Balancing Acid-Base Reactions worksheet with chemical equations to solve.

Let’s go through each reaction one by one and balance them. Remember, in acid-base reactions, we’re usually making a salt and water. The key is to make sure the number of atoms of each element is the same on both sides.

We’ll start with simple ones and build up.

---

i. HBr + NaOH → NaBr + H₂O

Left: H=1+1=2, Br=1, Na=1, O=1
Right: Na=1, Br=1, H=2, O=1 → Already balanced!

→ 1 HBr + 1 NaOH → 1 NaBr + 1 H₂O

---

ii. H₂SO₄ + KOH → K₂SO₄ + H₂O

Left: H=2+1=3? Wait — let’s count properly.

H₂SO₄ has 2H, 1S, 4O
KOH has 1K, 1O, 1H

Right: K₂SO₄ has 2K, 1S, 4O
H₂O has 2H, 1O

We need 2K on right → so 2 KOH on left.

Try: H₂SO₄ + 2KOH → K₂SO₄ + ?

Now left: H = 2 (from acid) + 2 (from 2KOH) = 4H
O = 4 + 2 = 6
K = 2, S=1

Right: K₂SO₄ uses 2K, 1S, 4O → remaining: 2H and 2O → that’s 2 H₂O!

So: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Check:
Left: H=2+2=4, S=1, O=4+2=6, K=2
Right: K=2, S=1, O=4+2=6, H=4 → Balanced!

→ 1 H₂SO₄ + 2 KOH → 1 K₂SO₄ + 2 H₂O

---

iii. HCl + Ca(OH)₂ → CaCl₂ + H₂O

Ca(OH)₂ has 1Ca, 2O, 2H
HCl has 1H, 1Cl

Right: CaCl₂ has 1Ca, 2Cl → so we need 2 HCl on left.

Try: 2HCl + Ca(OH)₂ → CaCl₂ + ?

Left: H = 2 (from HCl) + 2 (from Ca(OH)₂) = 4H
Cl = 2, Ca=1, O=2

Right: CaCl₂ uses Ca and 2Cl → remaining: 4H and 2O → that’s 2 H₂O!

So: 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

Check:
Left: H=2+2=4, Cl=2, Ca=1, O=2
Right: Ca=1, Cl=2, H=4, O=2 → Balanced!

→ 2 HCl + 1 Ca(OH)₂ → 1 CaCl₂ + 2 H₂O

---

iv. Fe(OH)₃ + H₂SO₄ → Fe₂(SO₄)₃ + H₂O

Fe₂(SO₄)₃ means 2 Fe, 3 SO₄ groups → so 3 S, 12 O from sulfate, plus Fe.

Left: Fe(OH)₃ has 1Fe, 3O, 3H
H₂SO₄ has 2H, 1S, 4O

We need 2 Fe on right → so 2 Fe(OH)₃ on left.

We need 3 SO₄ → so 3 H₂SO₄ on left.

Try: 2Fe(OH)₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + ?

Left:
Fe = 2
O from hydroxide: 2×3=6, from acid: 3×4=12 → total O=18
H from hydroxide: 2×3=6, from acid: 3×2=6 → total H=12
S = 3

Right: Fe₂(SO₄)₃ → 2Fe, 3S, 12O
Remaining: H=12, O=6 → that’s 6 H₂O!

So: 2Fe(OH)₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 6H₂O

Check:
Left: Fe=2, S=3, O=6+12=18, H=6+6=12
Right: Fe=2, S=3, O=12+6=18, H=12 → Balanced!

→ 2 Fe(OH)₃ + 3 H₂SO₄ → 1 Fe₂(SO₄)₃ + 6 H₂O

---

v. H₂SO₄ + B(OH)₃ → B₂(SO₄)₃ + H₂O

B₂(SO₄)₃ → 2B, 3SO₄ → so 3 S, 12 O from sulfate.

Left: H₂SO₄ → 2H, 1S, 4O
B(OH)₃ → 1B, 3O, 3H

Need 2B → so 2 B(OH)₃
Need 3S → so 3 H₂SO₄

Try: 3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + ?

Left:
H = 3×2 + 2×3 = 6+6=12
S = 3
O = 3×4 + 2×3 = 12+6=18
B = 2

Right: B₂(SO₄)₃ → 2B, 3S, 12O
Remaining: H=12, O=6 → 6 H₂O

So: 3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + 6H₂O

Check:
Left: H=12, S=3, O=18, B=2
Right: B=2, S=3, O=12+6=18, H=12 → Balanced!

→ 3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + 6 H₂O

---

vi. Pb(OH)₂ + HCl → PbCl₂ + H₂O

PbCl₂ needs 2 Cl → so 2 HCl

Left: Pb(OH)₂ → 1Pb, 2O, 2H
2HCl → 2H, 2Cl

Total left: Pb=1, O=2, H=4, Cl=2

Right: PbCl₂ → Pb, 2Cl → remaining: 4H, 2O → 2 H₂O

So: Pb(OH)₂ + 2HCl → PbCl₂ + 2H₂O

Check:
Left: Pb=1, O=2, H=2+2=4, Cl=2
Right: Pb=1, Cl=2, H=4, O=2 → Balanced!

