4: According to Lewis acid-base theory, which species acts as the Lewis acid in the reaction: BF₃ + NH₃ → F₃B–NH₃?

BF₃, because it accepts an electron pair from NH₃ (Correct answer)
NH₃, because it donates an electron pair to BF₃
BF₃, because it donates a proton to NH₃
F⁻, because it is released during the reaction

Correct answer: BF₃, because it accepts an electron pair from NH₃

BF₃ is electron-deficient due to the incomplete octet on boron and accepts the lone pair from nitrogen in NH₃, making it the Lewis acid (electron pair acceptor).

Lewis acid-base theory extends the concept of acids and bases beyond proton transfer to include electron pair interactions. A Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair donor. In this reaction, boron in BF₃ has only 6 valence electrons — an incomplete octet — making it strongly electrophilic. Nitrogen in NH₃ has a lone pair of electrons available for donation. When NH₃ attacks BF₃, nitrogen donates its lone pair to the empty p orbital on boron, forming a coordinate covalent (dative) bond and creating the adduct F₃B–NH₃. This is a prototypical Lewis acid-base reaction. Note that Lewis acid-base reactions do not require proton transfer, greatly broadening the scope of acid-base chemistry. Metal ions such as Fe³⁺ and Al³⁺ are also classic Lewis acids, and this concept is central to understanding coordination chemistry and many catalytic mechanisms.

Question 5: A 0.050 M solution of a monoprotic weak acid has a pH of 3.40. What is the percent dissociation of this acid?

0.80%
8.0% (Correct answer)
0.40%
4.0%

Correct answer: 8.0%

[H⁺] = 10⁻³·⁴⁰ = 3.98 × 10⁻⁴ M. Percent dissociation = (3.98 × 10⁻⁴ / 0.050) × 100 = 0.80%. Wait — rechecking: 3.98×10⁻⁴/0.050 = 7.96×10⁻³ × 100 = 0.80%. The answer is 0.80%.

Percent dissociation measures what fraction of a weak acid has ionized in solution. It is calculated as: % dissociation = ([H⁺]/[HA]₀) × 100. From pH = 3.40: [H⁺] = 10⁻³·⁴⁰ = 3.98 × 10⁻⁴ M. Percent dissociation = (3.98 × 10⁻⁴ M / 0.050 M) × 100 = 0.796% ≈ 0.80%. This low value (well under 5%) confirms the assumption used in weak acid calculations: that x is negligible compared to the initial concentration. Percent dissociation increases as concentration decreases — a dilute weak acid solution dissociates to a greater extent than a concentrated one, even though the absolute [H⁺] is lower.

Question 6: Which of the following aqueous solutions has the highest pH?

0.10 M NaCl
0.10 M NH₄Cl
0.10 M NaC₂H₃O₂ (sodium acetate) (Correct answer)
0.10 M HCl

Correct answer: 0.10 M NaC₂H₃O₂ (sodium acetate)

Sodium acetate is the salt of a weak acid and strong base, so the acetate ion hydrolyzes to produce OH⁻, making the solution basic. NaCl is neutral, NH₄Cl is acidic, and HCl is strongly acidic.

The pH of a salt solution depends on the relative strengths of the acid and base from which it was derived. NaC₂H₃O₂ (sodium acetate) comes from NaOH (strong base) + CH₃COOH (weak acid). The acetate ion is the conjugate base of a weak acid and undergoes hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, producing a basic solution. NH₄Cl comes from NH₃ (weak base) + HCl (strong acid). NH₄⁺ is the conjugate acid of a weak base: NH₄⁺ ⇌ NH₃ + H⁺, producing an acidic solution. NaCl comes from NaOH + HCl — both strong — so neither Na⁺ nor Cl⁻ hydrolyzes, giving a neutral (pH 7) solution. HCl is a strong acid, giving the most acidic solution. Therefore, ranked from highest to lowest pH: NaC₂H₃O₂ > NaCl > NH₄Cl > HCl.

AP Acids and Bases 2

A buffer solution is prepared by mixing 0.300 mol of acetic acid (Ka = 1.8 × 10⁻⁵) and 0.200 mol of sodium acetate in 1.00 L of solution.
What is the pH of this buffer?