Solutions and Concentrations

This lesson covers the preparation of solutions, concentration calculations in both mol dm⁻³ and g dm⁻³, dilutions, and the preparation of standard solutions. These concepts are essential for volumetric analysis (titrations).

---

Key Definitions

• Solute: The substance being dissolved (e.g., NaOH).
• Solvent: The liquid doing the dissolving (usually water).
• Solution: A homogeneous mixture of solute and solvent.
• Concentration: The amount of solute dissolved in a given volume of solution.

---

Concentration in mol dm⁻³

Concentration is most commonly expressed in mol dm⁻³ (also written as M, meaning molar). This tells you the number of moles of solute per dm³ (per litre) of solution.

c = n / V

where:

• c = concentration (mol dm⁻³)
• n = number of moles (mol)
• V = volume of solution (dm³)

Rearranged:

• n = c × V (to find moles)
• V = n / c (to find volume)

Key Point: Volume must be in dm³ for this equation. To convert cm³ to dm³, divide by 1000.

Worked Example 1: Finding Concentration

0.400 g of sodium hydroxide (NaOH) is dissolved in water to make 250 cm³ of solution. Calculate the concentration in mol dm⁻³.

M(NaOH) = 40.0 g mol⁻¹

n(NaOH) = 0.400 / 40.0 = 0.0100 mol

V = 250 / 1000 = 0.250 dm³

c = n / V = 0.0100 / 0.250 = 0.0400 mol dm⁻³

Worked Example 2: Finding Moles from Concentration and Volume

Calculate the number of moles of H₂SO₄ in 25.0 cm³ of 0.100 mol dm⁻³ sulfuric acid.

V = 25.0 / 1000 = 0.0250 dm³

n = c × V = 0.100 × 0.0250 = 2.50 × 10⁻³ mol (or 0.00250 mol)

Worked Example 3: Finding Volume

What volume of 0.200 mol dm⁻³ HCl contains 0.0500 mol?

V = n / c = 0.0500 / 0.200 = 0.250 dm³ = 250 cm³

---

Concentration in g dm⁻³

Concentration can also be expressed as the mass of solute per dm³ of solution, in g dm⁻³.

concentration (g dm⁻³) = mass of solute (g) / volume of solution (dm³)

Converting Between mol dm⁻³ and g dm⁻³

To convert from mol dm⁻³ to g dm⁻³:

concentration (g dm⁻³) = concentration (mol dm⁻³) × molar mass (g mol⁻¹)

To convert from g dm⁻³ to mol dm⁻³:

concentration (mol dm⁻³) = concentration (g dm⁻³) / molar mass (g mol⁻¹)

Worked Example 4: Converting Units

A solution of sodium carbonate (Na₂CO₃) has a concentration of 0.150 mol dm⁻³. Express this in g dm⁻³.

M(Na₂CO₃) = 2(23.0) + 12.0 + 3(16.0) = 106.0 g mol⁻¹

Concentration = 0.150 × 106.0 = 15.9 g dm⁻³

Worked Example 5: From g dm⁻³ to mol dm⁻³

A glucose (C₆H₁₂O₆) solution has a concentration of 36.0 g dm⁻³. Calculate the concentration in mol dm⁻³.

M(C₆H₁₂O₆) = 6(12.0) + 12(1.0) + 6(16.0) = 180.0 g mol⁻¹

Concentration = 36.0 / 180.0 = 0.200 mol dm⁻³

---

Dilution

When you dilute a solution by adding more solvent, the number of moles of solute stays the same, but the volume increases and the concentration decreases.

c₁V₁ = c₂V₂

where:

• c₁ = initial concentration
• V₁ = initial volume
• c₂ = final concentration
• V₂ = final volume

Worked Example 6: Dilution Calculation

50.0 cm³ of 2.00 mol dm⁻³ HCl is diluted to 500 cm³. What is the new concentration?

c₁V₁ = c₂V₂

2.00 × 50.0 = c₂ × 500

c₂ = (2.00 × 50.0) / 500 = 0.200 mol dm⁻³

Worked Example 7: Volume for Dilution

What volume of 0.500 mol dm⁻³ NaOH is needed to prepare 250 cm³ of 0.100 mol dm⁻³ NaOH?

c₁V₁ = c₂V₂

0.500 × V₁ = 0.100 × 250