AP Acids and Bases 3

Question 1

What is the pH of a solution prepared by dissolving 5.30 g of Na₂CO₃ (molar mass = 106 g/mol) in enough water to make 500.0 mL of solution? (Ka₂ for H₂CO₃ = 4.7 × 10⁻¹¹)

Accepted Answer

11.48

Answer

10.82

Answer

8.33

Answer

12.00

Question 2

Which indicator would be most appropriate for a titration of a weak acid (pKa = 6.0) with a strong base, given that the equivalence point occurs at pH ≈ 9?

Accepted Answer

Phenolphthalein (pH range 8.2–10.0)

Answer

Methyl orange (pH range 3.1–4.4)

Answer

Bromothymol blue (pH range 6.0–7.6)

Answer

Litmus (pH range 6.0–8.0)

Question 3

A diprotic acid H₂A has Ka₁ = 1.0 × 10⁻³ and Ka₂ = 1.0 × 10⁻⁷. Which species predominates in a solution with pH = 5.0?

Accepted Answer

HA⁻

Answer

H₂A

Answer

A²⁻

Answer

Equal amounts of H₂A and A²⁻

Question 4

Which of the following best describes what happens when a small amount of strong acid is added to a buffer made from acetic acid and sodium acetate?

Accepted Answer

The added H⁺ reacts with acetate ion to form more acetic acid, causing a small decrease in pH

Answer

The added H⁺ reacts with water to form hydronium, raising the pH significantly

Answer

The added H⁺ reacts with acetic acid to form acetate, causing a small increase in pH

Answer

The buffer is destroyed, and the pH drops to the pH of the strong acid solution

Question 5

In a 0.10 M solution of HCN (Ka = 6.2 × 10⁻¹⁰), what is the approximate concentration of CN⁻?

Accepted Answer

7.9 × 10⁻⁶ M

Answer

6.2 × 10⁻⁹ M

Answer

7.9 × 10⁻⁵ M

Answer

6.2 × 10⁻⁵ M

Question 6

Sulfuric acid is classified as a strong acid with respect to its first ionization. What makes the second ionization (HSO₄⁻ → H⁺ + SO₄²⁻, Ka₂ = 0.012) weaker?

Accepted Answer

It is harder to remove a proton from an already negatively charged species due to increased electrostatic attraction

Answer

HSO₄⁻ has lower electronegativity than H₂SO₄

Answer

The second proton is more tightly bound in a covalent bond than the first

Answer

SO₄²⁻ is too large to effectively stabilize the negative charge