Answer to Question #291134 in Chemistry for thaiana

Question

1. Determine the volume of 0.7M hydrochloric acid needed to fully neutralize 60 mL of 0.5M Ca(OH)₂.
2. If 50 mL of 1.2M H₃PO_4 is neutralized by 75 mL of NaOH, what is the concentration of the NaOH?
3. What volume of 0.2M HNO_3 acid would be needed to fully neutralize 1L of 0.4M KOH?
4. 200 mL of 2M Sr(OH)₂ fully neutralized 60 mL of H₂SO₄. What is the concentration of the acid?

Answer 1

Solution:

Molarity (M) = Moles of solute / Liters of solution

Moles of solute = Molarity × Liters of solution

Determine the volume of 0.7 M hydrochloric acid needed to fully neutralize 60 mL of 0.5 M Ca(OH)₂.

Balanced chemical equation:

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

According to stoichiometry:

Moles of Ca(OH)₂ = Moles of HCl / 2

Molarity of Ca(OH)₂ × Volume of Ca(OH)₂ = Molarity of HCl × Volume of HCl / 2

Volume of HCl = (2 × Molarity of Ca(OH)₂ × Volume of Ca(OH)₂) / (Molarity of HCl)

Volume of HCl = (2 × 0.5 M × 0.06 L) / (0.7 M) = 0.0857 L = 85.7 mL

The volume of hydrochloric acid (HCl) is 85.7 mL

If 50 mL of 1.2 M H₃PO₄ is neutralized by 75 mL of NaOH, what is the concentration of the NaOH?

Balanced chemical equation:

H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

According to stoichiometry:

Moles of H₃PO₄ = Moles of NaOH / 3

Molarity of H₃PO₄ × Volume of H₃PO₄ = Molarity of NaOH × Volume of NaOH / 3

Molarity of NaOH = (3 × Molarity of H₃PO₄ × Volume of H₃PO₄) / (Volume of NaOH)

Molarity of NaOH = (3 × 1.2 M × 50 mL) / (75 mL) = 2.4 M

The concentration of NaOH is 2.4 M

What volume of 0.2 M HNO₃ acid would be needed to fully neutralize 1 L of 0.4 M KOH?

Balanced chemical equation:

HNO₃ + KOH → KNO₃ + H₂O

According to stoichiometry:

Moles of HNO₃ = Moles of KOH

Molarity of HNO₃ × Volume of HNO₃ = Molarity of KOH × Volume of KOH

Volume of HNO₃ = (Molarity of KOH × Volume of KOH) / (Molarity of HNO₃)

Volume of HNO₃ = (0.4 M × 1 L) / (0.2 M) = 2 L

The volume of HNO₃ is 2 L

200 mL of 2 M Sr(OH)₂ fully neutralized 60 mL of H₂SO₄. What is the concentration of the acid?

Balanced chemical equation:

H₂SO₄ + Sr(OH)₂ → SrSO₄ + 2H₂O

According to stoichiometry:

Moles of H₂SO₄ = Moles of Sr(OH)₂

Molarity of H₂SO₄ × Volume of H₂SO₄ = Molarity of Sr(OH)₂ × Volume of Sr(OH)₂

Molarity of H₂SO_4 = (Molarity of Sr(OH)₂ × Volume of Sr(OH)₂) / (Volume of H₂SO₄)

Molarity of H₂SO_4 = (2 M × 200 mL) / (60 mL) = 6.667 M = 6.7 M

The concentration of the acid (H₂SO₄) is 6.7 M