Dilution and Molarity

3.7 Solutions and Mixtures

Dilution and Molarity

A graduated cylinder and a volumetric flask drawn to one real scale. Solve M₁V₁ = M₂V₂ for any unknown and watch a fixed amount of solute spread through more solution, with every answer rounded to the significant figures the inputs allow.

M = n ÷ LM₁V₁ = M₂V₂Particle CountsSignificant Figures

Topic 3.7

Solutions and Mixtures

Calculate the number of solute particles, volume, or molarity of solutions.

A solution (a homogeneous mixture) can be solid, liquid, or gas. Its defining feature is that macroscopic properties do not vary throughout the sample. In a heterogeneous mixture, the macroscopic properties depend on where in the mixture you look.

Composition can be expressed several ways, but the CED is explicit that molarity is the most common method used in the laboratory:

M = n_solute / L_solution

Note "L of solution", not "L of solvent". A 1.0 M solution is made by dissolving 1.0 mol of solute and then adding solvent up to the 1.00 L mark — not by adding 1.0 mol to 1.00 L.

Dilution. Adding solvent changes the volume but not the number of moles of solute, so n = MV is conserved:

M₁V₁ = M₂V₂

Counting ions. For a strong electrolyte, dissociation multiplies the particle concentration. A 0.10 M solution of Na₂SO₄ is 0.20 M in Na⁺ and 0.10 M in SO₄²⁻. Getting this right is essential for net ionic equations (4.2), K_sp (7.11), and conductivity questions.

Key points

Molarity is moles of solute per litre of *solution*, not of solvent.
Dilution conserves moles: M₁V₁ = M₂V₂.
Ionic compounds multiply the ion concentration — always check the formula subscripts.
n = MV is the standard entry point from a solution into a stoichiometry problem.

Equations

$M = \frac{n _{so l u t e}}{L _{so l u t i o n}}$
on the exam sheet
- $M$ molarity (mol/L)
- $n$ moles of solute
$M_{1} V_{1} = M_{2} V_{2}$
not on the sheetNot printed on the sheet — it follows directly from n = MV being conserved.

Common mistakes

Molarity uses solution volume. Adding solute changes the total volume.
Forgetting to multiply ion concentration. 0.10 M CaCl₂ is 0.20 M in Cl⁻.
Mixing volumes in mL and L. Convert before dividing.
Molarity is temperature dependent (volume expands with heat), unlike a mass-based unit.

Worked example

How many millilitres of 12.0 M HCl are needed to prepare 500. mL of 0.150 M HCl? What is the concentration of chloride ion in the final solution?

Dilution: M₁V₁ = M₂V₂
(12.0)(V₁) = (0.150)(500.)
V₁ = 75.0 / 12.0 = 6.25 mL of the concentrated acid

(Procedure note: add that 6.25 mL to some water in a 500 mL volumetric flask, then dilute to the mark — always acid to water.)

Chloride concentration: HCl is a strong acid and ionizes completely, HCl → H⁺ + Cl⁻, in a 1 : 1 ratio.
[Cl⁻] = 0.150 M

Full notes for topic 3.7 →