Titrations
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Questions and Answers

What is the concentration of ethanoic acid in mol/dm³?

  • 0.124 (correct)
  • 0.150
  • 0.375
  • 0.300

    • How many moles of CH3COOH are present when 0.00375 mol of NaOH is used?

  • 0.00125
  • 0.01250
  • 0.00375 (correct)
  • 0.00750

    • Calculate the concentration of ethanoic acid in g/dm³.

  • 5.20
  • 7.43 (correct)
  • 4.12
  • 9.60

    • What is the molar mass of CH3COOH?

    <p>60 g/mol</p> Signup and view all the answers

    If the volume of CH3COOH is doubled, what happens to its concentration?

    <p>It halves.</p> Signup and view all the answers

    What is the final product when CH3COOH reacts with NaOH?

    <p>Both A and B</p> Signup and view all the answers

    Which of the following is needed to calculate the concentration of a solution?

    <p>Moles and volume</p> Signup and view all the answers

    If you start with 0.150 M NaOH, what can you deduce about the concentration of CH3COOH after neutralization?

    <p>It will be 0.150 M.</p> Signup and view all the answers

    What is the formula for calculating moles of a substance?

    <p>Moles = Concentration x Volume</p> Signup and view all the answers

    What volume of 0.150 mol/dm RbOH is needed to react with 25.0 cm of 0.240 mol/dm HNO3?

    <p>0.0400 dm</p> Signup and view all the answers

    In the reaction between NaOH and H2SO4, what is the mole ratio of NaOH to H2SO4?

    <p>2:1</p> Signup and view all the answers

    How many moles of H2SO4 are produced when 0.00500 mol of NaOH reacts?

    <p>0.00250 mol</p> Signup and view all the answers

    What is the concentration of H2SO4 if 0.00250 mol reacts with a volume of 28.7 cm?

    <p>0.0871 mol/dm</p> Signup and view all the answers

    How do you convert the volume from cm to dm?

    <p>Divide the volume in cm by 100</p> Signup and view all the answers

    What is the first step in determining the moles of an acid in a reaction?

    <p>Convert the volume to dm</p> Signup and view all the answers

    If 25.0 cm of 0.200 mol/dm NaOH is reacted, what is the mole of NaOH?

    <p>0.00500 mol</p> Signup and view all the answers

    What is the concentration of hydrochloric acid in the reaction with barium hydroxide?

    <p>0.439 mol/dm</p> Signup and view all the answers

    How many moles of hydrochloric acid are produced in the reaction with barium hydroxide?

    <p>0.0100 mol</p> Signup and view all the answers

    What is the total volume of hydrochloric acid used in the reaction with sodium hydroxide?

    <p>25.0 cm</p> Signup and view all the answers

    In the reaction between sodium hydroxide and hydrochloric acid, how many moles of sodium hydroxide are present?

    <p>0.00250 mol</p> Signup and view all the answers

    What is the concentration of sodium hydroxide solution if 22.5 cm reacted with 0.100 mol/dm hydrochloric acid?

    <p>0.111 mol/dm</p> Signup and view all the answers

    What is the relation between the moles of hydrochloric acid and sodium hydroxide in the reaction?

    <p>One mole of NaOH reacts with two moles of HCl</p> Signup and view all the answers

    Which of the following is TRUE about the volume used for calculating the concentration of HCl?

    <p>The reaction only considers the volume of hydrochloric acid used.</p> Signup and view all the answers

    How is the concentration of hydrochloric acid presented in the final answer?

    <p>To 3 significant figures</p> Signup and view all the answers

    Study Notes

    Titrations

    • Titration 1: 25.0 cm³ of 0.200 mol/dm³ barium hydroxide solution reacted with 22.8 cm³ of hydrochloric acid. Calculate the concentration of the hydrochloric acid in mol/dm³. The answer should be to 3 significant figures using the balanced chemical equation: Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2 H₂O(l).

