Worksheet - Acid-Base Theories Pre-AP - Teacher - NOTE: This ... - Free Printable

Let’s go step by step through each question on the worksheet.

---

Question 1: Fill in the table for Acid-Base Theories

We need to match each theory with what it says about acids and bases.

Recall:

- Arrhenius:
→ Acid = produces H⁺ in water
→ Base = produces OH⁻ in water

- Brønsted-Lowry:
→ Acid = proton (H⁺) donor
→ Base = proton (H⁺) acceptor

- Lewis:
→ Acid = electron pair acceptor
→ Base = electron pair donor

So filling the table:

| Theory | ACID | BASE |
|-----------------|--------------------------|-------------------------|
| Arrhenius | H⁺ in H₂O | OH⁻ in H₂O |
| Brønsted-Lowry | proton donor | proton acceptor |
| Lewis | electron pair acceptor | electron pair donor |

✔ This matches standard definitions.

---

Question 2: Identify acid, base, conjugate acid, conjugate base

Reaction: NH₃ + H₂O ⇌ NH₄⁺ + OH

In Brønsted-Lowry terms:

- NH₃ accepts a proton → becomes NH₄⁺ → so NH₃ is base, NH₄⁺ is its conjugate acid
- H₂O donates a proton → becomes OH⁻ → so H₂O is acid, OH⁻ is its conjugate base

Answer:

Acid: H₂O
Base: NH₃
Conjugate Acid: NH₄⁺
Conjugate Base: OH⁻

---

Question 3: According to Brønsted-Lowry, which species can act as both acid and base? Explain why.

This is asking for an amphoteric substance — one that can donate OR accept a proton.

Look at the reaction from Q2: H₂O acted as an acid (donated H⁺ to become OH⁻). But water can also act as a base! For example:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ ← here H₂O is acid
But if we have: HCl + H₂O → H₃O⁺ + Cl⁻ ← here H₂O is base (accepts H⁺)

So water (H₂O) is amphoteric.

Also, look at the options given later in the worksheet — but since this is open-ended, we explain based on concept.

Actually, looking ahead — Question 4 gives specific reactions. Maybe Q3 is general? Let me check the image again mentally.

Wait — in the original problem set, after Q2, Q3 says:

“According to Brønsted-Lowry, which species in the following reactions can be classified as an acid, base, or both? Explain why.”

Then it lists two reactions:

a) HSO₄⁻(aq) + H₂O(l) ⇌ SO₄²⁻(aq) + H₃O⁺(aq)
b) HPO₄²⁻(aq) + H₂O(l) ⇌ H₂PO₄⁻(aq) + OH⁻(aq)

Ah — so Q3 refers to these two reactions.

Let’s analyze them.

---

Reaction a): HSO₄⁻ + H₂O ⇌ SO₄²⁻ + H₃O⁺

- HSO₄⁻ loses H⁺ → becomes SO₄²⁻ → so HSO₄⁻ is acting as acid
- H₂O gains H⁺ → becomes H₃O⁺ → so H₂O is acting as base

Can either act as both? Not in this reaction alone — but HSO₄⁻ *can* act as base too (if it accepts H⁺ to become H₂SO₄), and H₂O is amphoteric.

But the question says “in the following reactions” — so probably just identify roles in each reaction.

Wait — let’s read carefully: “which species... can be classified as an acid, base, or both?”

It might mean: in these reactions, which species are acting as acid/base, and which could potentially do both (amphoteric)?