→ 1 Pb(OH)₂ + 2 HCl → 1 PbCl₂ + 2 H₂O

---

vii. H₂SO₄ + NH₄OH → (NH₄)₂SO₄ + H₂O

(NH₄)₂SO₄ → 2 NH₄, 1 SO₄ → so 2N, 8H, 1S, 4O

Left: H₂SO₄ → 2H, 1S, 4O
NH₄OH → 1N, 5H, 1O (since NH₄ is N+4H, OH is O+H → total N, 5H, O)

Need 2 NH₄ → so 2 NH₄OH

Try: H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + ?

Left:
H = 2 (acid) + 2×5 = 2+10=12
S = 1
O = 4 + 2×1 = 6
N = 2

Right: (NH₄)₂SO₄ → 2N, 8H, 1S, 4O
Remaining: H=4, O=2 → 2 H₂O

So: H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O

Check:
Left: H=12, S=1, O=6, N=2
Right: N=2, H=8+4=12, S=1, O=4+2=6 → Balanced!

→ 1 H₂SO₄ + 2 NH₄OH → 1 (NH₄)₂SO₄ + 2 H₂O

---

viii. H₂CO₃ + CsOH → Cs₂CO₃ + H₂O

Cs₂CO₃ → 2Cs, 1C, 3O

Left: H₂CO₃ → 2H, 1C, 3O
CsOH → 1Cs, 1O, 1H

Need 2Cs → so 2 CsOH

Try: H₂CO₃ + 2CsOH → Cs₂CO₃ + ?

Left:
H = 2 + 2×1 = 4
C = 1
O = 3 + 2×1 = 5
Cs = 2

Right: Cs₂CO₃ → 2Cs, 1C, 3O
Remaining: H=4, O=2 → 2 H₂O

So: H₂CO₃ + 2CsOH → Cs₂CO₃ + 2H₂O

Check:
Left: H=4, C=1, O=5, Cs=2
Right: Cs=2, C=1, O=3+2=5, H=4 → Balanced!

→ 1 H₂CO₃ + 2 CsOH → 1 Cs₂CO₃ + 2 H₂O

---

ix. HF + Mg(OH)₂ → MgF₂ + H₂O

MgF₂ → needs 2 F → so 2 HF

Left: 2HF → 2H, 2F
Mg(OH)₂ → 1Mg, 2O, 2H

Total left: H=4, F=2, Mg=1, O=2

Right: MgF₂ → Mg, 2F → remaining: 4H, 2O → 2 H₂O

So: 2HF + Mg(OH)₂ → MgF₂ + 2H₂O

Check:
Left: H=2+2=4, F=2, Mg=1, O=2
Right: Mg=1, F=2, H=4, O=2 → Balanced!

→ 2 HF + 1 Mg(OH)₂ → 1 MgF₂ + 2 H₂O

---

x. HNO₃ + Al(OH)₃ → Al(NO₃)₃ + H₂O

Al(NO₃)₃ → 1Al, 3NO₃ → so 3N, 9O from nitrate

Left: HNO₃ → 1H, 1N, 3O
Al(OH)₃ → 1Al, 3O, 3H

Need 3 NO₃ → so 3 HNO₃

Try: 3HNO₃ + Al(OH)₃ → Al(NO₃)₃ + ?

Left:
H = 3×1 + 3 = 6
N = 3
O = 3×3 + 3 = 9+3=12
Al = 1

Right: Al(NO₃)₃ → Al, 3N, 9O
Remaining: H=6, O=3 → 3 H₂O

So: 3HNO₃ + Al(OH)₃ → Al(NO₃)₃ + 3H₂O

Check:
Left: H=6, N=3, O=12, Al=1
Right: Al=1, N=3, O=9+3=12, H=6 → Balanced!

→ 3 HNO₃ + 1 Al(OH)₃ → 1 Al(NO₃)₃ + 3 H₂O

---

xi. HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + H₂O

Zn(NO₃)₂ → 1Zn, 2NO₃ → so 2N, 6O from nitrate

Left: HNO₃ → 1H, 1N, 3O
Zn(OH)₂ → 1Zn, 2O, 2H

Need 2 NO₃ → so 2 HNO₃

Try: 2HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + ?

Left:
H = 2×1 + 2 = 4
N = 2
O = 2×3 + 2 = 6+2=8
Zn = 1

Right: Zn(NO₃)₂ → Zn, 2N, 6O
Remaining: H=4, O=2 → 2 H₂O

So: 2HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + 2H₂O

Check:
Left: H=4, N=2, O=8, Zn=1
Right: Zn=1, N=2, O=6+2=8, H=4 → Balanced!