    • Calculation:

      • Moles of Ba(OH)₂: 0.200 mol/dm³ * (25.0 cm³ / 1000 cm³/dm³) = 0.00500 mol
      • Moles of HCl: 2 * 0.00500 mol = 0.0100 mol
      • Concentration of HCl: 0.0100 mol / (22.8 cm³ / 1000 cm³/dm³) = 0.439 mol/dm³

    • Titration 2 (Part a): 22.5 cm³ of sodium hydroxide solution reacted with 25.0 cm³ of 0.100 mol/dm³ hydrochloric acid. Calculate the concentration of the sodium hydroxide solution in mol/dm³. The answer should be to 3 significant figures using the balanced chemical equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l).

    • Calculation:

      • Moles of HCl: 0.100 mol/dm³ * (25.0 cm³ / 1000 cm³/dm³) = 0.00250 mol
      • Moles of NaOH: 0.00250 mol
      • Concentration of NaOH: 0.00250 mol / (22.5 cm³ / 1000 cm³/dm³) = 0.111 mol/dm³

    • Titration 2 (Part b): Calculate the concentration of the sodium hydroxide solution in g/dm³. The answer should be to 3 significant figures.

    • Calculation:

      • Concentration of NaOH: 0.111 mol/dm³ x 40 g/mol = 4.44 g/dm³

    • Titration 3: Calculate the volume of 0.150 mol/dm³ rubidium hydroxide that reacts with 25.0 cm³ of 0.240 mol/dm³ nitric acid. The answer should be to 3 significant figures using the balanced chemical equation : RbOH(aq) + HNO₃(aq) → RbNO₃(aq) + H₂O(l).

    • Calculation:

      • Moles of HNO₃: 0.240 mol/dm³ * (25.0 cm³ / 1000 cm³/dm³) = 0.00600 mol
      • Moles of RbOH: 0.00600 mol
      • Volume of RbOH: 0.00600 mol / 0.150 mol/dm³ = 0.0400 dm³ (or 40.0 cm³)

    • Titration 4: 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide solution reacted with 28.7 cm³ sulfuric acid. Calculate the concentration of the sulfuric acid in mol/dm³. The answer should be to 3 significant figures using the balanced chemical equation: 2 NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2 H₂O(l).

    • Calculation:

      • Moles of NaOH: 0.200 mol/dm³ * (25.0 cm³ / 1000 cm³/dm³) = 0.00500 mol
      • Moles of H₂SO₄: 0.00500 mol / 2 = 0.00250 mol
      • Concentration of H₂SO₄: 0.00250 mol / (28.7 cm³ / 1000 cm³/dm³) = 0.0871 mol/dm³

    Ethanoic Acid Titration

    • Titration 5 (Part a): 25.0 cm³ of 0.150 mol/dm³ sodium hydroxide reacted with 30.3 cm³ of ethanoic acid. Calculate the concentration of the ethanoic acid in mol/dm³. Answer should be to 3 significant figures using the equation CH₃COOH(aq) + NaOH(aq) → CH₃COONa (aq) + H₂O(l).

    • Calculation:

      • Moles of NaOH: 0.150 mol/dm³ * (25.0 cm³ / 1000 cm³/dm³) = 0.00375 mol
      • Moles of CH₃COOH: 0.00375 mol
      • Concentration of CH₃COOH: 0.00375 mol / (30.3 cm³ / 1000 cm³/dm³) = 0.124 mol/dm³

    • Titration 5 (Part b): Calculate the concentration in g/dm³. Answer should be to 3 significant figures.

    • Calculation:

      • Concentration of CH₃COOH: 0.124 mol/dm³ * 60 g/mol = 7.43 g/dm³

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    Titrations 1 Past Paper PDF

    Description

    Test your understanding of titration calculations with this quiz. Solve problems involving the concentration of acids and bases using balanced chemical equations. Perfect for chemistry students looking to reinforce their skills.

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