The Foundations of Chemistry

Multiplication of the volume of a solution, in liters, by its molar concentration gives

the amount of solute in the solution.

When we dilute a solution by mixing it with more solvent, the amount of solute pres-

ent does not change. But the volume and the concentration of the solution dochange.

Because the same number of moles of solute is divided by a larger number of liters of

solution, the molarity decreases. Using a subscript 1 to represent the original concen-

trated solution and a subscript 2 to represent the dilute solution, we obtain

volume 1 molarity 1 number of moles of solutevolume 2 molarity 2

or

V 1 M 1 V 2 M 2 (for dilution only)

This expression can be used to calculate any one of four quantities when the other

three are known (Figure 3-3). We frequently need a certain volume of dilute solution of

a given molarity for use in the laboratory, and we know the concentration of the initial

solution available. Then we can calculate the amount of initial solution that must be used

to make the dilute solution.

A can of frozen orange juice contains a
certain mass (or moles) of vitamin C.
After the frozen contents of the can
are diluted by addition of water, the
amount of vitamin C in the resulting
total amount of solution will be
unchanged. The concentration, or
amount per a selected volume, will be
less in the final solution, however.

We could use any volume unit as long
as we use the same unit on both sides
of the equation. This relationship also
applies when the concentration is
changed by evaporating some solvent.

108 CHAPTER 3: Chemical Equations and Reaction Stoichiometry

EXAMPLE 3-20 Dilution

How many milliliters of 18.0 MH 2 SO 4 are required to prepare 1.00 L of a 0.900 Msolution

of H 2 SO 4?

Plan

The volume (1.00 L) and molarity (0.900 M) of the final solution, as well as the molarity (18.0

M) of the original solution, are given. Therefore, the relationship V 1 M 1 V 2 M 2 can be used,

with subscript 1 for the initial acid solution and subscript 2 for the dilute solution. We solve

V 1 M 1 V 2 M 2 for V 1

Solution

V 1 0.0500 L 50.0 mL

1.00 L0.900 M



18.0 M

V 2 M 2



M 1

Dilution of a concentrated solution, especially of a strong acid or

base, frequently liberates a great deal of heat. This can vaporize

drops of water as they hit the concentrated solution and can cause dangerous spattering.

As a safety precaution, concentrated solutions of acids or bases are always poured slowly into

water,allowing the heat to be absorbed by the larger quantity of water. Calculations are

usually simpler to visualize, however, by assuming that the water is added to the con-

centrated solution.

CAUTION!