→ 2 HNO₃ + 1 Zn(OH)₂ → 1 Zn(NO₃)₂ + 2 H₂O

---

xii. H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O

Ca₃(PO₄)₂ → 3Ca, 2PO₄ → so 2P, 8O from phosphate

Left: H₃PO₄ → 3H, 1P, 4O
Ca(OH)₂ → 1Ca, 2O, 2H

Need 2 P → so 2 H₃PO₄
Need 3 Ca → so 3 Ca(OH)₂

Try: 2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + ?

Left:
H = 2×3 + 3×2 = 6+6=12
P = 2
O = 2×4 + 3×2 = 8+6=14
Ca = 3

Right: Ca₃(PO₄)₂ → 3Ca, 2P, 8O
Remaining: H=12, O=6 → 6 H₂O

So: 2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O

Check:
Left: H=12, P=2, O=14, Ca=3
Right: Ca=3, P=2, O=8+6=14, H=12 → Balanced!

→ 2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO₄)₂ + 6 H₂O

---

xiii. HClO₃ + Al(OH)₃ → Al(ClO₃)₃ + H₂O

Al(ClO₃)₃ → 1Al, 3ClO₃ → so 3Cl, 9O from chlorate

Left: HClO₃ → 1H, 1Cl, 3O
Al(OH)₃ → 1Al, 3O, 3H

Need 3 ClO₃ → so 3 HClO₃

Try: 3HClO₃ + Al(OH)₃ → Al(ClO₃)₃ + ?

Left:
H = 3×1 + 3 = 6
Cl = 3
O = 3×3 + 3 = 9+3=12
Al = 1

Right: Al(ClO₃)₃ → Al, 3Cl, 9O
Remaining: H=6, O=3 → 3 H₂O

So: 3HClO₃ + Al(OH)₃ → Al(ClO₃)₃ + 3H₂O

Check:
Left: H=6, Cl=3, O=12, Al=1
Right: Al=1, Cl=3, O=9+3=12, H=6 → Balanced!

→ 3 HClO₃ + 1 Al(OH)₃ → 1 Al(ClO₃)₃ + 3 H₂O

---

xiv. HF + Ba(OH)₂ → BaF₂ + H₂O

BaF₂ → needs 2 F → so 2 HF

Left: 2HF → 2H, 2F
Ba(OH)₂ → 1Ba, 2O, 2H

Total left: H=4, F=2, Ba=1, O=2

Right: BaF₂ → Ba, 2F → remaining: 4H, 2O → 2 H₂O

So: 2HF + Ba(OH)₂ → BaF₂ + 2H₂O

Check:
Left: H=2+2=4, F=2, Ba=1, O=2
Right: Ba=1, F=2, H=4, O=2 → Balanced!

→ 2 HF + 1 Ba(OH)₂ → 1 BaF₂ + 2 H₂O

---

xv. HCl + Al(OH)₃ → AlCl₃ + H₂O

AlCl₃ → needs 3 Cl → so 3 HCl

Left: 3HCl → 3H, 3Cl
Al(OH)₃ → 1Al, 3O, 3H

Total left: H=6, Cl=3, Al=1, O=3

Right: AlCl₃ → Al, 3Cl → remaining: 6H, 3O → 3 H₂O

So: 3HCl + Al(OH)₃ → AlCl₃ + 3H₂O

Check:
Left: H=3+3=6, Cl=3, Al=1, O=3
Right: Al=1, Cl=3, H=6, O=3 → Balanced!

→ 3 HCl + 1 Al(OH)₃ → 1 AlCl₃ + 3 H₂O

---

Final Answer:

i. 1 HBr + 1 NaOH → 1 NaBr + 1 H₂O
ii. 1 H₂SO₄ + 2 KOH → 1 K₂SO₄ + 2 H₂O
iii. 2 HCl + 1 Ca(OH)₂ → 1 CaCl₂ + 2 H₂O
iv. 2 Fe(OH)₃ + 3 H₂SO₄ → 1 Fe₂(SO₄)₃ + 6 H₂O
v. 3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + 6 H₂O
vi. 1 Pb(OH)₂ + 2 HCl → 1 PbCl₂ + 2 H₂O
vii. 1 H₂SO₄ + 2 NH₄OH → 1 (NH₄)₂SO₄ + 2 H₂O
viii. 1 H₂CO₃ + 2 CsOH → 1 Cs₂CO₃ + 2 H₂O
ix. 2 HF + 1 Mg(OH)₂ → 1 MgF₂ + 2 H₂O
x. 3 HNO₃ + 1 Al(OH)₃ → 1 Al(NO₃)₃ + 3 H₂O
xi. 2 HNO₃ + 1 Zn(OH)₂ → 1 Zn(NO₃)₂ + 2 H₂O
xii. 2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO₄)₂ + 6 H₂O
xiii. 3 HClO₃ + 1 Al(OH)₃ → 1 Al(ClO₃)₃ + 3 H₂O
xiv. 2 HF + 1 Ba(OH)₂ → 1 BaF₂ + 2 H₂O
xv. 3 HCl + 1 Al(OH)₃ → 1 AlCl₃ + 3 H₂O