Acids and Bases Worksheet 2 - A page of chemistry problems covering topics such as Bronsted-Lowry acids and bases, conjugate pairs, buffer systems, and pH calculations.

Let’s go through each question one by one, with detailed explanations.

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**1. In the reaction: CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻
The carbonate ion is acting as a(n):**

(A) Arrhenius base
(B) Arrhenius acid
(C) Bronsted-Lowry base
(D) Bronsted-Lowry acid

✔ Correct Answer: (C) Bronsted-Lowry base

Explanation:
In this reaction, CO₃²⁻ accepts a proton (H⁺) from water to become HCO₃⁻, and water donates the proton to become OH⁻.
According to the Bronsted-Lowry definition, a base is a proton acceptor — so CO₃²⁻ is acting as a Bronsted-Lowry base.
It is *not* an Arrhenius base because Arrhenius bases are defined as substances that produce OH⁻ in water — while OH⁻ is produced here, it's a *result* of the reaction, not because CO₃²⁻ itself dissociated to give OH⁻. The base here is accepting a proton, which fits Bronsted-Lowry.

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2. Which of the following represents a Bronsted-Lowry conjugate acid-base pair?

(A) SO₃²⁻ and SO₂
(B) CO₃²⁻ and CO
(C) H₃O⁺ and H₂
(D) NH₄⁺ and NH₃

✔ Correct Answer: (D) NH₄⁺ and NH₃

Explanation:
A conjugate acid-base pair consists of two species that differ by only one proton (H⁺).
- NH₃ (base) + H⁺ → NH₄⁺ (conjugate acid) → ✔️ Valid pair.
- SO₃²⁻ and SO₂: SO₂ is not the conjugate acid of SO₃²⁻; the conjugate acid would be HSO₃⁻.
- CO₃²⁻ and CO: CO is not related by proton transfer.
- H₃O⁺ and H₂: H₂ is not the conjugate base of H₃O⁺; the conjugate base is H₂O.

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3. Which of the following would not make a good buffering system?

(A) SO₄²⁻ and H₂SO₄
(B) HCO₃⁻ and H₂CO₃
(C) NH₃ and NH₄⁺
(D) CH₃COO⁻ and CH₃COOH

✔ Correct Answer: (A) SO₄²⁻ and H₂SO₄

Explanation:
A good buffer consists of a weak acid and its conjugate base (or weak base and its conjugate acid).
- H₂SO₄ is a strong acid — it completely dissociates, so its conjugate base (HSO₄⁻, not SO₄²⁻) is too weak to effectively buffer. Also, SO₄²⁻ is the conjugate base of HSO₄⁻, not H₂SO₄ directly. So pairing SO₄²⁻ with H₂SO₄ doesn’t make sense chemically for buffering.
- (B), (C), (D) are all classic weak acid/base pairs: carbonic/bicarbonate, ammonia/ammonium, acetic/acetate — all excellent buffers.

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4. A water solution of which of the following compounds will turn blue litmus red?

(A) K₂CO₃
(B) NH₄Cl
(C) NaOH
(D) NaCl

✔ Correct Answer: (B) NH₄Cl

Explanation:
Blue litmus turns red in acidic solutions.
- K₂CO₃ → basic (carbonate hydrolyzes to give OH⁻)
- NH₄Cl → acidic (NH₄⁺ hydrolyzes to give H⁺)
- NaOH → strongly basic
- NaCl → neutral (salt of strong acid + strong base)

So only NH₄Cl gives an acidic solution → turns blue litmus red.

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5. 10.0 mL of a solution of HCl required 12.5 mL of 0.400 M Ba(OH)₂ for complete neutralization. How many moles of HCl were present in the sample?

✔ Correct Answer: (B) 1.00 × 10⁻²

Explanation:
Balanced equation:
2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

Moles of Ba(OH)₂ used = M × V (in L) = 0.400 mol/L × 0.0125 L = 0.00500 mol

From stoichiometry:
1 mol Ba(OH)₂ reacts with 2 mol HCl
→ Moles of HCl = 2 × 0.00500 = 0.0100 mol = 1.00 × 10⁻² mol

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6. If 1.0 L of 1.00 M CH₃COOH is mixed with 0.25 mole of solid NaOH (assume no volume change), what will be the pH of the resulting solution? (Kₐ = 1.8 × 10⁻⁵)

✔ Correct Answer: (B) 4.27

Explanation:
This is a buffer formation problem.

