Concentration of solutions

Unit 6 Concentration of solutions

Key unit competence: Determine the concentrations of solutions from data obtained by simple acid-base titration.

Key ideas

• Concentration is the amount of solute in a given volume of solution. Mass concentration in g/dm³ = mass of solute (g) ÷ volume (dm³). Molar concentration (molarity) in mol/dm³ = moles of solute ÷ volume (dm³). 1 dm³ = 1000 cm³.
• Useful relations: n = C × V (V in dm³); mass concentration = molar concentration × molar mass.
• A standard solution has an accurately known concentration. To prepare it: weigh the solute accurately, dissolve in a little distilled water in a beaker, transfer to a volumetric flask with rinsings, make up to the mark and shake.
• Dilution: adding water does not change the moles of solute, so C1V1 = C2V2.
• Titration: a known volume of one solution (pipette) is reacted with a solution from a burette until the end point, shown by an indicator (phenolphthalein or methyl orange). Repeat until concordant titres (within 0.10 cm³).
• Calculation: moles of the known solution = C × V; use the mole ratio from the balanced equation to find moles of the unknown; concentration = moles ÷ volume. Example: NaOH + HCl → NaCl + H₂O is 1 : 1; 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O is 2 : 1.

Common misconceptions

• "Volume can be used in cm³ directly in n = C × V." No: Convert cm³ to dm³ by dividing by 1000 first.
• "Diluting a solution reduces the number of moles of solute." No: Dilution adds only water; moles stay the same, concentration falls.
• "The first (rough) titre should be included in the average." No: Average only concordant accurate titres; the rough titre is usually excluded.

Worked examples

Practice questions

1. Define molar concentration.

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   The number of moles of solute dissolved in 1 dm³ of solution.

   Common mistake: Saying 'per 1 dm³ of water'.

2. Convert 250 cm³ to dm³.

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   0.250 dm³.

   Common mistake: Writing 2.5 dm³.

3. Find the moles in 50.0 cm³ of 0.20 mol/dm³ HCl.

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   0.20 × 0.050 = 0.010 mol.

   Common mistake: Using 50 instead of 0.050.

4. Calculate the concentration of a solution with 0.15 mol in 300 cm³.

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   0.15 ÷ 0.300 = 0.50 mol/dm³.

   Common mistake: Dividing by 300.

5. What is a standard solution?

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   A solution of accurately known concentration.

   Common mistake: Saying 'a solution of 1 mol/dm³'.

6. Name the apparatus used to add acid gradually during a titration.

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   Burette.

   Common mistake: Pipette.

7. Why is the conical flask swirled during titration?

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   To mix the reactants so the end point is accurate.

   Common mistake: To cool the solution.

8. Find the mass concentration of 0.2 mol/dm³ H₂SO₄ (Mr 98).

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   0.2 × 98 = 19.6 g/dm³.

   Common mistake: Dividing 98 by 0.2.

9. 50 cm³ of 1.0 mol/dm³ NaOH is diluted to 250 cm³. Find the new concentration.

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   1.0 × 50 ÷ 250 = 0.20 mol/dm³.

   Common mistake: Multiplying instead of dividing.

10. Explain why titrations are repeated.

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    To obtain concordant results and a reliable mean titre.

    Common mistake: Saying to use up the solution.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. A standard solution is one with:	a. 1000
   2. How many cm3 are in 1 dm3?	b. pink in alkali
   3. Phenolphthalein is colourless in acid and:	c. an accurately known concentration
   4. Convert 0.25 mol/dm3 NaOH (Mr 40) to g/dm3.	d. 10 g/dm3

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   1 → c (an accurately known concentration) · 2 → a (1000) · 3 → b (pink in alkali) · 4 → d (10 g/dm³)

   • It is used to find unknown concentrations.
   • 1 dm³ = 1 litre = 1000 cm³.
   • It turns pink above about pH 9.
   • 0.25 × 40 = 10 g/dm³.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   20 g/dm³  ·  0.5 mol/dm³  ·  end point  ·  boiling point
   1. The mass concentration of a solution with 10 g of solute in 500 cm³ is .
   2. 40 g of NaOH (Mr 40) dissolved to make 2 dm³ of solution gives a concentration of .
   3. The point at which the indicator changes colour in a titration is the .
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   1: 20 g/dm³ · 2: 0.5 mol/dm³ · 3: end point

   • 10 g ÷ 0.5 dm³ = 20 g/dm³.
   • 1 mol ÷ 2 dm³ = 0.5 mol/dm³.
   • It shows the reaction is complete.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. In 2NaOH + H2SO4 → Na2SO4 + 2H2O, 0.01 mol of H2SO4 reacts with:	a. 0.02 mol NaOH
   2. How many moles of NaOH are in 25.0 cm3 of 0.100 mol/dm3 solution?	b. agree within 0.10 cm3
   3. A solution contains 0.5 mol of NaCl in 1 dm3. Its concentration is:	c. 0.5 mol/dm3
   4. Concordant titres are those that:	d. 0.00250 mol

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   1 → a (0.02 mol NaOH) · 2 → d (0.00250 mol) · 3 → c (0.5 mol/dm³) · 4 → b (agree within 0.10 cm³)

   • Ratio NaOH : H₂SO₄ = 2 : 1.
   • n = 0.100 × 25.0/1000 = 0.00250.
   • C = n ÷ V = 0.5 ÷ 1.
   • Close agreement shows reliable results.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   pipette  ·  mol/dm³  ·  see the colour change clearly  ·  burette
   1. A white tile is placed under the conical flask in a titration to .
   2. Molar concentration is expressed in .
   3. In a titration, the solution of known volume is measured with a .
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   1: see the colour change clearly · 2: mol/dm³ · 3: pipette

   • The colour change is easier to see.
   • Moles of solute per cubic decimetre of solution.
   • A pipette delivers a fixed accurate volume.

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