Acid Base Neutralization Reaction Worksheet | PDF - Free Printable

Let's solve each of these neutralization reactions step by step. A neutralization reaction occurs when an acid reacts with a base to produce a salt and water.

The general form is:

> Acid + Base → Salt + Water

We will balance each equation, making sure the number of atoms on both sides is equal.

---

1. HCl + NaOH →

- Acid: HCl (hydrochloric acid)
- Base: NaOH (sodium hydroxide)
- Products: NaCl (sodium chloride) + H₂O

Balanced Equation:
> HCl + NaOH → NaCl + H₂O
✔ Already balanced.

---

2. HNO₃ + KOH →

- Acid: HNO₃ (nitric acid)
- Base: KOH (potassium hydroxide)
- Products: KNO₃ (potassium nitrate) + H₂O

Balanced Equation:
> HNO₃ + KOH → KNO₃ + H₂O
✔ Balanced.

---

3. HBr + KOH →

- Acid: HBr (hydrobromic acid)
- Base: KOH
- Products: KBr (potassium bromide) + H₂O

Balanced Equation:
> HBr + KOH → KBr + H₂O
✔ Balanced.

---

4. HCl + LiOH →

- Acid: HCl
- Base: LiOH (lithium hydroxide)
- Products: LiCl (lithium chloride) + H₂O

Balanced Equation:
> HCl + LiOH → LiCl + H₂O
✔ Balanced.

---

5. HF + Ba(OH)₂ →

- Acid: HF (hydrofluoric acid) – monoprotic
- Base: Ba(OH)₂ (barium hydroxide) – provides 2 OH⁻ ions
- So we need 2 HF to neutralize one Ba(OH)₂

Products:
- Salt: BaF₂ (barium fluoride)
- Water: 2 H₂O

Balanced Equation:
> 2HF + Ba(OH)₂ → BaF₂ + 2H₂O
✔ Balanced.

---

6. H₂SO₄ + Ca(OH)₂ →

- Acid: H₂SO₄ (sulfuric acid) – diprotic
- Base: Ca(OH)₂ – provides 2 OH⁻
- One molecule of each can neutralize completely.

Products:
- Salt: CaSO₄ (calcium sulfate)
- Water: 2 H₂O

Balanced Equation:
> H₂SO₄ + Ca(OH)₂ → CaSO₄ + 2H₂O
✔ Balanced.

---

7. H₂CO₃ + NaOH →

- Acid: H₂CO₃ (carbonic acid) – diprotic
- Base: NaOH – monobasic
- Need 2 NaOH for one H₂CO₃

Products:
- Salt: Na₂CO₃ (sodium carbonate)
- Water: 2 H₂O

Balanced Equation:
> H₂CO₃ + 2NaOH → Na₂CO₃ + 2H₂O
✔ Balanced.

---

8. H₃PO₄ + Ca(OH)₂ →

- Acid: H₃PO₄ (phosphoric acid) – triprotic
- Base: Ca(OH)₂ – provides 2 OH⁻ per molecule

To neutralize fully, we need:
- 3 H⁺ from H₃PO₄
- Each Ca(OH)₂ gives 2 OH⁻ → so we need 3/2 = 1.5 molecules of Ca(OH)₂

But we want whole numbers → multiply by 2:

- 2 H₃PO₄ → 6 H⁺
- 3 Ca(OH)₂ → 6 OH⁻

Products:
- Salt: Ca₃(PO₄)₂ (calcium phosphate)
- Water: 6 H₂O

Balanced Equation:
> 2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O
✔ Balanced.

---

9. HC₂H₃O₂ + KOH →

- Acid: HC₂H₃O₂ (acetic acid) – weak acid, monoprotic
- Base: KOH
- Products: KC₂H₃O₂ (potassium acetate) + H₂O

Balanced Equation:
> HC₂H₃O₂ + KOH → KC₂H₃O₂ + H₂O
✔ Balanced.

---

10. H₂CO₃ + Sr(OH)₂ →

- Acid: H₂CO₃ (diprotic)
- Base: Sr(OH)₂ (strontium hydroxide) – provides 2 OH⁻
- One-to-one ratio

Products:
- Salt: SrCO₃ (strontium carbonate)
- Water: 2 H₂O

Balanced Equation:
> H₂CO₃ + Sr(OH)₂ → SrCO₃ + 2H₂O
✔ Balanced.

---

11. HNO₃ + Zn(OH)₂ →

- Acid: HNO₃ (monoprotic)
- Base: Zn(OH)₂ – provides 2 OH⁻
- So need 2 HNO₃ to neutralize one Zn(OH)₂

Products:
- Salt: Zn(NO₃)₂ (zinc nitrate)
- Water: 2 H₂O

Balanced Equation:
> 2HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + 2H₂O
✔ Balanced.

---

12. HCl + Al(OH)₃ →

- Acid: HCl (monoprotic)
- Base: Al(OH)₃ – aluminum hydroxide, has 3 OH⁻ groups
- So need 3 HCl for one Al(OH)₃

Products:
- Salt: AlCl₃ (aluminum chloride)
- Water: 3 H₂O

Balanced Equation:
> 3HCl + Al(OH)₃ → AlCl₃ + 3H₂O
✔ Balanced.

---

## ✔ Final Answers:

1. HCl + NaOH → NaCl + H₂O
2. HNO₃ + KOH → KNO₃ + H₂O
3. HBr + KOH → KBr + H₂O
4. HCl + LiOH → LiCl + H₂O
5. 2HF + Ba(OH)₂ → BaF₂ + 2H₂O
6. H₂SO₄ + Ca(OH)₂ → CaSO₄ + 2H₂O
7. H₂CO₃ + 2NaOH → Na₂CO₃ + 2H₂O
8. 2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O
9. HC₂H₃O₂ + KOH → KC₂H₃O₂ + H₂O
10. H₂CO₃ + Sr(OH)₂ → SrCO₃ + 2H₂O
11. 2HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + 2H₂O
12. 3HCl + Al(OH)₃ → AlCl₃ + 3H₂O

---

🔍 Key Concepts Recap:

- Neutralization: acid + base → salt + water
- Balance the H⁺ and OH⁻ ions
- Use stoichiometry based on number of protons (H⁺) and hydroxide ions (OH⁻)
- Polyatomic ions like PO₄³⁻, CO₃²⁻, etc., stay intact in salts

Let me know if you'd like this in a printable format or with explanations for naming the salts!