Initial moles of CH₃COOH = 1.00 M × 1.0 L = 1.00 mol
NaOH added = 0.25 mol → reacts with CH₃COOH to form CH₃COO⁻

Reaction:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

After reaction:
CH₃COOH left = 1.00 - 0.25 = 0.75 mol
CH₃COO⁻ formed = 0.25 mol

Since volume is still 1.0 L, concentrations are:
[CH₃COOH] = 0.75 M, [CH₃COO⁻] = 0.25 M

Use Henderson-Hasselbalch equation:
pH = pKa + log([A⁻]/[HA])
pKa = -log(1.8 × 10⁻⁵) ≈ 4.74

pH = 4.74 + log(0.25 / 0.75) = 4.74 + log(1/3) = 4.74 - 0.477 ≈ 4.26

Closest answer: (B) 4.27

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7. What is the molarity of CH₃COOH in vinegar containing 4.0% CH₃COOH by mass and having a density of 1.02 g/mL?

✔ Correct Answer: (B) 0.68 M

Explanation:
Assume 1 L (1000 mL) of solution.

Mass of solution = density × volume = 1.02 g/mL × 1000 mL = 1020 g

Mass of CH₃COOH = 4.0% of 1020 g = 0.04 × 1020 = 40.8 g

Molar mass of CH₃COOH = 60.05 g/mol ≈ 60 g/mol

Moles of CH₃COOH = 40.8 g / 60 g/mol = 0.68 mol

Molarity = moles / volume in L = 0.68 mol / 1 L = 0.68 M

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8. If 5.00 mL of 15.4 M HNO₃ is diluted to 250 mL, what is the pH of the resulting solution?

✔ Correct Answer: (C) 1.45

Explanation:
First, find new concentration after dilution:

M₁V₁ = M₂V₂
15.4 M × 5.00 mL = M₂ × 250 mL
M₂ = (15.4 × 5) / 250 = 77 / 250 = 0.308 M

HNO₃ is a strong acid, so [H⁺] = 0.308 M

pH = -log[H⁺] = -log(0.308) ≈ 0.511? Wait — that’s not matching any option.

Wait — let me recalculate:

15.4 × 5.00 = 77.0
77.0 / 250 = 0.308 → yes.

-log(0.308) = ?

Calculate:
log(0.308) ≈ log(3.08 × 10⁻¹) = log(3.08) + log(10⁻¹) ≈ 0.489 - 1 = -0.511
→ -log(0.308) = 0.511

But that’s not among the options. Options are:
(A) 0.51 → very close!
(B) 0.76
(C) 1.45
(D) 2.89

Wait — did I misread?

Check calculation again:

M₂ = (15.4 × 5.00) / 250 = 77 / 250 = 0.308 M → correct.

pH = -log(0.308) = ?

Let me compute numerically:

0.308 → log₁₀(0.308) = -0.511 → so pH = 0.511 → rounds to 0.51

✔ So Answer should be (A) 0.51

But the given answer choices and your listed “correct” might have a typo? Let me check if you meant 15.4 M diluted to 250 mL — maybe it's 5.00 mL to 250 mL — yes.

Wait — perhaps the question says “diluted to 250 mL” meaning final volume is 250 mL — yes.

Then why is (C) 1.45 there? That would be if [H⁺] = 0.035 M or something.

Wait — maybe I misread the volume?

Recheck: 5.00 mL of 15.4 M → diluted to 250 mL.

M₂ = (15.4)(5)/250 = 0.308 M → pH = -log(0.308) = 0.511 → Answer (A) 0.51

But you listed the options as:

> (A) 0.51
> (B) 0.76
> (C) 1.45
> (D) 2.89

So (A) 0.51 is correct.

However, in some contexts, people may think HNO₃ is not fully dissociated? No — it’s strong acid.

Perhaps the question meant 5.00 mL diluted with 250 mL water? That would be total volume ~255 mL, but usually “diluted to 250 mL” means final volume is 250 mL.

So unless there’s a trick, answer is (A) 0.51

But since you listed (C) 1.45 as an option and maybe expected, let me see if there’s a mistake.

Wait — perhaps the original concentration is 1.54 M? No, it says 15.4 M.

Alternatively, maybe they want pH after dilution assuming something else?

No — calculation is correct.

✔ Final decision: (A) 0.51

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## ✔ Final Answers:

1. C
2. D
3. A
4. B
5. B
6. B
7. B
8. A *(Note: Some sources may expect 0.51, though 1.45 is incorrect based on math)*

If your worksheet expects 8 to be (C), there may be an error — but mathematically, 0.51 is correct.

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Let me know if you want step-by-step calculations for any specific one!