AP - Advanced Placement Acids and Bases 1 — Questions and Answers
Question 1: A 0.150 M solution of methylamine (CH₃NH₂) is prepared at 25°C. Given that Kb = 4.4 × 10⁻⁴ for methylamine, what is the pH of this solution?
- 11.90 (Correct answer)
- 2.10
- 11.18
- 12.66
Correct answer: 11.90
Setting up the Kb expression: x² / (0.150 – x) = 4.4 × 10⁻⁴. Solving (or using the quadratic) gives x = [OH⁻] ≈ 7.91 × 10⁻³ M. pOH = –log(7.91 × 10⁻³) ≈ 2.10, so pH = 14.00 – 2.10 = 11.90.
Question 2: Which of the following species is amphiprotic?
- SO₄²⁻
- NH₄⁺
- HCO₃⁻ (Correct answer)
- CO₃²⁻
Correct answer: HCO₃⁻
HCO₃⁻ is amphiprotic because it can donate a proton (acting as an acid: HCO₃⁻ → H⁺ + CO₃²⁻) or accept a proton (acting as a base: HCO₃⁻ + H⁺ → H₂CO₃). SO₄²⁻ and CO₃²⁻ are only bases; NH₄⁺ is only an acid.
Question 3: A 0.050 M solution of a weak acid HA has a measured pH of 3.17 at 25°C. What is the percent dissociation of HA?
- 1.35% (Correct answer)
- 6.34%
- 0.68%
- 3.17%
Correct answer: 1.35%
[H⁺] = 10⁻³·¹⁷ = 6.76 × 10⁻⁴ M. Percent dissociation = ([H⁺] / [HA]₀) × 100 = (6.76 × 10⁻⁴ / 0.050) × 100 = 1.35%.
Question 4: For carbonic acid, Ka1 = 4.3 × 10⁻⁷ and Ka2 = 4.7 × 10⁻¹¹. Which statement best explains why Ka1 is approximately 10,000 times larger than Ka2?
- H₂CO₃ has more hydrogen atoms available for donation than HCO₃⁻
- The second proton must be removed from HCO₃⁻, which is already negatively charged, requiring greater energy to overcome the electrostatic attraction (Correct answer)
- CO₃²⁻ is a stronger base than HCO₃⁻, shifting equilibrium left
- The O–H bond in HCO₃⁻ is longer and therefore weaker than in H₂CO₃
Correct answer: The second proton must be removed from HCO₃⁻, which is already negatively charged, requiring greater energy to overcome the electrostatic attraction
Removing the first proton from the neutral H₂CO₃ molecule is easier. The second proton must be pulled away from the already-negative HCO₃⁻ ion, and the increased charge–charge repulsion (and stronger electrostatic attraction holding that proton) makes the second dissociation far less favorable.
Question 5: A buffer solution has maximum resistance to pH change (maximum buffer capacity) when which condition is met?
- The ratio [A⁻]/[HA] = 10
- The concentration of the weak acid greatly exceeds the conjugate base concentration
- The molar concentrations of weak acid and conjugate base are equal (Correct answer)
- The solution pH equals twice the pKa of the weak acid
Correct answer: The molar concentrations of weak acid and conjugate base are equal
According to the Henderson–Hasselbalch equation, pH = pKa + log([A⁻]/[HA]). When [A⁻] = [HA], pH = pKa and the log term is zero. At this ratio the buffer can absorb equal amounts of added acid or base before the ratio shifts dramatically, giving maximum buffer capacity.
Question 6: 10.0 mL of 0.200 M HCl is diluted to a final volume of 500.0 mL with distilled water. What is the pH of the resulting solution?
- 2.40 (Correct answer)
- 1.70
- 3.10
- 2.70
Correct answer: 2.40
Moles of HCl = 0.0100 L × 0.200 mol/L = 2.00 × 10⁻³ mol. New concentration = 2.00 × 10⁻³ mol / 0.500 L = 4.00 × 10⁻³ M. Since HCl is a strong acid, [H⁺] = 4.00 × 10⁻³ M, and pH = –log(4.00 × 10⁻³) = 2.40.
AP - Advanced Placement Acids and Bases 1
A 0.150 M solution of methylamine (CH₃NH₂) is prepared at 25°C.
Given that Kb = 4.4 × 10⁻⁴ for methylamine, what is the pH of this solution?