Unit 8 practice test: Acids and Bases
12 exam-style questions from the unit, free, with an explanation for every option. Four answer choices each, and data sets that share one table, graph or particle diagram, as on the real exam.
Covers: Introduction to Acids and Bases; pH and pOH of Strong Acids and Bases; Weak Acid and Base Equilibria; Acid-Base Reactions and Buffers; Acid-Base Titrations; Molecular Structure of Acids and Bases; pH and pKa; Properties of Buffers; Henderson-Hasselbalch Equation; Buffer Capacity; pH and Solubility.
Graph
A titration with two steep rises
20.0 mL of 0.100 M of an acid, H₂A, is titrated with 0.100 M NaOH.
Data table
| Volume of NaOH added (mL) | pH |
|---|---|
| 0 | 2.51 |
| 2.5 | 3.18 |
| 5 | 3.53 |
| 7.5 | 3.78 |
| 10 | 4 |
| 12.5 | 4.22 |
| 15 | 4.48 |
| 17.5 | 4.85 |
| 19 | 5.27 |
| 20 | 6.25 |
| 21 | 7.23 |
| 22.5 | 7.66 |
| 25 | 8.02 |
| 27.5 | 8.28 |
| 30 | 8.5 |
| 32.5 | 8.72 |
| 35 | 8.98 |
| 37.5 | 9.34 |
| 39 | 9.76 |
| 40 | 10.51 |
| 41 | 11.23 |
| 42.5 | 11.6 |
| 45 | 11.89 |
| 50 | 12.16 |
1. What do the two steep rises, at 20.0 mL and 40.0 mL, show?
- H₂A gives up two protons, one at a time
- The solution contains two different acids
- The NaOH ran out halfway and was refilled
- The first rise is the half-equivalence point and the second the equivalence point
Show the answer
A diprotic acid has one equivalence point per proton: H₂A → HA⁻ (first 20.0 mL), then HA⁻ → A²⁻ (next 20.0 mL). Equal volumes, because each step removes the same moles of protons.
- Correct: H₂A gives up two protons, one at a time: Right: one equivalence point per ionizable proton.
- The solution contains two different acids: Two acids could also give two rises, but the equal 20.0 mL steps fit one diprotic acid with 2.00 × 10⁻³ mol of each proton.
- The NaOH ran out halfway and was refilled: A titration curve plots the pH of the flask; refilling a buret does not show up.
- The first rise is the half-equivalence point and the second the equivalence point: A half-equivalence point lies in a flat buffer region, not at a steep rise.
2. Which reading gives pKa₂, the pKa of HA⁻?
- pH at 30.0 mL, 8.50
- pH at 40.0 mL, 10.51
- pH at 10.0 mL, 4.00
- pH at 20.0 mL, 6.25
Show the answer
Between the first equivalence point (20.0 mL) and the second (40.0 mL), OH⁻ converts HA⁻ to A²⁻. Halfway, at 30.0 mL, [HA⁻] = [A²⁻], so pH = pKa₂.
- Correct: pH at 30.0 mL, 8.50: Right: the second half-equivalence point.
- pH at 40.0 mL, 10.51: That is the second equivalence point, where all HA⁻ has become A²⁻.
- pH at 10.0 mL, 4.00: That is pKa₁, halfway to the first equivalence point.
- pH at 20.0 mL, 6.25: That is the first equivalence point, where HA⁻ is the main species.
3. Apart from Na⁺ and water, which species is present in the greatest amount at 20.0 mL?
- HA⁻
- H₂A
- A²⁻
- OH⁻
Show the answer
At the first equivalence point exactly one proton per H₂A has been removed: all 2.00 × 10⁻³ mol H₂A has become HA⁻.
- Correct: HA⁻: Right: the first equivalence point holds HA⁻.
- H₂A: H₂A is used up at the first equivalence point.
- A²⁻: A²⁻ forms only after 20.0 mL.
- OH⁻: No excess OH⁻ is present until after 40.0 mL.
Particle view
Three buffers of the same acid
Key: HA weak acid; A⁻ conjugate base. Spectator ions and water are not drawn.
4. Which two boxes have the same pH?
- X and Y
- Y and Z
- X and Z
- Each box has a different pH, because each holds a different number of particles
Show the answer
X and Y both have a 1 : 1 ratio, so both are at pH = pKa.
- Correct: X and Y: Right: same ratio.
- Y and Z: Y is 1 : 1 and Z is 5 : 1 acid to base.
- X and Z: X is 1 : 1, Z is 5 : 1.
- Each box has a different pH, because each holds a different number of particles: The pH depends on the ratio, not the count: X and Y share a 1 : 1 ratio.
5. Three OH⁻ ions are added to each box. Which box can no longer act as a buffer afterward?
- X
- Y
- Z
- Y and Z
Show the answer
X has only 2 HA, so the third OH⁻ is left free and no HA remains: the buffer is exhausted.
