8-4.
PREPARING SOLUTIONS FROM CONCENTRATED LIQUIDS
When preparing a solution in the laboratory, one would normally think of taking a
chemical in powder or crystal form (such as NaCl) and calculate the amount to be
weighed. When preparing such a solution, one would ask, "How many grams of this
substance do I need to make the desired concentration?" Using appropriate factors,
you caculate the desired weight required. With concentrated liquids, this cannot be
done since it is impractical and too dangerous to try to weigh these liquids. We do
know, however, that each concentrated liquid contains a certain mass of solute per
milliliter of solution and that this can be determined using the information found on the
label of the bottle.
Example. A bottle of nitric acid (HNO3) has a S.G. of 1.42 and an assay of
70.0 percent. How many grams of HNO3 are contained within 1.0 mL of HNO3?
Solution. The given volume times the density of the concentrated liquid times
the % A yields the desired results.
1.42 g
70
1.0 mL X --------X ------ = 0.994 g
mL
100
8-5.
PROBLEM SOLVING
You have seen in earlier lessons that the mass of a solute needed to prepare a
solution of a certain concentration may be determined by the use of various
expressions, dependent upon the desired concentration unit. These various problem-
solving expressions all yielded the mass in grams needed to prepare a solution of
known concentration and volume. Similar problem solving techniques may be used to
determine the amount of concentrated liquid needed to yield a desired amount of solute.
The examples that follow will illustrate several methods needed to solve problems
involving concentrated liquids.
a. Example 1. How much concentrated HNO3 with a S.G. of 1.42 and an assay
of 70% would you need to prepare 1.0 liter of a 1.0 mol/L HNO3 solution?
Solution. As before, with problems that involve molarity, calculating the gram
molecular weight of the compound involved is a good starting point.
Calculate the GMW of the compound.
HNO3
H
1.0 X 1 =
1.0
N
14.0 X 1 = 14.0
O
16.0 X 3 = + 48.0
63.0 g/mol
Determine the amount of HNO3 needed to prepare the solution.
MD0837
8-4