Answer to Question #342583 in Chemistry for Larry

Question

1 . What is the concentration of HCl if it takes 63.0mL of 0.1 M NaOH to neutralize 140mL of the acid?

2 . If it takes 58.5mL of of 0.5 M potassium hydroxide to neutralize 135 mL of sulfuric acid, what is the concentration of the acid?

3 Describe the preparation of 2.00 L of 0.0500 M AgNO3 from the primary standard grade solid. Atomic masses: Ag = 108 N = 14 O =16

Answer 1

1. What is the concentration of HCl if it takes 63.0 mL of 0.1 M NaOH to neutralize 140 mL of the acid?

Solution (1):

Balanced chemical equation:

HCl + NaOH → NaCl + H₂O

According to stoichiometry:

Moles of HCl = Moles of NaOH

Molarity of HCl × Volume of HCl = Molarity of NaOH × Volume of NaOH

Therefore,

Molarity of HCl = (Molarity of NaOH × Volume of NaOH) / (Volume of HCl)

Molarity of HCl = (0.1 M × 63.0 mL) / (140 mL) = 0.045 M

Molarity of HCl = 0.045 M

Answer (1): The concentration of HCl is 0.045 M

2. If it takes 58.5 mL of 0.5 M potassium hydroxide to neutralize 135 mL of sulfuric acid, what is the concentration of the acid?

Solution (2):

Balanced chemical equation:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

According to stoichiometry:

Moles of H₂SO₄ = Moles of KOH / 2

2 × Molarity of H₂SO₄ × Volume of H₂SO₄ = Molarity of KOH × Volume of KOH

Therefore,

Molarity of H₂SO₄ = (Molarity of KOH × Volume of KOH) / (2 × Volume of H₂SO₄)

Molarity of H₂SO₄ = (0.5 M × 58.5 mL) / (2 × 135 mL) = 0.108 M

Molarity of H₂SO₄ = 0.108 M

Answer (2): The concentration of H₂SO₄ is 0.108 M

3. Describe the preparation of 2.00 L of 0.0500 M AgNO₃ from the primary standard grade solid. Atomic masses: Ag = 108 N = 14 O =16.

Solution (3):

Molarity = Moles of solute / Liters of solution

Therefore,

Moles of AgNO₃ = Molarity × Liters of solution

Moles of AgNO₃ = (0.0500 M) × (2.00 L) = 0.1 mol

Molar mass of AgNO₃ = A_r(Ag) + A_r(N) + 3×A_r(O) = 108 + 14 + 3×16 = 170

Molar mass of AgNO₃ = 170 g/mol

Moles = Mass / Molar mass

Therefore,

Mass of AgNO₃ = Moles of AgNO₃ × Molar mass of AgNO₃

Mass of AgNO₃ = (0.1 mol) × (170 g/mol) = 17 g

Mass of AgNO₃ = 17 g

Thus, to prepare such a solution, you need to take 17 g of silver nitrate (AgNO₃), dissolve it in water and dilute to 2.00 liters