Solution Chemistry and Molarity

Learning Objectives

Define solute, solvent, and solution.

Calculate the molarity of a solution.

Perform calculations involving dilutions using the formula M₁V₁ = M₂V₂.

The Chemistry of Mixtures: Solutions

A solution is a homogeneous mixture composed of two or more substances.

Solvent: The substance present in the largest amount, which does the dissolving.

Solute: The substance(s) present in smaller amounts, which is/are dissolved in the solvent.

Concentration: Molarity

The concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution. The most common unit of concentration in chemistry is molarity (M).

Molarity (M) = Moles of Solute / Liters of Solution

Example: What is the molarity of a solution made by dissolving 23.4 g of NaCl in enough water to make 100 mL of solution? (Molar Mass NaCl = 58.5 g/mol)

1.Moles = 23.4 g / 58.5 g/mol = 0.40 mol

2.Liters = 100 mL = 0.100 L

3.Molarity = 0.40 mol / 0.100 L = 4.0 M

Dilution

Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by adding more solvent.

When a solution is diluted, the amount of solvent is increased, but the number of moles of solute remains the same.

This leads to a simple relationship known as the dilution equation:

M₁V₁ = M₂V₂

M₁ and V₁ are the molarity and volume of the initial, concentrated solution (the 'stock' solution).

M₂ and V₂ are the molarity and volume of the final, diluted solution.

Example: How would you prepare 200 mL of a 0.5 M HCl solution from a 10 M stock solution?

1.Identify knowns: M₁=10 M, M₂=0.5 M, V₂=200 mL. We need to find V₁.

2.Rearrange the equation: V₁ = (M₂V₂) / M₁

3.Calculate: V₁ = (0.5 M * 200 mL) / 10 M = 100 / 10 = 10 mL.

4.Procedure: You would take 10 mL of the 10 M stock solution and add enough water to bring the total volume up to 200 mL.

Key Terms