Solutions and Dilution Calculations in Chemistry

Solutions

Definition and Properties of Solutions

A solution is a homogeneous mixture composed of two or more substances. In chemistry, solutions are commonly formed when a solute is dissolved in a solvent. The concentration of a solution refers to the amount of solute present in a given quantity of solvent or solution.

• Solute: The substance that is dissolved in the solvent.

• Solvent: The substance that dissolves the solute, usually present in greater amount.

• Concentration: Expressed in units such as molarity (mol/L), it quantifies the amount of solute per unit volume of solution.

Dilution of Solutions

Dilution is the process of adding more solvent to a solution, resulting in a decrease in the concentration of the solute. Importantly, dilution does not change the total number of solute particles present; it only increases the volume of the solution, thereby lowering the concentration.

• When a solution is diluted, the number of moles of solute remains constant.

• For example, if 50 mL of solution is diluted to 100 mL, both solutions contain the same number of dissolved particles, but the concentration is halved.

Calculating Dilution

Mathematical Relationship for Dilution

The calculation of dilution is based on the principle that the number of moles of solute before and after dilution remains unchanged. The relationship is given by:

• Let $V_1$ be the initial volume and $c_1$ the initial concentration.

• Let $V_2$ be the final volume after dilution and $c_2$ the final concentration.

The number of moles of solute before dilution: $n_1 = c_1 \times V_1$ The number of moles of solute after dilution: $n_2 = c_2 \times V_2$ Since $n_1 = n_2$, $c_1 V_1 = c_2 V_2$

Rearranging the Dilution Equation

• To solve for the new concentration after dilution: $c_2 = \frac{c_1 V_1}{V_2}$

• To solve for the volume of stock solution needed: $V_1 = \frac{c_2 V_2}{c_1}$

Worked Example: Dilution Calculation

Example 1: Diluting Hydrochloric Acid

Calculate the concentration of the solution formed when 10.0 mL of water is added to 5.00 mL of 1.2 mol L-1 HCl.

Step	Thinking	Working
1	Write down the value of $c_1$ and $V_1$ (original solution).	$c_1 = 1.2$ mol L-1, $V_1 = 5.00$ mL
2	Write down the value of $V_2$ (total volume after dilution).	$V_2 = 10.0 + 5.00 = 15.0$ mL
3	Transpose the equation $c_1 V_1 = c_2 V_2$ to solve for $c_2$.	$c_2 = \frac{c_1 V_1}{V_2}$
4	Calculate the concentration of the diluted solution.	$c_2 = \frac{1.2 \times 5.00}{15.0} = 0.40$ mol L-1

Example 2: Student Practice Problem

Calculate the concentration of the solution formed when 95.0 mL of water is added to 5.00 mL of 0.500 mol L-1 HCl.

• Total volume after dilution: $V_2 = 95.0 + 5.00 = 100$ mL

• Use the dilution equation: $c_2 = \frac{c_1 V_1}{V_2}$

• Calculation: $c_2 = \frac{0.500 \times 5.00}{100} = 0.0250$ mol L-1

Advanced Dilution Problem

Example 3: Preparing a Sulfuric Acid Solution

Calculate the volume of 18.4 mol L-1 sulfuric acid solution required to prepare 5.00 L of 1.50 mol L-1 solution. What volume of water must be added to the sulfuric acid solution to give the required volume of the diluted solution?

• Use the dilution equation: $c_1 V_1 = c_2 V_2$

• Given: $c_1 = 18.4$ mol L-1, $c_2 = 1.50$ mol L-1, $V_2 = 5.00$ L

• Calculate $V_1$: $V_1 = \frac{c_2 V_2}{c_1} = \frac{1.50 \times 5.00}{18.4} = 0.408$ L

• Volume of water to add: $V_{water} = V_2 - V_1 = 5.00 - 0.408 = 4.59$ L

Key Concepts and Summary Table

Term	Definition
Solution	Homogeneous mixture of solute and solvent
Solute	Substance dissolved in the solvent
Solvent	Substance that dissolves the solute
Concentration	Amount of solute per unit volume of solution
Dilution	Process of adding solvent to decrease concentration
Stock Solution	Concentrated solution used to prepare diluted solutions

Summary

• Dilution calculations are essential for preparing solutions of desired concentrations in laboratory settings.

• The key equation $c_1 V_1 = c_2 V_2$ allows for the calculation of unknown volumes or concentrations during dilution.

• Always ensure units are consistent when performing calculations (e.g., all volumes in mL or L).

Additional info: These concepts are foundational for laboratory work in both general and organic chemistry, especially when preparing reagents for reactions or analytical procedures.