Concentration defines the amount of a dissolved substance, known as the solute, relative to the total volume of the solution. This ratio is important in various fields, from preparing chemical reagents in a laboratory to manufacturing beverages or medicines. Dilution is the process of reducing this concentration by adding more of the solvent, typically water, to the existing solution. Calculating the final concentration after this process is necessary to ensure the solution is prepared correctly for its intended purpose.
The Fundamental Dilution Equation
The relationship between the initial, concentrated solution and the final, diluted solution is captured by a straightforward mathematical expression. This formula is based on the principle that the total amount of solute, measured in moles or mass, does not change during the dilution process. Only the volume of the solvent is increased, which spreads the fixed amount of solute over a larger volume, thus lowering the concentration.
The equation used is \(C_1V_1 = C_2V_2\), which represents the equality of the solute amount before and after the addition of solvent. \(C_1\) and \(V_1\) describe the initial, or stock, solution (concentration and volume measured out for dilution). \(C_2\) and \(V_2\) describe the final, diluted solution (final concentration and total final volume). Because the amount of solute (\(C \times V\)) remains constant, knowing any three of these four variables allows for the calculation of the fourth.
Step-by-Step Calculation Procedure
To find the concentration of a diluted solution, the fundamental dilution equation must first be algebraically rearranged to isolate the unknown variable, \(C_2\). Since the equation is \(C_1V_1 = C_2V_2\), the final concentration \(C_2\) can be solved by dividing both sides of the equation by the final volume, \(V_2\). This manipulation results in the working formula: \(C_2 = (C_1V_1) / V_2\).
This rearranged formula requires three pieces of information: the initial concentration of the stock solution (\(C_1\)), the exact volume of the stock solution used for the dilution (\(V_1\)), and the total volume of the solution after the solvent has been added (\(V_2\)). The units used for the concentration and volume must be consistent throughout the calculation to ensure a correct result.
The concentration unit for \(C_1\) will directly determine the concentration unit for the calculated \(C_2\). For instance, if the initial concentration is expressed in Molarity (moles per liter), the final concentration will also be in Molarity. Similarly, the volumes \(V_1\) and \(V_2\) must be in the same units, such as both in milliliters or both in liters. Maintaining this consistency prevents calculation errors and ensures that the final concentration value is accurate for further use.
Working Through a Dilution Example
Consider a common scenario where a researcher needs to dilute a highly concentrated stock solution for an experiment. Imagine a stock solution of a chemical is prepared with an initial concentration (\(C_1\)) of 5.0 M (Molar). The researcher decides to take a measured volume (\(V_1\)) of 100 mL of this stock solution and add enough water to bring the total final volume (\(V_2\)) to 500 mL. The goal is to determine the resulting final concentration (\(C_2\)).
First, the known values are identified: \(C_1 = 5.0 \text{ M}\), \(V_1 = 100 \text{ mL}\), and \(V_2 = 500 \text{ mL}\). The volumes are already in consistent units of milliliters, so no conversion is necessary before calculation. The working formula, \(C_2 = (C_1V_1) / V_2\), is used to plug in these values.
The calculation becomes \(C_2 = (5.0 \text{ M} \times 100 \text{ mL}) / 500 \text{ mL}\). Multiplying the initial concentration by the initial volume yields \(500 \text{ M} \cdot \text{mL}\). Dividing this result by the final volume of \(500 \text{ mL}\) gives a final concentration of 1.0 M. The units of milliliters cancel out during the division, leaving the final concentration in the desired unit of Molarity. This result confirms that taking 100 mL of the 5.0 M stock solution and diluting it to a total volume of 500 mL produces a new solution with a final concentration of 1.0 M.