Fundamental Chemistry and Hydraulics - Mathematics and Statistics for the Environment - page 357

To demonstrate the final dilution factor calculation, a typical lab experiment involves a three-step
1:100 serial dilution of a bacterial culture. The initial step combines 1 unit volume culture (10 µL)
with 99 unit volumes of broth (990 µL) for a 1:100 dilution. In the next step, 1 unit volume of the
1:100 dilution is combined with 99 unit volumes of broth to yield a total dilution of 1:100 × 100 =
1:10,000 dilution. Repeated again (the third step), the total dilution would now be 1:100 × 10,000
= 1:1,000,000 total dilution. The concentration of bacteria is now 1 million times less than in the
original sample.

14.1.2.8.3 C i V i = C f V f Method (Fixed Volumes of Specific
Concentrations from Liquid Reagents)
When we are diluting solutions, the product of the concentration and volume of the initial solution
must be equal to the product of the concentration and volume of the diluted solution when the same
system of units is used in both solutions. This can be expressed as the following relationship:

C i × V i = C f × V f (14.9)

where
C i = Concentration of initial solution.
V i = Volume of initial solution.
C f = Concentration of final solution.
V f = Volume of final solution.

■ EXAMPLE 14.11
Problem: How much water must be added to 60 mL of 1.3- M HCl solution to produce a 0.5- M HCl
solution?

Solution:

CVCV ×= ×

i i f f

13 . ×=× 60 05 . x

13 60
05

. ×

x = = 156 mL

.

The volume of the final solution is 156 mL. The volume of water to be added is then the difference
between the volumes of the two solutions:

156 - 60 = 96 mL of water

14.1.2.8.4 Molar Solutions
Sometimes it may be more efficient to use molarity when calculating concentrations. A 1.0- M solu-
tion is equivalent to 1 formula weight (FW) of chemical dissolved in 1 liter of solvent (usually
water). Formula weight is always given on the label of a chemical bottle (use molecular weight if it
is not given), and is expressed in grams per mole.

■ EXAMPLE 14.12
Problem: Given the following data, determine how many grams of reagent to use: chemical FW =
195 g/mole; making a 0.15- M solution.

Solution:

195 g/mol × 0.15 mol/L = 29.25 g/L