Acids and Bases: Unit Diagnostic (Chemistry Unit 8)

What this diagnostic is for

Use this page to find which Unit 8 topics to revisit. Answer every question before opening any answer, then use the table at the end.

These are original Marlbridge practice questions, not past exam questions. They are not calibrated, so your total gives no predicted score.

Calculator and data. Scientific calculator allowed. Data are at 25 °C (K_w = 1.0 × 10⁻¹⁴, pH + pOH = 14.00) unless a question says otherwise. Lettered acids and bases are fictional and monoprotic. Volumes add on mixing.

Question 1 (multiple choice · 8.1)

At a temperature where pK_w = 13.60, a solution has pH 6.95. Which statement is correct?

(A) The solution is acidic, because its pH is below 7.
(B) The solution is neutral, because its pH is close to 7.
(C) The solution is basic, because its pH is above 6.80.
(D) It cannot be classified without measuring [OH⁻].

Answer and explanation

Answer: (C). Neutral means [H₃O⁺] = [OH⁻], so here neutral pH = pK_w ÷ 2 = 6.80. pOH = 6.65, so [OH⁻] = 2.2 × 10⁻⁷ M exceeds [H₃O⁺] = 1.1 × 10⁻⁷ M.

(A) pH 7 divides only when pK_w = 14.00.
(B) Neutral means equal ion concentrations, not “close to 7”.
(D) pK_w and pH already give [OH⁻].

If you missed this: read “Neutral means equal, not pH 7” in the Topic 8.1 study guide.

Question 2 (multiple choice · 8.2)

Which solution has the highest pH?

(A) 0.0050 M Ca(OH)₂
(B) 0.0080 M NaOH
(C) 0.012 M LiOH
(D) 0.0070 M Ba(OH)₂

Answer and explanation

Answer: (D). Each is a strong base; a group 2 hydroxide gives two OH⁻ per formula unit. [OH⁻]: (A) 0.010 M, (B) 0.0080 M, (C) 0.012 M, (D) 0.014 M, giving pH 12.00, 11.90, 12.08 and 12.15.

(A), (B) give less OH⁻.
(C) has the largest formula concentration, but one OH⁻ per unit.

If you missed this: read “Strong bases” in the Topic 8.2 study guide.

Question 3 (multiple choice · 8.3)

A 0.050 M solution of the weak acid HV is 2.0% ionized. What is K_a for HV?

(A) 1.0 × 10⁻³
(B) 2.0 × 10⁻⁵
(C) 4.0 × 10⁻⁴
(D) 1.0 × 10⁻⁶

Answer and explanation

Answer: (B). [H₃O⁺] = [V⁻] = 0.020 × 0.050 = 1.0 × 10⁻³ M, and [HV] = 0.050 − 0.0010 = 0.049 M. K_a = (1.0 × 10⁻³)² ÷ 0.049 = 2.0 × 10⁻⁵.

(A) is [H₃O⁺], not K_a.
(C) squares the fraction ionized (0.020²).
(D) forgets to divide by [HV].

If you missed this: read “Percent ionization” in the Topic 8.3 study guide.

Question 4 (short answer · 8.3)

The weak base D has pK_b = 4.30.

(a) Calculate the pH and percent ionization of 0.080 M D. (b) Calculate K_a for HD⁺, the conjugate acid of D. Is HD⁺ stronger or weaker than an acid with K_a = 1.0 × 10⁻⁹?

Worked answer

(a) D + H₂O ⇌ HD⁺ + OH⁻. K_b = 10⁻⁴·³⁰ = 5.0 × 10⁻⁵.

[OH⁻] = x ≈ √(5.0 × 10⁻⁵ × 0.080) = 2.0 × 10⁻³ M, which is 2.5% ionized (under 5%, so the shortcut is fine). pOH = 2.70, so pH = 11.30.

(b) K_a = K_w ÷ K_b = 1.0 × 10⁻¹⁴ ÷ 5.0 × 10⁻⁵ = 2.0 × 10⁻¹⁰ (pK_a = 9.70). Smaller K_a, so HD⁺ is weaker.

If you missed this: read “Weak bases work the same way” and “Linking a conjugate pair” in the Topic 8.3 study guide.

Question 5 (multiple choice · 8.4)

20.0 mL of 0.100 M HK (pK_a = 4.20) is mixed with 25.0 mL of 0.100 M NaOH. What is the pH of the mixture?

(A) 4.20
(B) 8.42
(C) 12.05
(D) 12.30

Answer and explanation

Answer: (C). React first: 2.00 × 10⁻³ mol HK and 2.50 × 10⁻³ mol OH⁻ leave 5.0 × 10⁻⁴ mol OH⁻ in 45.0 mL. [OH⁻] = 0.0111 M, pOH = 1.95, pH = 12.05.