- Correct: X: Right: its acid component runs out.
- Y: Y goes to 3 HA and 9 A⁻, still a buffer.
- Z: Z goes to 7 HA and 5 A⁻, still a buffer (now closer to 1 : 1).
- Y and Z: Both Y and Z keep both components.
6. Two H₃O⁺ ions are added to Box Z. Which statement is correct?
- Its A⁻ is used up, so Z stops being a buffer
- Z resists well, because it has the most HA
- The pH of Z rises
- Z is unchanged, because it has more acid than base
Show the answer
H₃O⁺ reacts with A⁻. Z has only 2 A⁻, so two H₃O⁺ convert both, leaving 12 HA and no conjugate base.
- Correct: Its A⁻ is used up, so Z stops being a buffer: Right: the base component runs out.
- Z resists well, because it has the most HA: HA does not react with added acid; A⁻ does.
- The pH of Z rises: Added acid lowers the pH.
- Z is unchanged, because it has more acid than base: Added acid reacts with the A⁻ that Z has.
7. Solution P has pH 4.0 and Solution Q has pH 6.0. How does [H₃O⁺] in P compare with Q?
- 100 times greater in P
- 2 times greater in P
- 1.5 times greater in P
- 100 times smaller in P
Show the answer
Two pH units apart is a factor of 10² = 100, and the lower pH has more H₃O⁺.
- Correct: 100 times greater in P: Right: each pH unit is a factor of ten.
- 2 times greater in P: This subtracts the pH values and treats the result as a ratio. The difference is an exponent: 10².
- 1.5 times greater in P: This divides 6 by 4. pH values are logarithms, so their ratio has no meaning.
- 100 times smaller in P: The direction is reversed: a lower pH means more H₃O⁺.
8. What is the pH of 2.5 × 10⁻⁴ M Sr(OH)₂ at 25 °C?
Type a number.
Show the answer
[OH⁻] = 2 × 2.5 × 10⁻⁴ = 5.0 × 10⁻⁴ M; pOH = 3.301; pH = 14.00 − 3.30 = 10.70.
- Answer: 10.70
9. A 0.085 M solution of a weak acid HA has pH 3.25. What is Ka? Give your answer in scientific notation.
Type a number.
Show the answer
[H₃O⁺] = [A⁻] = 10^−3.25 = 5.62 × 10⁻⁴ M. [HA] = 0.085 − 0.00056 ≈ 0.0844 M. Ka = (5.62 × 10⁻⁴)² / 0.0844 = 3.7 × 10⁻⁶.
- Answer: 3.7 × 10-6
10. A solution contains 0.30 M HNO₂ and 0.20 M NaNO₂ (Ka of HNO₂ = 4.0 × 10⁻⁴). What is its pH?
Type a number.
Show the answer
Rearrange Ka = [H₃O⁺][NO₂⁻]/[HNO₂]: [H₃O⁺] = 4.0 × 10⁻⁴ × 0.30/0.20 = 6.0 × 10⁻⁴ M. pH = −log(6.0 × 10⁻⁴) = 3.22. Both amounts barely change, because the NO₂⁻ already present pushes the ionization of HNO₂ even further left.
- Answer: 3.22
11. HCl, HBr and HI are all strong in water, but HF is weak. Which factor best explains why HF holds its proton?
- The H–F bond is short and strong, and F⁻ is small
- F is the least electronegative halogen
- HF molecules are nonpolar
- F⁻ is the conjugate base of a strong acid, so it holds H⁺
Show the answer
F is a small atom, so the H–F bond is short and strong (567 kJ/mol), and F⁻ holds its charge in a tiny volume. Both oppose giving up H⁺, so HF is the one weak hydrohalic acid.
- Correct: The H–F bond is short and strong, and F⁻ is small: Right: bond strength and the small anion.
- F is the least electronegative halogen: F is the most electronegative element.
- HF molecules are nonpolar: H–F is very polar; polarity alone would suggest a strong acid, which is why bond strength matters here.
- F⁻ is the conjugate base of a strong acid, so it holds H⁺: F⁻ is the conjugate base of a weak acid; that is the point to explain.
12. A weak acid has pKa 4.2. In a solution at pH 2.2, which form predominates?
- HA, by about 100 to 1
- A⁻, by about 100 to 1
- HA and A⁻ in equal amounts
- A⁻, by about 2 to 1
Show the answer
pH is 2 units below pKa: [A⁻]/[HA] = 10⁻², so HA outnumbers A⁻ about 100 to 1.
- Correct: HA, by about 100 to 1: Right: pH below pKa, protonated form.
- A⁻, by about 100 to 1: Below pKa, the protonated form predominates, not the deprotonated one.
- HA and A⁻ in equal amounts: Equal amounts happen only at pH = pKa.
- A⁻, by about 2 to 1: This reads the pH difference as a ratio; the exponent is −2, so the factor is 100 in favor of HA.
Keep going
Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8 · Unit 9.