When working with chemical solutions, concentration refers to the amount of solute—the substance dissolved—relative to the amount of solvent—the substance doing the dissolving. Dilution is the process of lowering this concentration by adding more solvent, such as water, without changing the total amount of solute present. This action increases the overall volume of the solution, spreading the original amount of solute over a larger space, resulting in a less concentrated mixture. Calculating the exact concentration after a dilution is a fundamental requirement in laboratory work.
Understanding the Dilution Equation
The conceptual basis for dilution calculations rests on the principle of conservation of the solute. Because no solute is added or removed during the dilution process, the total quantity of solute remains unchanged. This constant amount of solute is the foundation of the standard dilution equation: \(C_1V_1 = C_2V_2\).
The product of the initial concentration (\(C_1\)) and initial volume (\(V_1\)) equals the product of the final concentration (\(C_2\)) and the final volume (\(V_2\)). \(C_1V_1\) represents the total amount of solute before dilution, and \(C_2V_2\) represents the total amount of solute after dilution. Concentration (\(C\)) can be expressed in various units like molarity or percent volume.
The volume (\(V\)) must be in a consistent unit, such as milliliters or liters, on both sides of the equation. By knowing any three of the four variables, the unknown fourth variable, including the final concentration, can be determined through simple algebraic rearrangement.
Step-by-Step Guide to Calculating the Final Concentration
To calculate the final concentration, \(C_2\), the dilution equation \(C_1V_1 = C_2V_2\) is algebraically rearranged to isolate \(C_2\): \(C_2 = (C_1V_1) / V_2\). The first step is to identify the three known variables: the initial concentration (\(C_1\)), the initial volume (\(V_1\)), and the final volume (\(V_2\)).
Consider diluting 10 milliliters (mL) of a 5 molar (M) stock solution to a final volume of 50 mL. Here, \(C_1\) is 5 M, \(V_1\) is 10 mL, and \(V_2\) is 50 mL. Substituting these values into the formula gives \(C_2 = (5 \text{ M} \times 10 \text{ mL}) / 50 \text{ mL}\).
Multiplying the numerator yields \(50 \text{ M} \cdot \text{mL}\). The calculation is \(C_2 = 50 \text{ M} \cdot \text{mL} / 50 \text{ mL}\).
The final division results in a final concentration (\(C_2\)) of 1 M. The volume units (mL) cancel out, leaving the concentration unit (M). This confirms that increasing the solution volume five times decreases the concentration by the same factor.
Ensuring Accuracy: Unit Consistency and Common Errors
Maintaining consistency in units is necessary for accurate dilution calculations. The unit used for \(C_1\) must match \(C_2\), and the unit for \(V_1\) must match \(V_2\). For example, if the initial volume is measured in milliliters, the final volume must also be expressed in milliliters.
A frequent source of error is confusing the volume of solvent added with the final total volume, \(V_2\). The final volume (\(V_2\)) is the sum of the initial volume (\(V_1\)) and the volume of the solvent added. For instance, if 40 mL of water are added to 10 mL of a solution, \(V_2\) is 50 mL, not 40 mL. Double-checking that the calculated final concentration (\(C_2\)) is lower than the initial concentration (\(C_1\)) is a practical way to verify the calculation was correctly executed.