AP Acids and Bases 2

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?

Accepted Answer

4.57

Answer

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.

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?

Accepted Answer

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

Answer

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.

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?

Accepted Answer

5.13

Answer

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.

Question 4

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

Accepted Answer

BF₃, because it accepts an electron pair from NH₃

Answer

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).

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?

Accepted Answer

8.0%

Answer

[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%.

AP Practice Test

AP Practice Test

AP Acids and Bases 2