Page 48 - Modern Analytical Chemistry

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Chapter 2 Basic Tools of Analytical Chemistry 31

(c) The concentration of this solution is only approximate, so volumes do not
need to be measured exactly. The necessary volume of glacial acetic acid is

4 mL CH 3 COOH
´ 2000 mL = 80 mL CH COOH
3
100 mL
To prepare the solution we use a graduated cylinder to transfer 80 mL of
glacial acetic acid to a container that holds approximately 2 L, and we then
add sufficient water to bring the solution to the desired volume.

2E.2 Preparing Solutions by Dilution
Solutions with small concentrations are often prepared by diluting a more concen-
trated stock solution. A known volume of the stock solution is transferred to a new
container and brought to a new volume. Since the total amount of solute is the
same before and after dilution, we know that dilution
The process of preparing a less
2.4
C o ´V o =C d ´V d concentrated solution from a more
concentrated solution.
where C o is the concentration of the stock solution, V o is the volume of the stock
solution being diluted, C d is the concentration of the dilute solution, and V d is the
volume of the dilute solution. Again, the type of glassware used to measure V o and
V d depends on how exact the solution’s concentration must be known.

EXAMPLE 2.10

A laboratory procedure calls for 250 mL of an approximately 0.10 M solution
of NH 3 . Describe how you would prepare this solution using a stock solution of
concentrated NH 3 (14.8 M).
SOLUTION
Substituting known volumes in equation 2.4

14.8 M ´V o =0.10 M ´0.25 L
–3
and solving for V o gives 1.69 ´10 L, or 1.7 mL. Since we are trying to make a
solution that is approximately 0.10 M NH 3 , we can measure the appropriate
amount of concentrated NH 3 using a graduated cylinder, transfer the NH 3 to a
beaker, and add sufficient water to bring the total solution volume to
approximately 250 mL.

As shown in the following example, equation 2.4 also can be used to calculate a
solution’s original concentration using its known concentration after dilution.

EXAMPLE 2.11

2+
A sample of an ore was analyzed for Cu as follows. A 1.25-g sample of the ore
was dissolved in acid and diluted to volume in a 250-mL volumetric flask. A
20-mL portion of the resulting solution was transferred by pipet to a 50-mL
volumetric flask and diluted to volume. An analysis showed that the
concentration of Cu 2+ in the final solution was 4.62 ppm. What is the weight
percent of Cu in the original ore?