Neutralization Reactions Problems
2Al(NO₃)₃ + 3Na₂CO₃ → Al₂(CO₃)₃ + 6NaNO₃
1. 0.2 M HCl is used to neutralize 20 mL of 0.35 M NaOH.
HCl + NaOH → NaCl + H₂O
a. How many moles of OH⁻ are present?
1 mol.
b. How many moles of H⁺ are needed to neutralize the OH⁻?
1 mol.
c. What volume of 0.2 M HCl is needed?
nₐ x Mₐ x Vₐ = nb x Mb x Vb
Vₐ = ( nb x Mb x Vb ) ÷ ( nₐ x Mₐ)
Vₐ = ( 1 x 0.35 x 0.020 ) ÷ ( 1 x 0.2)
= 0.035
∴ 0.035 L or 35 mL of 0.2 M HCl is needed.
2. If 358 mL of H₂SO₄ is neutralized with 60.0 mL of 1.50 M NaOH, what is the initial
concentration of the acid?
, H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
nₐ x Mₐ x Vₐ = nb x Mb x Vb
Mₐ = ( nb x Mb x Vb ) ÷ ( nₐ x Vₐ)
Mₐ = ( 1 x 1.5 x 0.060 ) ÷ ( 2 x 0.358)
= 0.126
∴ 0.126 M is the initial concentration of H₂SO₄.
3. 78 mL of 2.5 M phosphoric acid is neutralized with 500 mL of potassium
hydroxide. What is the concentration of the base?
H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
nₐ x Mₐ x Vₐ = nb x Mb x Vb
Mb = ( nₐ x Mₐ x Vₐ ) ÷ ( nb x Vb)
Mb = ( 3 x 2.5 x 0.078 ) ÷ ( 1 x 0.5)
= 1.17
∴ 1.17 M is the concentration of KOH.
4. When titrating 0.15 M HCl with a calcium hydroxide solution of unknown
concentration, 35 mL of acid are required to neutralize 25 mL of the base.
2Al(NO₃)₃ + 3Na₂CO₃ → Al₂(CO₃)₃ + 6NaNO₃
1. 0.2 M HCl is used to neutralize 20 mL of 0.35 M NaOH.
HCl + NaOH → NaCl + H₂O
a. How many moles of OH⁻ are present?
1 mol.
b. How many moles of H⁺ are needed to neutralize the OH⁻?
1 mol.
c. What volume of 0.2 M HCl is needed?
nₐ x Mₐ x Vₐ = nb x Mb x Vb
Vₐ = ( nb x Mb x Vb ) ÷ ( nₐ x Mₐ)
Vₐ = ( 1 x 0.35 x 0.020 ) ÷ ( 1 x 0.2)
= 0.035
∴ 0.035 L or 35 mL of 0.2 M HCl is needed.
2. If 358 mL of H₂SO₄ is neutralized with 60.0 mL of 1.50 M NaOH, what is the initial
concentration of the acid?
, H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
nₐ x Mₐ x Vₐ = nb x Mb x Vb
Mₐ = ( nb x Mb x Vb ) ÷ ( nₐ x Vₐ)
Mₐ = ( 1 x 1.5 x 0.060 ) ÷ ( 2 x 0.358)
= 0.126
∴ 0.126 M is the initial concentration of H₂SO₄.
3. 78 mL of 2.5 M phosphoric acid is neutralized with 500 mL of potassium
hydroxide. What is the concentration of the base?
H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
nₐ x Mₐ x Vₐ = nb x Mb x Vb
Mb = ( nₐ x Mₐ x Vₐ ) ÷ ( nb x Vb)
Mb = ( 3 x 2.5 x 0.078 ) ÷ ( 1 x 0.5)
= 1.17
∴ 1.17 M is the concentration of KOH.
4. When titrating 0.15 M HCl with a calcium hydroxide solution of unknown
concentration, 35 mL of acid are required to neutralize 25 mL of the base.