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 (D)</p> Signup and view all the answers

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

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

What is the final product when CH3COOH reacts with NaOH?

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

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

<p>Moles and volume (A)</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. (B)</p> Signup and view all the answers

What is the formula for calculating moles of a substance?

<p>Moles = Concentration x Volume (A)</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 (B)</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 (C)</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 (A)</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 (B)</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 (C)</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 (D)</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 (D)</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 (D)</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 (D)</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 (C)</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 (D)</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 (C)</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 (D)</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. (B)</p> Signup and view all the answers

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

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

Flashcards

Moles of HNO3

Calculated by multiplying the concentration (mol/dm³) of nitric acid by the volume (dm³).

Moles of RbOH

Equal to the moles of HNO3 in a neutralization reaction.

Volume of RbOH

Determined by dividing the moles of RbOH by its concentration.

Moles of NaOH

Calculated by the product of concentration (mol/dm³) and volume (dm³) of sodium hydroxide.

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Moles of H2SO4

Calculated from moles of NaOH, using the mole ratio from the balanced equation. Half the moles of NaOH.

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Concentration of H2SO4

Determined by dividing the moles of H2SO4 by its volume in dm³.

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Converting cm³ to dm³

Divide the volume in cm³ by 1000 to get the volume in dm³.

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Solution stoichiometry

Calculating amounts of substances in solutions using balanced chemical equations.

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Moles of Ba(OH)2

Calculated by multiplying the concentration of Barium Hydroxide by its volume in dm^3

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Moles of HCl (relative to Ba(OH)2)

In the reaction Ba(OH)2 + 2HCl → BaCl2 + 2H2O, the mole ratio between HCl and Ba(OH)2 is 2:1. Calculate the moles of HCl by multiplying the moles of Ba(OH)2 by 2.

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Concentration of HCl

Calculated by dividing the moles of HCl by the volume of HCl in dm^3.

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Moles of HCl (NaOH example)

Calculated by multiplying the concentration of HCl by its volume in dm^3.

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Moles of NaOH in NaOH example

In a 1:1 reaction like NaOH + HCl, the moles of NaOH are the same as moles of HCl.

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Concentration of NaOH (mol/dm^3)

Calculated by dividing the moles of NaOH by the volume of NaOH in dm^3.

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Concentration of NaOH (g/dm^3)

Calculated by multiplying the concentration in mol/dm^3 by the molar mass of NaOH.

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Titration

A technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration(the standardization process).

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Calculating moles of NaOH

Moles of NaOH = concentration of NaOH × volume of NaOH in dm³

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Moles of CH3COOH

Moles of CH3COOH are equal to moles of NaOH in a neutralization reaction.

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Concentration of CH3COOH (mol/dm³)

Concentration = moles of CH3COOH / volume of CH3COOH (dm³)

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Unit conversion (mol/dm³ to g/dm³)

Multiply the concentration in mol/dm³ by the molar mass of CH3COOH in g/mol

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Concentration of CH3COOH (g/dm³)

Concentration is calculated as the product of molar concentration and molar mass.

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Significant Figures

A way to express a value to a specific degree of accuracy.

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Volume in dm³

Volume converted from cm³ by dividing by 1000.

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Neutralization reaction

A reaction between an acid and a base, producing salt and water.

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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)â‚‚(aq) + 2HCl(aq) → BaClâ‚‚(aq) + 2 Hâ‚‚O(l).

  • Calculation:

    • Moles of Ba(OH)â‚‚: 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) + Hâ‚‚O(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) + Hâ‚‚O(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) + Hâ‚‚SOâ‚„(aq) → Naâ‚‚SOâ‚„(aq) + 2 Hâ‚‚O(l).

  • Calculation:

    • Moles of NaOH: 0.200 mol/dm³ * (25.0 cm³ / 1000 cm³/dm³) = 0.00500 mol
    • Moles of Hâ‚‚SOâ‚„: 0.00500 mol / 2 = 0.00250 mol
    • Concentration of Hâ‚‚SOâ‚„: 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) + Hâ‚‚O(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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