(A) assumes a buffer, but no HK is left.
(B) ignores the excess OH⁻ and uses K⁻ alone.
(D) divides by 25.0 mL, not the total 45.0 mL.

If you missed this: read “Weak acid with strong base: three possible outcomes” in the Topic 8.4 study guide.

Question 6 (multiple choice · 8.5)

25.00 mL of a monoprotic acid is titrated with 0.120 M NaOH. The pH starts at 2.86, is 4.65 after 9.00 mL, and is 8.68 at the steepest rise, at 18.00 mL. Which conclusion is correct?

(A) Strong acid; concentration 0.0864 M
(B) Weak acid; concentration 0.0864 M; pK_a 4.65
(C) Weak acid; concentration 0.0432 M; pK_a 4.65
(D) Weak acid; concentration 0.0864 M; pK_a 8.68

Answer and explanation

Answer: (B). At equivalence n(acid) = 18.00 mL × 0.120 M = 2.160 mmol, so c = 2.160 ÷ 25.00 = 0.0864 M. At 9.00 mL (half-equivalence) pH = pK_a = 4.65. A basic equivalence point means a weak acid.

(A) A strong acid would start near pH 1.06 and reach pH 7.00 at equivalence.
(C) uses the half-equivalence volume for the moles.
(D) reads pK_a at equivalence, not halfway.

If you missed this: read “The equivalence point: moles, not pH” and “The half-equivalence point” in the Topic 8.5 study guide.

Question 7 (short answer · 8.5)

Two 20.00 mL samples of the weak acid HE (pK_a = 5.20) are titrated: sample 1 is 0.100 M HE with 0.100 M NaOH; sample 2 is 0.0100 M HE with 0.0100 M NaOH.

(a) Compare the equivalence volumes and the pH values at the half-equivalence points. (b) Calculate the pH at each equivalence point and explain the difference.

Worked answer

(a) Both need 20.00 mL: base and acid have equal concentrations. Both half-equivalence points (10.00 mL) have pH 5.20, because pH = pK_a whenever [HE] = [E⁻], at any concentration.

(b) Only E⁻ is left, in 40.00 mL. K_b = K_w ÷ K_a = 1.6 × 10⁻⁹.

Sample 1: [E⁻] = 0.0500 M; [OH⁻] = √(1.6 × 10⁻⁹ × 0.0500) = 8.9 × 10⁻⁶ M; pH = 8.95.
Sample 2: [E⁻] = 0.00500 M; [OH⁻] = 2.8 × 10⁻⁶ M; pH = 8.45.

E⁻ takes protons from water, so both are basic; sample 2’s E⁻ is ten times more dilute, so makes less OH⁻.

If you missed this: read “What sets the pH at the equivalence point” in the Topic 8.5 study guide.

Question 8 (multiple choice · 8.6)

Which ion is the weakest base?

(A) ClO⁻
(B) ClO₂⁻
(C) ClO₄⁻
(D) OH⁻

Answer and explanation

Answer: (C). ClO₄⁻ is the conjugate base of the strong acid HClO₄. Resonance spreads its charge over four oxygens, and they pull electron density away (inductive effect), so it barely takes a proton.

(A) ClO⁻ keeps its charge on one oxygen: the strongest base of the three chlorine ions.
(B) ClO₂⁻ spreads its charge over only two oxygens.
(D) OH⁻ is a strong base.

If you missed this: read “Resonance” in the Topic 8.6 study guide.

Question 9 (multiple choice · 8.7)

The weak acid HD (pK_a = 4.4) is added in small amounts to a solution held at pH 1.4 and to one held at pH 7.4. Which statement is correct?

(A) At pH 1.4 about 99.9% is HD; at pH 7.4 about 99.9% is D⁻.
(B) HD predominates in both, because a weak acid is mostly un-ionized.
(C) At pH 1.4 D⁻ predominates, because a low pH means more ionization.
(D) At pH 7.4 HD and D⁻ are equal, because the solution is neutral.

Answer and explanation

Answer: (A). [D⁻]/[HD] = 10^(pH − pK_a): 10⁻³ at pH 1.4, 10³ at pH 7.4.

(B) “Mostly un-ionized” applies to a weak acid alone, not at a fixed pH.
(C) reverses the rule: pH < pK_a favours the protonated form.
(D) The forms are equal at pH = pK_a (4.4), not pH 7.

If you missed this: read “The rule: compare pH with pK_a” in the Topic 8.7 study guide.

Question 10 (short answer · 8.8)

A buffer contains 0.080 mol NaHCO₃ and 0.050 mol Na₂CO₃.

