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 V_{1},
M_{1}, V_{2}, and M_{2}.

Original

Final

M_{1} = 12.0 M

M_{2} = 2.00 M

V_{1} is the unknown that we need to find.

V_{2} = 250.0 mL

Step 2: Substitute the values into
the equation.

V_{1} x 12.0 M = (250.0 mL )( 2.0 M)

V_{1} = 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.