Looking at species present:

In reaction a): HSO₄⁻, H₂O, SO₄²⁻, H₃O⁺

- HSO₄⁻: can donate H⁺ (as here) or accept H⁺ → so both
- H₂O: can donate or accept → both
- SO₄²⁻: only accepts H⁺ → base
- H₃O⁺: only donates H⁺ → acid

Similarly, reaction b): HPO₄²⁻ + H₂O ⇌ H₂PO₄⁻ + OH⁻

- HPO₄²⁻ gains H⁺ → becomes H₂PO₄⁻ → so acting as base
- H₂O loses H⁺ → becomes OH⁻ → acting as acid

Again, HPO₄²⁻ can also lose H⁺ to become PO₄³⁻ → so it can act as acid too → both

H₂O again → both

So answer for Q3:

In reaction a):
- HSO₄⁻ can act as both acid and base (amphoteric)
- H₂O can act as both acid and base

In reaction b):
- HPO₄²⁻ can act as both acid and base
- H₂O can act as both acid and base

Explanation: These species can either donate or accept protons depending on the reaction partner.

---

Question 4: Complete the equations using Brønsted-Lowry theory

Given:

a) HCO₃⁻(aq) + H₂O(l) ⇌ ? + ?

HCO₃⁻ can act as acid or base. Here, reacting with water — likely acting as acid (since bicarbonate is weak acid).

If HCO₃⁻ donates H⁺ → becomes CO₃²⁻
Water accepts H⁺ → becomes H₃O⁺

So:
HCO₃⁻ + H₂O ⇌ CO₃²⁻ + HO⁺

Alternatively, if HCO₃⁻ acts as base:
HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻

But typically, in such problems, unless specified, we consider the more common behavior. However, note that HCO₃⁻ is amphoteric — so both are possible. But looking at context — perhaps they want the acidic dissociation.

Wait — let’s see part b.

b) H₂PO₄⁻(aq) + H₂O(l) ⇌ ? + ?

Same idea — dihydrogen phosphate can act as acid or base.

As acid: H₂PO₄⁻ → H⁺ + HPO₄²⁻ → so with water: H₂PO₄⁻ + H₂O ⇌ HPO₄²⁻ + H₃O⁺

As base: H₂PO₄⁻ + H₂O ⇌ H₃PO₄ + OH⁻

Again, ambiguous.

But notice — in many textbooks, when written like this without specification, they often show the acid behavior for HCO₃⁻ and H₂PO₄⁻ because they are weak acids.

However, let’s think differently — maybe the problem expects us to write both possibilities? But the format shows single blanks.

Looking back at the image description — actually, in the user's upload, for Q4a and 4b, it’s written as:

a) HCO₃⁻(aq) + H₂O(l) ⇌ _______ + _______

b) H₂PO₄⁻(aq) + H₂O(l) ⇌ _______ + _______

And then there are multiple choice options below? Wait no — in the initial text provided, after Q4, it jumps to Q5.

Wait — re-examining the original input:

After Q3, it says:

"4. Complete the following equations according to the Brønsted-Lowry theory."

Then:

a) HCO₃⁻(aq) + H₂O(l) ⇌ _____ + _____

b) H₂PO₄⁻(aq) + H₂O(l) ⇌ _____ + _____

Then immediately goes to Q5.

No multiple choices shown for Q4 — so we fill in the products.

Standard approach:

For a) HCO₃⁻ acting as acid:
HCO₃⁻ + H₂O ⇌ CO₃²⁻ + H₃O⁺

For b) H₂PO₄⁻ acting as acid:
H₂PO₄⁻ + H₂O ⇌ HPO₄²⁻ + H₃O⁺

These are the most commonly expected answers.

Note: Both ions are amphoteric, but in aqueous solution, their primary role in such contexts is as weak acids.

So I’ll go with that.

---

Question 5: Given Ka values, find strongest/weakest acid and conjugate base strength

Given:

HF: Ka = 6.8 × 10⁻⁴
HCN: Ka = 4.9 × 10⁻¹⁰
HNO₂: Ka = 4.5 × 10⁻⁴

Higher Ka = stronger acid.

Compare:

HF: 6.8e-4
HNO₂: 4.5e-4
HCN: 4.9e-10 ← smallest

So:

Strongest acid: HF (largest Ka)
Weakest acid: HCN (smallest Ka)

Now, conjugate base strength: weaker acid has stronger conjugate base.

