A solution is a homogeneous mixture — every part of it looks and behaves the same. It has two components:
- Solute — the substance that is dissolved (the minor component)
- Solvent — the dissolving medium (the major component)
An aqueous solution is one where water is the solvent. Most reactions studied in chemistry are aqueous.
| Term | Definition | Example |
|---|---|---|
| Solute | Substance being dissolved | NaCl in saltwater |
| Solvent | Dissolving medium | Water in saltwater |
| Solution | Homogeneous mixture of solute + solvent | Saltwater, HCl(aq) |
| Aqueous | Water is the solvent (denoted (aq)) | NaOH(aq) |
| Concentration | Amount of solute per volume of solution | 0.5 mol/L NaCl |
Molarity (symbol C) is the most important concentration unit in chemistry. It connects volume of solution directly to moles of solute — which is exactly what stoichiometry needs.
Where: C = concentration (mol/L), n = moles of solute (mol), V = volume of solution in litres.
Worked Example
C = n / V = 0.50 / 2.0 = 0.25 mol/L
To prepare an accurate molar solution you must control the total volume of the solution, not just the volume of solvent added. A volumetric flask is used because it has a precise graduation mark at an exact volume.
Procedure
- Calculate the mass of solute required: m = n × M, where n = C × V
- Weigh out the calculated mass
- Dissolve in a small amount of solvent (e.g., 50–100 mL) in a beaker — stir until fully dissolved
- Transfer the solution to a volumetric flask of the desired volume
- Rinse the beaker with solvent and add rinsings to the flask
- Add solvent up to the graduation mark — this ensures the total volume is exact
Worked Example
n = C × V = 0.200 × 0.500 = 0.100 mol
m = n × M = 0.100 × 58.5 = 5.85 g
Weigh 5.85 g NaCl. Dissolve in ~200 mL water. Transfer to 500 mL volumetric flask. Dilute to mark.
Dilution means adding solvent to a solution to reduce its concentration. The key insight: the number of moles of solute does not change — only the volume (and therefore the concentration) changes.
| Symbol | Meaning |
|---|---|
| C₁ | Initial (concentrated stock) concentration |
| V₁ | Initial (stock) volume taken |
| C₂ | Final (diluted working) concentration |
| V₂ | Final total volume of diluted solution |
Worked Example
C₂ = C₁V₁ / V₂ = (2.0 × 0.100) / 0.500 = 0.200 / 0.500 = 0.40 mol/L
Check: n before = 2.0 × 0.100 = 0.20 mol. n after = 0.40 × 0.500 = 0.20 mol. ✓
Solubility is the maximum amount of a substance that can dissolve in a given amount of solvent at a specific temperature. It is usually expressed in g/100 mL or mol/L.
Saturation States
| State | Description |
|---|---|
| Unsaturated | Less solute than the maximum — more can still dissolve |
| Saturated | Maximum amount of solute is dissolved — equilibrium between dissolving and crystallising |
| Supersaturated | More solute than normal maximum — unstable; crystallises readily on disturbance |
Solubility Rules for Ionic Compounds in Water
| Rule | Ions involved | Exceptions |
|---|---|---|
| Always soluble | Group 1 ions (Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺), NH₄⁺ | None |
| Always soluble | NO₃⁻, CH₃COO⁻ (most) | None |
| Usually soluble | Cl⁻, Br⁻, I⁻ | Insoluble with Ag⁺, Pb²⁺, Hg₂²⁺ |
| Usually soluble | SO₄²⁻ | Insoluble with Ba²⁺, Pb²⁺, Ca²⁺ |
| Usually insoluble | OH⁻ | Soluble with Group 1, Ba²⁺ |
| Usually insoluble | CO₃²⁻, PO₄³⁻, S²⁻ | Soluble with Group 1, Group 2 |
When ionic compounds and strong acids/bases dissolve in water, they dissociate (break apart) into ions. These free ions allow the solution to conduct electricity. Substances that produce ions in solution are called electrolytes.
| Type | Behaviour | Examples |
|---|---|---|
| Strong electrolyte | Fully dissociates into ions | NaCl, HCl, NaOH, H₂SO₄, KNO₃ |
| Weak electrolyte | Partially dissociates — equilibrium between molecules and ions | CH₃COOH (acetic acid), NH₃ |
| Nonelectrolyte | Does not dissociate — no ions formed | C₆H₁₂O₆ (glucose), C₂H₅OH (ethanol) |
Strong Acids and Bases (fully dissociate)
- Strong acids: HCl, HNO₃, H₂SO₄, HBr, HI, HClO₄
- Strong bases: NaOH, KOH, Ca(OH)₂, Ba(OH)₂
Dissociation Equations
In solution reactions, concentrations replace masses. You use C × V = n to convert from litres of solution to moles of solute, then follow the usual stoichiometry roadmap: moles of A → moles of B → grams or concentration of B.
Roadmap
- Step 1: Write the balanced equation
- Step 2: Find moles of given substance using n = C × V
- Step 3: Use the mole ratio from the equation to find moles of target
- Step 4: Convert moles of target to grams (m = n × M) or concentration (C = n / V)
Worked Example
Equation: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
n(H₂SO₄) = 0.150 × 0.0250 = 0.00375 mol
n(NaOH) = 0.00375 × 2 = 0.00750 mol (ratio 1:2)
V(NaOH) = n / C = 0.00750 / 0.200 = 0.0375 L = 37.5 mL
| Mistake | What to do instead |
|---|---|
| Using mL instead of L in C = n/V | Always convert to litres first: divide mL by 1000. |
| Confusing C₁ and C₂ in dilution | C₁ is the concentrated stock; C₂ is the diluted working solution. C₂ < C₁ always. |
| Using volume of water added as V₂ | V₂ is the total final volume, not just the water added. |
| Forgetting to balance equation before stoichiometry | Always write and balance the equation — the mole ratio comes from coefficients. |
| Applying solubility rules incorrectly | Check both the anion AND cation — a compound is insoluble only if it matches a specific exception. |
| Calling all acids strong electrolytes | Only the six strong acids (HCl, HNO₃, H₂SO₄, HBr, HI, HClO₄) fully dissociate. All others are weak. |