Solutions and concentration

Most chemical reactions take place in solution: processes in living organisms, laboratory experiments and industrial production. So expressing the composition of a solution quantitatively and calculating with it is one of chemistry’s most practical skills. Solution problems are the most common type in the exam: mixing, diluting, evaporating, dissolving a hydrated crystal, and the concentration after a reaction. They all rest on two formulae — mass fraction and molar concentration — and on the principle that "the mass of solute is conserved". The topic is closely linked to "Electrolytic dissociation" and "Calculation problems".

§1. Solutions and the dissolving process

A solution is a homogeneous system made of a solvent and one or more solutes. The most important solvent is water; ethanol, benzene and propanone can also act as solvents. A solution must be distinguished from a suspension (solid particles in a liquid, such as muddy water) and an emulsion (droplets of one liquid in another, such as milk): these are not uniform and separate with time.

According to D. I. Mendeleev’s theory, dissolving is a physico-chemical process. The physical side is the spreading of solute particles through the whole volume (diffusion); the chemical side is their combination with solvent molecules (hydration, in water). So dissolving can absorb heat (the solution cools when NH₄NO₃ dissolves) or release heat (the solution warms when H₂SO₄ or NaOH dissolves). When diluting concentrated sulfuric acid, always add the acid slowly to water, never the reverse — otherwise the water boils and acid spatters.

§2. Solubility and hydrated crystals

Solubility is the greatest mass of a substance that can dissolve in 100 g of water at a given temperature. The solubility of most solids increases with temperature, whereas that of gases decreases (which is why a fizzy drink goes flat quickly when warm); gas solubility increases with pressure. Substances may be very soluble, slightly soluble or practically insoluble.

Kinds of solution: saturated — no more solute dissolves at that temperature; unsaturated — more can dissolve; supersaturated — contains more solute than its solubility allows; it is unstable, and dropping in a crystal makes the excess crystallise.

Some substances crystallise from solution together with water molecules — hydrated crystals (crystalline hydrates): CuSO₄·5H₂O (blue vitriol), FeSO₄·7H₂O (green vitriol), Na₂SO₄·10H₂O (Glauber’s salt), Na₂CO₃·10H₂O (washing soda). The water of crystallisation joins the solution when the crystal dissolves and is driven off on heating.

Worked example

At a certain temperature the solubility of a salt is 25 g. Find the mass fraction of the saturated solution at that temperature.

1. 25 g of salt dissolves in 100 g of water, so the saturated solution has a mass of 100 + 25 = 125 g.
2. ω = 25/125 = 0.2.
3. Answer: 20 %.

§3. Ways of expressing concentration

On dilution or evaporation the mass (and amount) of solute does not change — the key idea of every calculation. For molar concentration: C₁·V₁ = C₂·V₂. Mass fraction has no units and does not depend on temperature, whereas molar concentration depends on volume, so for accurate analysis solutions are made up in a volumetric flask at a fixed temperature. Physiological saline used in medicine is a 0.9 % NaCl solution.

Worked example

50 g of water was added to 200 g of a 20 % solution. Find the new mass fraction.

1. Solute: 200·0.2 = 40 g — unchanged.
2. Mass of the new solution: 200 + 50 = 250 g.
3. ω = 40/250 = 0.16, i.e. 16 %.

§4. Mixing and the "cross" rule

When two solutions are mixed, the mass ratio is m₁/m₂ = (ω₂ − ω₃)/(ω₃ − ω₁) (the "cross" or diagonal method). Pure water counts as ω = 0 % and pure solute as ω = 100 %. The concentration of the mixture always lies between the two starting values.

Worked example

300 g of a 20 % solution must be made from 40 % and 10 % solutions. What mass of each is needed?

1. m₁/m₂ = (20 − 10)/(40 − 20) = 10/20 = 1/2.
2. 3 parts in total; 1 part = 300/3 = 100 g.
3. 100 g of the 40 % solution and 200 g of the 10 % solution.

§5. Problems involving reactions

Mass of the solution after a reaction: m = m(solutions) + m(added solid) − m(gas) − m(precipitate). Gas given off and precipitate formed leave the solution, so they are subtracted.

Worked example

A solution of H₂O₂ partly decomposed; its mass fell by 8 % and a 10 % H₂O₂ solution remained. Find the original mass fraction.

1. Take 100 g of solution; the loss of mass is the O₂ released: 8 g = 0.25 mol.
2. 2H₂O₂ → 2H₂O + O₂: H₂O₂ decomposed 0.5 mol = 17 g.
3. 92 g of solution remains, containing 0.1·92 = 9.2 g of H₂O₂.
4. Original H₂O₂: 9.2 + 17 = 26.2 g → 26.2 %.

Worked example

50 mL of 1 M HCl (0.05 mol) was added to 200 mL of 0.5 M Na₃PO₄ (0.1 mol). Which salt is absent from the solution?

1. Na₃PO₄ + HCl → Na₂HPO₄ + NaCl — 0.05 mol of Na₃PO₄ is used.
2. The solution contains Na₃PO₄ 0.05, Na₂HPO₄ 0.05 and NaCl 0.05 mol.
3. The absent salt is NaH₂PO₄.

§6. Calculations with hydrated crystals

When a solution is made from a hydrate, the anhydrous salt is only part of the hydrate’s mass: in CuSO₄·5H₂O (M = 250) the fraction of CuSO₄ is 160/250 = 0.64.

Worked example

What masses of blue vitriol and water are needed to make 500 g of an 8 % CuSO₄ solution?

1. m(CuSO₄) = 500·0.08 = 40 g, n = 40/160 = 0.25 mol.
2. The hydrate is also 0.25 mol: m = 0.25·250 = 62.5 g.
3. Water: 500 − 62.5 = 437.5 g.

Check yourself

1. What mass of alkali is in 250 mL of 0.2 M NaOH?

2. 100 g of water was evaporated from 400 g of a 15 % solution. What is the new mass fraction?

3. What is the molar concentration of 98 % H₂SO₄ (ρ = 1.84 g/mL)?

4. 10 g more NaCl was added to a solution of 10 g NaCl in 90 g water. What is the mass fraction?

5. In what mass ratio should 5 % and 25 % solutions be mixed to give a 15 % solution?

6. Excess CaCO₃ was added to 100 g of 7.3 % HCl. What is the mass fraction of CaCl₂ in the solution?

7. 100 mL of a 2 M solution was diluted to 500 mL. What is the new concentration?

8. How does the solubility of gases in water change as temperature rises?

Common mistakes

• Confusing the mass of the solution with the mass of water: solution = solute + water.
• Not subtracting gas or precipitate from the mass of the solution.
• Ignoring the water in a hydrated crystal.
• Adding volumes to get the volume of a mixture: work with masses via density.
• Thinking the mass of solute changes on dilution.