Dilution calculation

Dilution calculations are a central part of practical chemistry, especially when working with solutions. Understanding how dilution works allows chemists to prepare solutions of desired concentrations from more concentrated stock solutions. This concept is fundamental in a variety of fields such as chemistry, biology, medicine, and environmental science.

Diluting a solution means reducing its concentration by adding more solvent to it. The amount of solute remains constant during this process. This relationship is described by the equation:

C1 × V1 = C2 × V2

Where:

• C1 is the initial concentration of the solution.
• V1 is the initial volume of the solution.
• C2 is the final concentration of the solution.
• V2 is the final volume of the solution.

This equation is known as the dilution equation and is derived from the idea that the moles of solute before dilution (C1 × V1) are equal to the moles of solute after dilution (C2 × V2).

Understanding concentrations and units

Concentration refers to how much solute is present in a given volume of solution. It can be expressed in a variety of units, including molarity (M), percent composition, parts per million (ppm), and more.

Molarity is one of the most common units used in dilution problems and is defined as the moles of solute per liter of solution:

M = moles of solute / liters of solution

Before diving deeper into dilution calculations, it is important to understand these units and make sure they are consistent when performing calculations.

Looking at dilution with examples

Let us understand the dilution process with some simple examples:

Example 1: Simple dilution

Suppose you have a 2 M solution of hydrochloric acid (HCl), and you want to prepare 500 mL of a 0.5 M HCl solution. Using the dilution equation:

C1 = 2 M, C2 = 0.5 M, V2 = 500 mL (0.5 L), V1 = ?

Insert the values into the equation:

(2 M) × V1 = (0.5 M) × (0.5 L)

Solve for V1:

V1 = (0.5 M × 0.5 L) / 2 M = 0.125 L or 125 mL

Therefore, you will need 125 mL of the original 2 M solution, and you will add enough water to it to reach a total volume of 500 mL to achieve the desired concentration.

Example 2: Multi-step dilution

Sometimes, a solution may require more than one dilution step, especially if the starting solution is highly concentrated. Suppose you have a 10 M stock solution and you need to prepare 1 L of a 0.1 M solution.

Step 1: First, dilute the 10 M solution to 1 M.

C1 = 10 M, C2 = 1 M, V2 = 1 L, V1 = ?

Insert the values into the equation:

(10 M) × V1 = (1 M) × (1 L)

Solve for V1:

V1 = (1 M × 1 L) / 10 M = 0.1 L or 100 mL

Step 2: Dilute the 1 M solution to 0.1 M.

C1 = 1 M, C2 = 0.1 M, V2 = 1 L, V1 = ?

Insert the values into the equation:

(1 M) × V1 = (0.1 M) × (1 L)

Solve for V1:

V1 = (0.1 M × 1 L) / 1 M = 0.1 L or 100 mL

Thus, to obtain the desired concentration, dilute the initial concentrated solution in stages.

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