(a) Identify the acid member and the base member. (b) Write the net ionic equation for the reaction with 0.020 mol of added HCl, and give the amounts of both members afterwards. (c) Write the net ionic equation for the reaction with added NaOH.

Worked answer

(a) The pair is HCO₃⁻ / CO₃²⁻: HCO₃⁻ is the acid member (it has the extra proton) and CO₃²⁻ the base member. Na⁺ is a spectator.

(b) CO₃²⁻(aq) + H₃O⁺(aq) → HCO₃⁻(aq) + H₂O(l). Afterwards: CO₃²⁻ 0.030 mol, HCO₃⁻ 0.100 mol. Both remain, so pH falls only slightly.

(c) HCO₃⁻(aq) + OH⁻(aq) → CO₃²⁻(aq) + H₂O(l)

If you missed this: read “The two reactions that do the work” in the Topic 8.8 study guide.

Question 11 (multiple choice · 8.9)

A buffer is 0.30 M HW and 0.20 M NaW (pK_a of HW = 6.10). A sample is diluted tenfold with water. What is the pH of the diluted buffer?

(A) 5.92
(B) 6.28
(C) 6.92
(D) 7.00

Answer and explanation

Answer: (A). pH = 6.10 + log(0.20 ÷ 0.30) = 6.10 − 0.18 = 5.92. Dilution divides both concentrations by 10, so the ratio, and the pH, stay the same.

(B) inverts the ratio.
(C) treats the buffer like a strong acid, whose pH rises 1 unit per tenfold dilution.
(D) assumes dilution makes a solution neutral.

If you missed this: read “Reading the equation” and “Moles or concentrations?” in the Topic 8.9 study guide.

Question 12 (multiple choice · 8.10)

A buffer made from the weak acid HL (pK_a = 5.10) and its salt NaL has pH 5.40. Which statement about its buffer capacity is correct?

(A) It can neutralize more added acid than added base, because it has more L⁻ than HL.
(B) It can neutralize more added base than added acid, because pH > pK_a means more HL.
(C) Its capacity is equal both ways, because only total concentration matters.
(D) Nothing can be said without the volume.

Answer and explanation

Answer: (A). [L⁻]/[HL] = 10^(5.40 − 5.10) ≈ 2.0: twice as much L⁻, the member that reacts with added acid.

(B) pH above pK_a means more of the base form.
(C) Total concentration sets overall capacity; the ratio sets which direction is stronger.
(D) Volume scales both directions alike.

If you missed this: read “Unequal amounts: capacity is lopsided” in the Topic 8.10 study guide.

Question 13 (multiple choice · 8.11)

Calcium oxalate, CaC₂O₄, is slightly soluble; oxalic acid is a weak acid. Which substance, added to a saturated solution with excess solid, increases the amount that dissolves?

(A) Dilute nitric acid, HNO₃
(B) Sodium oxalate, Na₂C₂O₄
(C) Sodium nitrate, NaNO₃
(D) Calcium chloride, CaCl₂

Answer and explanation

Answer: (A). CaC₂O₄(s) ⇌ Ca²⁺(aq) + C₂O₄²⁻(aq). Oxalate is a weak base, so H₃O⁺ converts it to HC₂O₄⁻. [C₂O₄²⁻] falls, Q < K_sp, and more solid dissolves.

(B) and (D) add a common ion, which lowers the solubility.
(C) Na⁺ and NO₃⁻ react with neither ion.

If you missed this: read “Which salts are sensitive to pH?” in the Topic 8.11 study guide.

Your next step

A short answer counts as missed if any part was wrong.

Topic	Question(s)	If you missed it, read
8.1 Introduction to Acids and Bases	1	Study guide
8.2 pH and pOH of Strong Acids and Bases	2	Study guide
8.3 Weak Acid and Base Equilibria	3, 4	Study guide
8.4 Acid-Base Reactions and Buffers	5	Study guide
8.5 Acid-Base Titrations	6, 7	Study guide
8.6 Molecular Structure of Acids and Bases	8	Study guide
8.7 pH and pK_a	9	Study guide
8.8 Properties of Buffers	10	Study guide
8.9 Henderson-Hasselbalch Equation	11	Study guide
8.10 Buffer Capacity	12	Study guide
8.11 pH and Solubility	13	Study guide

How to use your result

Look at topics, not the total. Count a guess as missed.
Fix gaps in order. Topics 8.1 to 8.3 support everything else; Topics 8.4 and 8.5 add “react first, then equilibrium”; Topics 8.7 to 8.11 compare pH with pK_a. Start with the earliest topic you missed.
Then practise. After each study guide, try that topic’s practice questions, then the Unit 8 mixed review, where each question combines topics.