IB Colourful Solutions in Chemistry

Solutions are one of the commonest ways of performing reactions, as the particles of solute are free to move and collide. It is, however, important for the chemist to know how many particles there are per unit volume of a solution and for this the concept of concentration and molarity are used

Syllabus reference S1.4.5

Structure 1.4.5 - The molar concentration is determined by the amount of solute and the volume of solution.

Solve problems involving the molar concentration, amount of solute and volume of solution.

Guidance

The use of square brackets to represent molar concentration is required.
Units of concentration should include g dm^–3 and mol dm^–3 and conversion between these.
The relationship n = CV is given in the data booklet.

Tools and links

Tool 1 - What are the considerations in the choice of glassware used in preparing a standard solution and a serial dilution?
Tool 1, Inquiry 2 - How can a calibration curve be used to determine the concentration of a solution?

Concentration
Molarity
Molality
Worked examples
Now test yourself

Concentration

The concentration of a solution is the quantity of solute that it contains per unit volume.

This may be given in grams per 100cm³ or grams per litre, but it is usually given in terms of molarity as this gives a direct measure of the number of solute particles contained by the solution.

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Molarity

The concept of molarity arises from the need to know the amount of solute present in a solution in moles. 1 mole of any substance contains an Avogadro number of particles of that substance = 6.02 x 10²³. A 1 molar solution contains 1 mole of solute, dissolved in 1 litre of solution.

Note: the definition is not per 1 litre of solvent, but per 1 litre of solution. This allows us to measure a volume of solution and work out the number of moles, and hence the number of particles, that it contains.

Molarity = number of moles of solute per litre of solution (1 litre = 1000cm³)

The molarity is denoted by the capital letter M, and given the units mol dm^-3

Example: Calculate the molarity of a solution containing 0.15 moles of potassium nitrate in 100cm³ of solution.

Molarity = moles/litres

100cm³ = 0.1dm³

Molarity = 0.15/0.1 = 1.5 M

The molarity of a specific ion within an ionic solution may also be considered separately. In a 1M solution of copper sulfate (CuSO₄) the copper ions are separate from the sulfate ions. The solution may be said to be both 1 molar in terms of copper 2+ ions and 1 molar in terms of sulfate 2- ions.

A 1 molar (1M) solution of copper nitrate (Cu(NO₃)₂), however, is 1 molar with respect to copper 2+ ions but 2M with respect to nitrate ions.

Example: Calculate the molarity of hydrogen ions in a 0.15 molar solution of sulfuric acid.

The formula of sulfuric acid is H₂SO₄. It dissociates in solution according to the following equation:

H₂SO₄ ⇋ 2H⁺ + SO₄^2-

Hence, if a solution is 1 molar in sulfuric acid, it must be double that in hydrogen ions.

Molarity of the solution = 0.15M in sulfuric acid,

Therefore the molarity in hydrogen ions = 0.15 x 2 = 0.3 mol dm^-3

Using:

In conjunction with

We can calculate the mass needed to prepare solutions or the mass contained in solutions of known concentration.

Example: Calculate the mass of iron(II) sulfate in 100 cm³ of 0.1 mol dm^-3 solution.

The formula of sulfuric acid is FeSO₄. It has a relative formula mass = 56 + 32 + 64 = 152

Number of moles in 100cm³ of 0.1 mol dm^-3 solution = 0.1 x 0.1 = 0.01 moles

Therefore mass of iron /(II) sulfate = moles x relative formula mass = 0.01 x 152

Therefore mass of iron(II) sulfate = 1.52 g

The first of the above mathematical formulae can be manipulated by rearrangement to obtain any of the three factors, moles, molarity or volume of solution.

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Molality

The molality of a solution is the mass of solute per kiologram of solvent. It has limited use.

Worked examples

Calculate the molarity of the following solutions.