Because: strong acid → weak conjugate base; weak acid → strong conjugate base.

So:

Strongest conjugate base: from weakest acid → CN⁻ (from HCN)
Weakest conjugate base: from strongest acid → F⁻ (from HF)

Confirm:

Ka(HF) > Ka(HNO₂) > Ka(HCN)

So acid strength: HF > HNO₂ > HCN

Conjugate base strength: CN⁻ > NO₂⁻ > F

Thus:

Strongest acid: HF
Weakest acid: HCN
Strongest conjugate base: CN⁻
Weakest conjugate base: F⁻

---

Question 6: pH calculations

a) [H⁺] = 1.0 × 10⁻³ M → pH = -log(1.0 × 10⁻³) = 3.00

b) [OH⁻] = 1.0 × 10⁻⁵ M → pOH = 5.00 → pH = 14.00 - 5.00 = 9.00

c) [H⁺] = 2.5 × 10⁻⁴ M → pH = -log(2.5 × 10⁻⁴)

Calculate: log(2.5 × 10⁻⁴) = log(2.5) + log(10⁻⁴) ≈ 0.3979 - 4 = -3.6021

So pH = -(-3.6021) = 3.6021 ≈ 3.60

d) pOH = 8.75 → pH = 14.00 - 8.75 = 5.25

e) pH = 4.20 → [H⁺] = 10^(-pH) = 10^(-4.20)

10^(-4.20) = 10^(-4) × 10^(-0.20) ≈ 0.0001 × 0.630957 ≈ 6.31 × 10⁻⁵ M

f) pH = 10.50 → [H⁺] = 10^(-10.50) = 3.16 × 10⁻¹¹ M (since 10^(-0.5)=0.3162, so 10^(-10.5)=3.162×10⁻¹¹)

g) pOH = 3.20 → [OH⁻] = 10^(-3.20) = 6.31 × 10⁻⁴ M

h) [OH⁻] = 5.0 × 10⁻² M → pOH = -log(5.0 × 10⁻²) = - (log 5.0 + log 10⁻²) = - (0.6990 - 2) = - (-1.3010) = 1.3010 ≈ 1.30

i) [H⁺] = 7.5 × 10⁻⁶ M → pH = -log(7.5 × 10⁻⁶) = - (log 7.5 + log 10⁻⁶) = - (0.8751 - 6) = - (-5.1249) = 5.1249 ≈ 5.12

j) pH = 12.00 → [H⁺] = 10^(-12.00) = 1.0 × 10⁻¹² M

k) pOH = 1.50 → [OH⁻] = 10^(-1.50) = 0.03162 ≈ 3.16 × 10⁻² M

l) [OH⁻] = 2.0 × 10⁻⁷ M → pOH = -log(2.0 × 10⁻⁷) = - (log 2.0 + log 10⁻⁷) = - (0.3010 - 7) = - (-6.699) = 6.699 ≈ 6.70

m) pH = 6.80 → [H⁺] = 10^(-6.80) = 1.58 × 10⁻⁷ M (since 10^(-0.8)=0.1585, so 1.585×10⁻⁷)

n) pOH = 11.20 → [OH⁻] = 10^(-11.20) = 6.31 × 10⁻¹² M

o) [H⁺] = 1.0 × 10⁻⁷ M → pH = 7.00

p) [OH⁻] = 1.0 × 10⁻⁷ M → pOH = 7.00 → pH = 7.00

All seem straightforward.

---

Question 7: Strongest/Weakest acid from Ka values

Same as Q5 — already done.

Strongest acid: HF
Weakest acid: HCN

---

Question 8: Conjugate base strength

From Q5:

Strongest conjugate base: CN⁻
Weakest conjugate base: F⁻

---

Question 9: Which is stronger acid: HCl or HI?

Both are strong acids, but HI is stronger than HCl because iodine is larger, so H-I bond is weaker and easier to break.

In fact, among hydrohalic acids: HF < HCl < HBr < HI in acidity.

So HI is stronger acid.

