Concentration
\( c = \frac{n}{V} \)
Unit Warning
Volume MUST be in dm³ (not cm³). To convert: divide cm³ by 1000.
Standard Solutions & Dilution Formula (c₁V₁ = c₂V₂)
Standard Solution: A solution of accurately known concentration.
Dilution Formula: \(c_1 V_1 = c_2 V_2\) (amount of solute remains constant during dilution with pure solvent).
Volumetric Titration Calculation
Problem: \(25.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\;\text{NaOH}\) is neutralized by \(20.0\text{ cm}^3\) of \(\text{HCl}\). Calculate the concentration of \(\text{HCl}\).
\(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\) (1:1 stoichiometric ratio)
1. Calculate Moles of \(\text{NaOH}\): \(n = c \times V = 0.100\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}\)
2. Use Mole Ratio: \(n(\text{HCl}) = n(\text{NaOH}) = 2.50 \times 10^{-3}\text{ mol}\)
3. Calculate Concentration of \(\text{HCl}\): \(c = \dfrac{n}{V} = \dfrac{2.50 \times 10^{-3}\text{ mol}}{0.0200\text{ dm}^3} = \mathbf{0.125\text{ mol dm}^{-3}}\)
Notation and Units
Square Bracket Concentration Notation [X]
In IB Chemistry, concentration is written using square brackets: [HCl] = 0.10 mol dm⁻³ means "the concentration of hydrochloric acid is 0.10 mol dm⁻³".
Converting Between mol dm⁻³ and g dm⁻³
\[ \text{concentration (g dm}^{-3}\text{)} = \text{concentration (mol dm}^{-3}\text{)} \times M \]
Example: [NaCl] = 0.50 mol dm⁻³. Molar mass of NaCl = 58.44 g mol⁻¹. Concentration = 0.50 × 58.44 = 29.2 g dm⁻³.
Parts Per Million (ppm) Calculations
The unit parts per million (ppm) is used for very dilute solutions: 1 ppm = 1 mg dm⁻³ = 1 mg kg⁻¹. Always check whether the exam question asks for mol dm⁻³, g dm⁻³, or ppm.
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