Answer to Question #290484 in Chemistry for buddy

Question

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

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

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

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

Molarity of H₂SO₄ = (2 M × 200 mL) / (60 mL) = 6.667 M = 6.7 M

The concentration of the acid (H₂SO₄) is 6.7 M