IMAT Chemistry · Solutions and Concentration

Preparing, Diluting, Mixing and Reacting Solutions

Adding water spreads the same moles of solute through more volume, and in a reaction between solutions the moles from n = cV follow the equation's ratio.

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Why this matters

Dilution appeared in 2021 and in the 2024 ministry paper, which asked how much water to add to reach a lower concentration. A titration with sulfuric acid, answered in g/L, appeared in 2020.

Concept 1 of 4: Preparing a solution of known concentration

To make a solution of a chosen concentration, work out the moles you need in the final volume, then the mass of those moles. The solid is dissolved first and then topped up to the mark, because the concentration is per litre of solution, not per litre of water added.

Definition

Mass of solute needed: $m = c \times V \times M$ .
Method: weigh the solid, dissolve it in less water than the final volume, pour it into a volumetric flask, rinse the beaker into the flask, add water up to the mark, then stopper and invert to mix.
The volumetric flask fixes the volume of the solution. Dissolving the solid in 1 L of water gives slightly more than 1 L of solution, so a slightly lower concentration.

Mass of solute needed

m = c \times V \times M

$m$ mass of solute to weigh, in g
$c$ concentration wanted, in mol/L
$V$ final volume of solution, in L
$M$ molar mass, in g/mol

Worked example

What mass of anhydrous sodium carbonate,

\mathrm{Na_2CO_3}

(

M = 106\ \text{g/mol}

), is needed to make 250 mL of a 0.100 mol/L solution?

Practice this conceptself-check · 3 quick reps

Made up to 1 L is not the same as dissolved in 1 L

Molarity is per litre of solution. Adding a solid to 1 L of water gives more than 1 L of solution, so the concentration is slightly below the target. Correct preparation always makes the solution up to the mark.

Concept 2 of 4: Dilution: c₁V₁ = c₂V₂

Adding water does not add or remove any solute. The same moles are spread through a larger volume, so the concentration falls by the same factor as the volume grows. Taking a sample out of a solution changes neither: the sample has the concentration of the solution it came from.

Definition

On dilution, the moles of solute stay the same: $c_1V_1 = c_2V_2$ .
Any volume unit works, as long as both volumes use the same one.
Water added = final volume minus starting volume (assuming the volumes add).
Dilution factor = $V_2/V_1$ = $c_1/c_2$ . In a two-step dilution, multiply the factors.
Taking a portion (an aliquot) with a pipette does not change the concentration.

Dilution

c_1 V_1 = c_2 V_2

$c_1, V_1$ concentration and volume before adding water
$c_2, V_2$ concentration and total volume after

Worked example

20 mL of 2.0 mol/L hydrochloric acid is diluted to 0.50 mol/L. What is the final volume, and how much water is added?

Practice this conceptself-check · 3 quick reps

Water added is not the final volume

c_1V_1 = c_2V_2

gives the final volume. If the question asks how much water to add, subtract the starting volume. The final volume itself is usually one of the wrong options.

Concept 3 of 4: Mixing two solutions of the same solute

When two solutions are poured together, nothing reacts and nothing is lost: the moles of solute add up, and so do the volumes. The new concentration is the total moles over the total volume, which always lies between the two starting concentrations.

Definition

Moles add: $n = c_1V_1 + c_2V_2$ . Volumes add: $V = V_1 + V_2$ (a good approximation for dilute solutions).
The result lies between $c_1$ and $c_2$ . It is their simple average only when the volumes are equal.
If two different solutes are mixed, each one is diluted by the total volume. For an ion that comes from both, add its moles from each source.

Concentration after mixing

c = \frac{c_1V_1 + c_2V_2}{V_1 + V_2}

$c_1, V_1$ first solution
$c_2, V_2$ second solution

Worked example

100 mL of 0.50 mol/L sodium chloride is mixed with 300 mL of 0.10 mol/L sodium chloride. What is the concentration of the mixture?

Practice this conceptself-check · 3 quick reps

Concentrations do not add

Mixing 0.1 mol/L with 0.1 mol/L gives 0.1 mol/L, not 0.2. Add moles and volumes separately, then divide.

Concept 4 of 4: Reacting volumes and titration

In a titration, one solution is added from a burette until it has exactly reacted with a measured volume of the other. At that point the moles have met in the ratio of the balanced equation. Knowing one concentration, both volumes and the ratio gives the unknown concentration.

Definition

Moles of the known solution: $n = cV$ .
Use the coefficients of the balanced equation to get the moles of the other substance.
Divide by its volume (in litres) for its concentration; multiply by its molar mass for g/L.
The equivalence point is where the reactants have exactly reacted; an indicator shows it by a colour change.
A diprotic acid such as $\mathrm{H_2SO_4}$ neutralises two moles of NaOH per mole.

At the equivalence point

\frac{c_A V_A}{a} = \frac{c_B V_B}{b}

$c_A, V_A$ concentration and volume of solution A
$a, b$ coefficients of A and B in the balanced equation

Worked example

25.0 mL of barium hydroxide solution is exactly neutralised by 20.0 mL of 0.150 mol/L hydrochloric acid:

\mathrm{Ba(OH)_2} + 2\mathrm{HCl} \rightarrow \mathrm{BaCl_2} + 2\mathrm{H_2O}

. What is the concentration of the barium hydroxide?

Practice this conceptself-check · 3 quick reps

Check the ratio before using c₁V₁ = c₂V₂

The dilution formula only works for reactions in a 1 : 1 ratio. Sulfuric acid needs twice its moles of sodium hydroxide, and a carbonate needs twice its moles of hydrochloric acid. Use the coefficients of the balanced equation.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Preparing a solution of known concentration
Mass of solute needed
$m = c \times V \times M$
Dilution: c₁V₁ = c₂V₂
Dilution
$c_1 V_1 = c_2 V_2$
Mixing two solutions of the same solute
Concentration after mixing
$c = \frac{c_1V_1 + c_2V_2}{V_1 + V_2}$
Reacting volumes and titration
At the equivalence point
$\frac{c_A V_A}{a} = \frac{c_B V_B}{b}$

Watch out for (4)

Made up to 1 L is not the same as dissolved in 1 L
→ Preparing a solution of known concentration
Water added is not the final volume
→ Dilution: c₁V₁ = c₂V₂
Concentrations do not add
→ Mixing two solutions of the same solute
Check the ratio before using c₁V₁ = c₂V₂
→ Reacting volumes and titration

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