Q351-01 3.65 g of HCl in 1000 cm³ of solution

Answer

Mr of HCl = 35.5 + 1 = 36.5

3.65g = 3.65/36.5 moles = 0.1 moles

1000 cm³ = 1 litre

Molarity of solution = moles/litres

Molarity of solution = 0.1/ 1 = 0.1 M

Q351-02 3.65 g of HCl in 100 cm³ of solution

Answer

Mr of HCl = 35.5 + 1 = 36.5

3.65g = 3.65/36.5 moles = 0.1 moles

100 cm³ = 0.1 litre

Molarity of solution = moles/litres

Molarity of solution = 0.1/ 0.1 = 1.0 M

Q351-03 1.00 g of NaOH in 250 cm³ of solution

Answer

Mr of NaOH = 23 + 16 + 1= 40

1.00 g = 1.00/40 moles = 0.025 moles

250 cm³ = 0.25 litre

Molarity of solution = moles/litres

Molarity of solution = 0.025 / 0.25 = 0.1 M

Q351-04 1.96 g of H₂SO₄ in 250 cm³ of solution

Answer

Mr of H₂SO₄ = [2 x 1]+ 32 + [4 x 16] = 98

1.96 g = 1.96/98 moles = 0.02 moles

250 cm³ = 0.25 litre

Molarity of solution = moles/litres

Molarity of solution = 0.02 / 0.25 = 0.08 M

Q351-05 25.0 g of Na₂S₂O₃.5H₂O in 250 cm³ of solution

Answer

Mr of Na₂S₂O₃.5H₂O = [2 x 23]+ [2 x 32] + [3 x 16] + [5 x 18] = 248

24.8 g = 24.8/248 moles = 0.1 moles

250 cm³ = 0.25 litre

Molarity of solution = moles/litres

Molarity of solution = 0.1 / 0.25 = 0.4 M

Q351-06 Calculate the concentration of hydrogen ions in a 2M solution of hydrochloric acid.

Answer

Hydrochloric acid is a strong acid and completely dissociates into ions in aqueous solution:

HCl → H⁺ + Cl^-

1 mole of HCl produces 1 mole of H⁺ ions

Concentration of hydrogen ions is therefore the same as the concentration of hydrochloric acid = 2 moles dm^-3

Q351-07 Calculate the concentration of hydrogen ions in a 2M solution of sulfuric acid.

Answer

sulfuric acid is a strong dibasic acid and dissociates 100% in solution.

H₂SO₄ → 2H⁺ + SO₄^2-

1 mole of sulfuric acid produces 2 moles of hydrogen ions

The 2 M solution of sulfuric acid is therefore 2 x 2 = 4 M with respect to hydrogen ions

Q351-08 Calculate the molarity of the sodium ions in solution when 10.6g of anhydrous sodium carbonate is dissolved in 50cm³ of water and the volume made up to 100cm³.

Answer

Relative formula mass of sodium carbonate, Na₂CO₃ = 106

10.6g of sodium carbonate = 10.6/106 moles = 0.1 moles

But when 1 mole of Na₂CO₃ dissolves in water it produces 2 moles of Na⁺ ions

Therefore moles of sodium ions = 0.1 x 2 = 0.2 moles

Total volume of solution = 100cm³

Molarity = moles/ volume(litres) = 0.2/0.1 = 2 M with respect to sodium ions

Q351-09 Calculate the molarity of the chloride ion solution produced, when 1.345 g of copper chloride is dissolved in 100cm³ of water and the solution volume made up to 250cm³.

Answer

Relative formula mass CuCl₂ = 63.5 + 2(35.5) = 134.5

Moles of copper chloride = 1.35/134.5 = 0.01 moles

Total volume of solution = 250cm³

Molarity of the copper chloride solution = moles/volume(litres) = 0.01/0.25 = 0.04 M

But 1 mole of copper chloride produces 2 moles of chloride ions

Therefore molarity with respect to chloride ions = 0.04 x 2 = 0.08 moles dm^-3

Q351-10 Calculate the molarity of the potassium carbonate solution formed by absorbing 4.4g of carbon dioxide in 500cm³ of 2M potassium hydroxide solution, according to the equation:

2KOH + CO₂ → K₂CO₃ + H₂O

Answer

Moles of carbon dioxide = 4.4/44 = 0.1 moles

From the equation for the reaction, 1 mole of carbon dioxide produces 1 mole of potassium carbonate

Therefore moles of potassium carbonate formed = 0.1 moles

Total volume of solution = 500cm³

Thefore molarity of potassium carbonate = 0.1/0.5 = 0.2 moles dm^-3

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Now test yourself

Exercise 3.51 - Solutions, solutes and concentrations
Answer to three significant figures.

Your answer		Questions	Correct

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