CLINICAL LAB SCIENEC

CHAPTER 9: LABORATORY MATHEMATICS 223
Problem:
A buffer is made by adding 2 parts of solution A to 5 parts of solution B. How
much of solution A and solution B would be required to make 70 mL of the buffer?
Formula:
Total volume required (C)
parts of A + parts of B
= volume of one part (V)
Solution:
70 mL required
2 parts of A + 5 parts of B
= V
70
= 10 mL = V
7
2 parts of solution A = 2 × 10 = 20 mL
5 parts of solution B = 5 × 10 = 50 mL
Answer:
The buffer would be made by mixing 20 mL of solution A with 50 mL of solution B
to give a total volume of 70 mL.
FIGURE 9-10 Preparing a solution using proportion.
Source: Delmar/Cengage Learning.
Problem:
Formula:
Solution:
Answer:
Prepare 100 mL of 0.1 M HCl using 1.0 M HCl.
C 1 × V 1 = C 2 × V 2
(1.0 M) (V 1 ) = (100 mL) (0.1 M)
100 mL × 0.1
V 1 =
1.0
V 1 = 10 mL
10 mL of 1.0 M HCl is added to 90 mL of H 2 O to make 100 mL of 0.1 M HCl solution.
FIGURE 9-11 Preparing a solution using the formula C 1
× V 1
= C 2
× V 2
, where C = concentration and V = volume.
Source: Delmar/Cengage Learning.
Dilutions
The term dilution refers to a concentrated material that may need to be diluted
to meet the requirements for a certain procedure. This would also be relevant if
a patient sample were of such a concentration for a constituent such as glucose,
that the method used for determining the level would be higher than the procedure
could accurately handle (called linearity). The stock solution provided
would need to be diluted with a solvent such as water where the total volume of
the concentrated solution along with the volume of the diluent (solvent) would
be compared with the concentration of the stock solution. This is called the
dilution factor. As the dilution factor increases, the concentration of the solute
(concentrated stock solution) decreases. An example for diluting a stock solution
would be as follows:
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