Unit #1Unit #2Unit #3Unit #4Unit #5Unit #6Unit #7Unit #8Unit #9Unit #10
spacerUNIT # 9  
spacerspacerIntroduction
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spacer9.1 What is a
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spacer9.2 Terminology
spacer9.3 Solubility of
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spacer9.4 Solubility of
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spacer9.5 Molarity
spacer9.6 Solution
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Unit #9 SOLUTIONS

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.

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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.

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Section 14.8
Molarity ..p377

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