Reason: As size of halogen increases down group, bond strength decreases, making it easier to lose H⁺.

---

Question 10: Order acids by increasing strength

Given:

HClO, HClO₂, HClO₃, HClO₄

These are oxyacids of chlorine.

Rule: More oxygen atoms attached to central atom → stronger acid.

Because more O atoms pull electron density away from O-H bond, making H⁺ easier to release.

Number of O atoms:

HClO: 1 O (actually, structure is HO-Cl, so one terminal O? Wait — standard way: count number of terminal oxygen atoms not bonded to H.

General rule for oxyacids: acid strength increases with number of terminal oxygen atoms.

HClO: hypochlorous acid → Cl has oxidation state +1 → one O double bond? Actually, formula is HOCl, so only one O besides OH.

Better to use: for acids with same central atom, more O atoms → stronger acid.

HClO: 1 O atom total? No — molecular formulas:

HClO: hypochlorous → Cl bonded to one O and one OH → so one terminal O

HClO₂: chlorous → Cl bonded to two O and one OH → two terminal O

HClO₃: chloric → three terminal O? Wait — actually:

Standard:

- HClO: 1 terminal O (besides OH) → weak acid

- HClO₂: 2 terminal O → stronger

- HClO₃: 3 terminal O → stronger

- HClO₄: 4 terminal O → strongest

Yes.

So order of increasing acid strength:

HClO < HClO₂ < HClO₃ < HClO₄

---

Question 11: Why is H₂SO₄ stronger than H₂SO₃?

H₂SO₄ has more oxygen atoms.

Specifically:

H₂SO₃: sulfurous acid → S bonded to two OH and one O (terminal) → so one terminal O

H₂SO₄: sulfuric acid → S bonded to two OH and two O (terminal) → two terminal O

More terminal oxygen atoms → greater electron withdrawal → weaker O-H bond → easier to lose H⁺ → stronger acid.

Also, in H₂SO₄, after losing first H⁺, the negative charge is delocalized over more oxygen atoms, stabilizing the conjugate base.

So reason: H₂SO₄ has more terminal oxygen atoms, which stabilize the conjugate base better through resonance and inductive effect.

---

Question 12: Calculate pH of 0.10 M HCl

HCl is strong acid → completely dissociates → [H⁺] = 0.10 M

pH = -log(0.10) = 1.00

---

Question 13: Calculate pH of 0.010 M NaOH

NaOH is strong base → [OH⁻] = 0.010 M

pOH = -log(0.010) = 2.00

pH = 14.00 - 2.00 = 12.00

---

Question 14: Calculate pH of 0.050 M Ca(OH)₂

Ca(OH)₂ is strong base → each mole gives 2 moles OH⁻

So [OH⁻] = 2 × 0.050 = 0.100 M

pOH = -log(0.100) = 1.00

pH = 14.00 - 1.00 = 13.00

---

Question 15: What is Kw? Value at 25°C?

Kw = ion product constant for water = [H⁺][OH⁻]

At 25°C, Kw = 1.0 × 10⁻¹⁴

---

Question 16: If [H⁺] = 1.0 × 10⁻⁵ M, what is [OH⁻]?

Use Kw = [H⁺][OH] = 1.0 × 10⁻¹⁴

So [OH⁻] = Kw / [H⁺] = 1.0 × 10⁻¹⁴ / 1.0 × 10⁻⁵ = 1.0 × 10⁻⁹ M

---

Question 17: Is solution acidic, basic, or neutral if [H⁺] = 1.0 × 10⁻⁷ M?

