Solution dilution
Solutions are often prepared by dilutions of a more concentrated solution. For example, hydrochloric acid, HCl, is a solution that is purchased as a 12.0 M solution. In the laboratory, we may only need 250.0 mL of a 2.00 M HCl solution. How do we prepare this dilute solution?
The following equation is useful when you need to do calculations involving dilutions.
Calculate the number of milliliters of 12.0 M HCl that is needed to prepare 250.0 mL of a 2.00 M HCl solution. Explain the steps that are needed to carry out this dilution.
Step 1: Identify V1, M1, V2, and M2.
| Original | Final |
| M1 = 12.0 M | M2 = 2.00 M |
| V1 is the unknown that we need to find. | V2 = 250.0 mL |
Step 2: Substitute the values into the equation.
V1 x 12.0 M = (250.0 mL )( 2.0 M)
V1 = 41.7 mL.
Step 3: To prepare the 2.00 M solution, withdraw 41.7 mL of the 12.0 M HCl into a 250.0 mL volumetric flask. Make up to the mark with water to give a total volume of 250.0 mL. Mix thoroughly.
External link
Preparing a solution by dilution
Content suitability
BCIT courses: CHEM 0011
