UNIT # 9 Introduction Objectives 9.1 What is a 9.2 Terminology 9.3 9.4 9.5 Molarity 9.6 Solution Problems 1 | 2

9.6 - Preparation of Solutions by 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.

 Click on this link to prepare a solution by dilution (Flash plugin required)

Section 14.8
Molarity ..p377

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