These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. All data are at 25 °C: K_w = 1.0 × 10⁻¹⁴ and pK_w = 14.00. Acids and bases named with capital letters (HX, HR, HW, HY, Y) are fictional and monoprotic.
Question 1 (multiple choice · foundation)
A 0.20 M solution of the weak acid HX (K_a = 1.0 × 10⁻⁵) is at equilibrium. Which ranking of concentrations is correct?
- (A) [HX] > [H₃O⁺] ≈ [X⁻] > [OH⁻]
- (B) [H₃O⁺] ≈ [X⁻] > [HX] > [OH⁻]
- (C) [HX] ≈ [H₃O⁺] ≈ [X⁻] > [OH⁻]
- (D) [HX] > [OH⁻] > [H₃O⁺] ≈ [X⁻]
Answer and explanation
Answer: (A). Only a small fraction of a weak acid ionizes, so un-ionized HX is the main species. Each ionization makes one H₃O⁺ and one X⁻, so those two are equal. The solution is acidic, so [OH⁻] is tiny. (Here [H₃O⁺] ≈ 1.4 × 10⁻³ M and [OH⁻] ≈ 7 × 10⁻¹² M.)
- (B) describes a strong acid, which ionizes completely.
- (C) would need [HX] to stay equal to [H₃O⁺], which means about half the acid ionized. A weak acid at this concentration ionizes far less.
- (D) puts [OH⁻] above [H₃O⁺], which would make the solution basic.
Question 2 (multiple choice · core)
What is the pH of 0.040 M HR, a weak acid with K_a = 2.5 × 10⁻⁶?
- (A) 1.40
- (B) 3.50
- (C) 5.60
- (D) 7.00
Answer and explanation
Answer: (B). x = [H₃O⁺] ≈ √(2.5 × 10⁻⁶ × 0.040) = √(1.0 × 10⁻⁷) = 3.16 × 10⁻⁴ M, so pH = 3.50. Check: x is 0.79% of 0.040, so the shortcut holds.
- (A) is −log 0.040: it treats HR as a strong acid.
- (C) is pK_a, not pH.
- (D) is −log(K_a × c): it forgets to take the square root.
Question 3 (multiple choice · core)
A weak acid has pK_a = 3.80. What is K_b for its conjugate base?
- (A) 1.6 × 10⁻⁴
- (B) 6.3 × 10⁻¹¹
- (C) 6.3 × 10³
- (D) 10.20
Answer and explanation
Answer: (B). pK_b = 14.00 − 3.80 = 10.20, so K_b = 10^(−10.20) = 6.3 × 10⁻¹¹. (Or: K_a = 1.58 × 10⁻⁴ and K_b = K_w ÷ K_a.)
- (A) is K_a of the acid itself.
- (C) is 1 ÷ K_a; it leaves out K_w.
- (D) is pK_b, not K_b.
Question 4 (multiple choice · core)
A 0.10 M solution of a weak acid (K_a = 1.0 × 10⁻⁵) is diluted to 0.0010 M at constant temperature. Which statement is correct?
- (A) The pH increases by 2 and the percent ionization stays the same.
- (B) The pH increases by about 1 and the percent ionization increases.
- (C) The pH increases by about 1 and the percent ionization decreases.
- (D) The pH stays the same because K_a does not change.
Answer and explanation
Answer: (B). [H₃O⁺] ≈ √(K_a × c), so a 100-fold dilution lowers [H₃O⁺] by about √100 = 10 times: pH goes from 3.00 to 4.02. Percent ionization rises from 1.0% to 9.5% (the exact quadratic is needed at 0.0010 M). Dilution lowers Q below K_a, so more acid ionizes.
- (A) is what happens to a strong acid, which is always fully ionized.
- (C) has the direction of percent ionization wrong.
- (D) confuses K_a, which is constant, with [H₃O⁺], which is not.
Question 5 (calculation · core)
A 0.120 M solution of the weak base Y has pH 10.85. Calculate (a) [OH⁻], (b) the percent ionization of Y, (c) K_b for Y, (d) pK_a for its conjugate acid HY⁺.
Worked solution
Y + H₂O ⇌ HY⁺ + OH⁻
- pOH = 14.00 − 10.85 = 3.15, so [OH⁻] = 10^(−3.15) = 7.08 × 10⁻⁴ M. This equals [HY⁺].
- Percent ionization = (7.08 × 10⁻⁴ ÷ 0.120) × 100% = 0.59%.
- [Y] = 0.120 − 0.000708 = 0.1193 M. K_b = (7.08 × 10⁻⁴)² ÷ 0.1193 = 4.2 × 10⁻⁶.
- pK_b = −log(4.20 × 10⁻⁶) = 5.38, so pK_a(HY⁺) = 14.00 − 5.38 = 8.62.
Suggested mark points (4): 1 for converting pH to pOH and [OH⁻]; 1 for percent ionization; 1 for K_b with [OH⁻] = [HY⁺]; 1 for pK_a from pK_a + pK_b = 14.00.
