Concentration of Solutions

Concentration tells you how much solute is packed into a given volume of solution. A strong squash and a weak squash may fill the same glass, but the strong one has more solute dissolved in it, so it is more concentrated.

The standard unit for volume in these calculations is the cubic decimetre (dm³), which is the same as one litre.

Concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³). This calculation is required at both Foundation and Higher Tier.

Two things change the concentration: the mass of solute dissolved and the volume of the solution. Dissolving more solute raises the concentration; adding more water increases the volume and lowers it.

(Higher Tier only) You must be able to explain how the mass of solute and the volume of solution together determine the concentration: more solute in the same volume, or the same solute in a smaller volume, both give a higher concentration.

The first job in almost every question is to get the volume into cubic decimetres. Since 1 dm³ = 1000 cm³, you divide a volume in cm³ by 1000.

Worked example — 20 g of sodium chloride is dissolved to make 500 cm³ of solution. Find the concentration in g/dm³.

Step 1 — convert the volume: 500 cm³ ÷ 1000 = 0.5 dm³.

Step 2 — apply the formula:

$\text{concentration}=\frac{20}{0.5}=40{\text{g/dm}}^3$

The concentration is 40 g/dm³.

Convert cm³ to dm³ before dividing. Forgetting to divide by 1000 is the most frequent mistake in concentration calculations and makes the answer 1000 times too small or too large.

Rearranging the formula lets you find the mass of solute in a given volume, which is what you need when making up a solution of a stated concentration.

$\text{mass (g)}={\text{concentration (g/dm}}^3)\times {\text{volume (dm}}^3)$

Worked example — how much solute is present in 200 cm³ of a solution with a concentration of 25 g/dm³?

Step 1 — convert the volume: 200 cm³ ÷ 1000 = 0.2 dm³.

Step 2 — apply the rearranged formula:

$\text{mass}=25\times 0.2=5\text{g}$

So 200 cm³ of this solution contains 5 g of dissolved solute. A formula triangle with mass on top and concentration × volume beneath helps you rearrange between the two forms without error.

(Separate Chemistry and Higher Tier only) Concentration in mol/dm³ and reacting-volume calculations are assessed only at Higher Tier and only in the separate Chemistry course, not in Combined Science.

Chemists more often quote concentration in moles per cubic decimetre (mol/dm³), because reactions happen in mole ratios. The formula mirrors the g/dm³ version but uses moles instead of mass.

${\text{concentration (mol/dm}}^3)=\frac{\text{moles of solute}}{{\text{volume (dm}}^3)}$

Worked example — 0.25 mol of sodium hydroxide is dissolved in 500 cm³ of water. Find the concentration in mol/dm³.

Volume = 500 ÷ 1000 = 0.5 dm³.

$\text{concentration}=\frac{0.25}{0.5}=0.5{\text{mol/dm}}^3$

To find a concentration in mol/dm³ from a mass, convert the mass to moles first (moles = mass ÷ Mr), then divide by the volume in dm³.

Worked example — 4 g of sodium hydroxide (Mr = 40) is dissolved to make 250 cm³ of solution.

• Moles NaOH = 4 ÷ 40 = 0.1 mol
• Volume = 250 ÷ 1000 = 0.25 dm³
• Concentration = 0.1 ÷ 0.25 = 0.4 mol/dm³

Because moles = mass ÷ Mr, the two concentration units are linked by the relative formula mass. Divide a concentration in g/dm³ by the Mr to get mol/dm³; multiply by the Mr to go the other way.

${\text{concentration (mol/dm}}^3)=\frac{{\text{concentration (g/dm}}^3)}{M_r}$

Worked example — a sodium hydroxide solution has a concentration of 40 g/dm³. What is its concentration in mol/dm³? (Mr of NaOH = 40.)

$\frac{40}{40}=1{\text{mol/dm}}^3$

Worked example — a solution of calcium carbonate equivalent has a concentration of 0.2 mol/dm³ of a solute with Mr = 100. Its concentration in g/dm³ is:

$0.2\times 100=20{\text{g/dm}}^3$

Keeping the units in each step guards against dividing when you should multiply. If the numbers look a thousand times out, check the cm³-to-dm³ conversion first.

(Separate Chemistry and Higher Tier only) When two solutions react and you know the reacting volumes plus one concentration, you can calculate the other concentration. This underpins titration calculations.

The method combines the mole ratio from the balanced equation with the concentration formula. Work in moles throughout.

Worked example — 25.0 cm³ of sodium hydroxide of concentration 0.100 mol/dm³ is exactly neutralised by 20.0 cm³ of hydrochloric acid. Find the concentration of the acid.

$\text{NaOH}+\text{HCl}\to \text{NaCl}+{\text{H}}_2\text{O}$

Step 1 — moles of NaOH. Volume = 25.0 ÷ 1000 = 0.0250 dm³.

$\text{moles NaOH}=0.100\times 0.0250=0.00250\text{mol}$

Step 2 — mole ratio. NaOH : HCl is 1 : 1, so moles of HCl = 0.00250 mol.

Step 3 — concentration of HCl. Volume = 20.0 ÷ 1000 = 0.0200 dm³.

$\text{concentration HCl}=\frac{0.00250}{0.0200}=0.125{\text{mol/dm}}^3$

The acid concentration is 0.125 mol/dm³.

1. Not converting cm³ to dm³

Divide any volume in cm³ by 1000 before using it. 25 cm³ is 0.025 dm³; using 25 straight into the formula gives an answer 1000 times too small.

2. Confusing g/dm³ and mol/dm³

These are different units. Convert a mass to moles with moles = mass ÷ Mr before finding a concentration in mol/dm³, and divide g/dm³ by the Mr to reach mol/dm³.

3. Skipping the mole ratio in reacting-volume calculations

The moles of the two reactants are equal only when the ratio is 1 : 1. For other ratios, apply the balanced equation before finding the second concentration.

4. Applying the mol/dm³ method at the wrong tier

Concentration in mol/dm³ and reacting-volume calculations are Higher Tier and separate Chemistry only. The g/dm³ calculation is the part required at Foundation.

5. Treating concentration as a fixed property of a substance

Concentration depends on both the mass of solute and the volume of solution. Adding water lowers the concentration without changing the amount of solute.