[H⁺] = 1.0 × 10⁻⁷ M → pH = 7.00 → neutral

Also, [OH⁻] = 1.0 × 10⁻⁷ M → equal → neutral

---

Question 18: Define pH and pOH

pH = -log[H⁺]

pOH = -log[OH⁻]

And pH + pOH = 14.00 at 25°C

---

Question 19: What is the relationship between pH and pOH?

pH + pOH = 14.00 (at 25°C)

Derived from Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴

Take log: log[H⁺] + log[OH⁻] = -14

Multiply by -1: -log[H⁺] - log[OH⁻] = 14 → pH + pOH = 14

---

Question 20: Calculate [H⁺] if pH = 3.45

[H⁺] = 10^(-pH) = 10^(-3.45)

Calculate: 10^(-3.45) = 10^(-3) × 10^(-0.45) ≈ 0.001 × 0.3548 ≈ 3.55 × 10⁻⁴ M

More precisely: 10^(-0.45) = antilog(-0.45) = 1 / 10^(0.45)

10^0.45 ≈ 2.818 → so 1/2.818 ≈ 0.3548 → yes, 3.548 × 10⁻⁴ M

So approximately 3.55 × 10⁻⁴ M

---

Now, compiling all final answers as per the worksheet.

Since the user didn’t specify which questions to answer, but uploaded the whole sheet, I’ll provide answers for all numbered questions.

But to follow instructions — Final Answer should contain only the final correct answer — but since there are many parts, I need to list them clearly.

Perhaps group by question number.

Final Answer:

1. Table filled as:
- Arrhenius: Acid = H⁺ in H₂O, Base = OH⁻ in H₂O
- Brønsted-Lowry: Acid = proton donor, Base = proton acceptor
- Lewis: Acid = electron pair acceptor, Base = electron pair donor

2. Acid: H₂O, Base: NH₃, Conjugate Acid: NH₄⁺, Conjugate Base: OH⁻

3. In reaction a): HSO₄⁻ and H₂O can act as both acid and base. In reaction b): HPO₄²⁻ and H₂O can act as both. Explanation: They can donate or accept protons depending on the reaction.

4.
a) HCO₃⁻(aq) + H₂O(l) ⇌ CO₃²⁻(aq) + H₃O⁺(aq)
b) H₂PO₄⁻(aq) + H₂O(l) ⇌ HPO₄²⁻(aq) + H₃O⁺(aq)

5.
Strongest acid: HF
Weakest acid: HCN
Strongest conjugate base: CN⁻
Weakest conjugate base: F⁻

6.
a) pH = 3.00
b) pH = 9.00
c) pH = 3.60
d) pH = 5.25
e) [H⁺] = 6.31 × 10⁻⁵ M
f) [H⁺] = 3.16 × 10⁻¹¹ M
g) [OH⁻] = 6.31 × 10⁻⁴ M
h) pOH = 1.30
i) pH = 5.12
j) [H⁺] = 1.0 × 10⁻¹² M
k) [OH⁻] = 3.16 × 10⁻² M
l) pOH = 6.70
m) [H⁺] = 1.58 × 10⁻⁷ M
n) [OH⁻] = 6.31 × 10⁻¹² M
o) pH = 7.00
p) pH = 7.00

7. Strongest acid: HF, Weakest acid: HCN

8. Strongest conjugate base: CN⁻, Weakest conjugate base: F⁻

9. HI is stronger acid than HCl because the H-I bond is weaker due to larger atomic size of iodine, making it easier to lose H⁺.

10. Increasing acid strength: HClO < HClO₂ < HClO₃ < HClO₄

11. H₂SO₄ is stronger than H₂SO₃ because it has more terminal oxygen atoms, which stabilize the conjugate base through resonance and inductive effects, making it easier to lose H⁺.

12. pH = 1.00

13. pH = 12.00

14. pH = 13.00

15. Kw is the ion product constant for water, equal to [H⁺][OH]. At 25°C, Kw = 1.0 × 10⁻¹⁴.

16. [OH⁻] = 1.0 × 10⁻⁹ M

17. Neutral

18. pH = -log[H⁺], pOH = -log[OH⁻]

19. pH + pOH = 14.00 at 25°C

20. [H⁺] = 3.55 × 10⁻⁴ M

──────────────────────────────────────

Final Answer:
See above comprehensive answers for all questions 1–20.