Common error: using 10^(−10.85) as [OH⁻]. That is [H₃O⁺], and it gives a K_b that is wildly too small.
Check. Put K_b back into the shortcut: √(4.20 × 10⁻⁶ × 0.120) = 7.10 × 10⁻⁴ M, very close to the measured 7.08 × 10⁻⁴ M. The small ionization (0.59%) also confirms that Y behaves as a weak base, and pK_a 8.62 tells you HY⁺ is a weak acid, weaker than acetic acid.
Question 6 (constructed response · core)
A student prepares 0.0200 M HW, a weak acid with K_a = 4.0 × 10⁻³.
(a) Use the approximation x ≈ √(K_a × c) to estimate [H₃O⁺], then show that the approximation is not valid here. (b) Solve for [H₃O⁺] without the approximation and calculate the pH. (c) Calculate the percent ionization and the equilibrium concentration of HW.
Worked solution and suggested Marlbridge rubric
(a) x ≈ √(4.0 × 10⁻³ × 0.0200) = √(8.0 × 10⁻⁵) = 8.94 × 10⁻³ M. This is (8.94 × 10⁻³ ÷ 0.0200) × 100% = 45% of c, far above 5%, so c − x ≈ c is not valid.
(b) x² ÷ (0.0200 − x) = 4.0 × 10⁻³ gives x² + (4.0 × 10⁻³)x − 8.0 × 10⁻⁵ = 0. x = [−4.0 × 10⁻³ + √((4.0 × 10⁻³)² + 4 × 8.0 × 10⁻⁵)] ÷ 2 = 7.17 × 10⁻³ M. pH = −log(7.17 × 10⁻³) = 2.14.
(c) Percent ionization = (7.17 × 10⁻³ ÷ 0.0200) × 100% = 36%. [HW] = 0.0200 − 0.00717 = 0.0128 M.
| Point | What earns it |
|---|---|
| 1 | Approximate x = 8.9 × 10⁻³ M with a percent (about 45%) compared against 5% |
| 1 | Correct quadratic set up from K_a = x² / (c − x) |
| 1 | x = 7.2 × 10⁻³ M and pH 2.14 |
| 1 | Percent ionization 36% |
| 1 | [HW] = 0.0128 M, subtracting x |
Successive approximation (feeding x back into c − x until it stops changing) is equally valid. The approximate pH (2.05) earns no credit in (b).
Why the shortcut overestimates. Writing c − x ≈ c makes the denominator too large: 0.0200 instead of 0.0128. To keep the same K_a, the shortcut must then make x² too large. The bigger K_a is compared with c, the worse this error. A quick sign: if K_a is more than about 1/400 of c, expect the 5% check to fail. Here K_a is 1/5 of c.
Question 7 (explanation · stretch)
Two solutions are each 0.100 M: one of HX (pK_a = 3.50) and one of HY (pK_a = 6.50).
(a) Without a full calculation, state which solution has the higher pH and explain why. (b) Which is the stronger base, X⁻ or Y⁻? Support your answer with pK_b values. (c) A student says: “HY is the weaker acid, so the HY solution contains fewer un-ionized acid molecules than the HX solution.” Evaluate the claim.
Model answer and suggested Marlbridge rubric
(a) The HY solution has the higher pH. HY has the larger pK_a, so the smaller K_a. At the same concentration, a smaller fraction of HY ionizes, giving a lower [H₃O⁺]. (Calculated: pH 3.75 for HY and 2.26 for HX.)
(b) Y⁻ is the stronger base. pK_b(X⁻) = 14.00 − 3.50 = 10.50; pK_b(Y⁻) = 14.00 − 6.50 = 7.50. The smaller pK_b means a larger K_b. The weaker the acid, the stronger its conjugate base, because K_a × K_b = K_w is fixed.
(c) The claim is wrong. Because HY ionizes less (about 0.18%, against about 5.5% for HX), more of it stays as un-ionized molecules: about 0.0998 M HY against 0.0945 M HX. The student has mixed up “weaker acid” with “less acid present”. Both solutions started with the same amount of acid; acid strength decides only how much of it ionizes.
Note: for HX the shortcut gives 5.6%, just over the 5% limit, so the HX values above come from the quadratic. That does not affect any of the reasoning.
| Point | What earns it |
|---|---|
| 1 | HY has the higher pH, linked to its smaller K_a (larger pK_a) giving lower [H₃O⁺] |
| 1 | Y⁻ is the stronger base, with pK_b values 10.50 and 7.50 |
| 1 | Links conjugate strength to K_a × K_b = K_w (or pK_a + pK_b = 14.00) |
| 1 | Claim is wrong because the weaker acid ionizes less, so more of it stays un-ionized |
How did you do?
- Q1 or Q7(c) wrong: look again at Figure 1 in the study guide.
- Q2, Q5 or Q6 wrong: re-work Worked examples 1 to 3, and always do the 5% check.
- Q3 or Q7(b) wrong: practise switching between pK_a and pK_b with pK_a + pK_b = 14.00.
- Q4 wrong: revisit the dilution table in the “Percent ionization” section.
Then tick off the topic checklist.