CK-12-Chemistry Intermediate

http://www.ck12.org Chapter 16. Solutions

FIGURE 16.8

Steps to follow in preparing a solution of

known molarity. (A) Weigh out the correct

mass of solute. (B) Dissolve the solute

into the desired solvent in a partially filled

volumetric flask. (C) Add more solvent

until the fill line on the flask is reached,

and then mix.

M 2 =

M 1 ×V 1

V 2

=

2 .0M× 100 .mL

500 .mL

= 0 .40 M HCl

The solution has been diluted by a factor of five, since the new volume is five times as great as the original volume.
Consequently, the molarity is one-fifth of its original value.

Another common dilution problem involves deciding how much of a highly concentrated solution is required to make
a desired quantity of solution with a lower concentration. The highly concentrated solution is typically referred to
as the stock solution.

Sample Problem 16.4: Dilution of a Stock Solution

Nitric acid (HNO 3 ) is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity
is 16 M. How much of the stock solution of nitric acid needs to be used to make 8.00 L of a 0.50 M solution?

Step 1: List the known quantities and plan the problem.

Known

• stock HNO 3 (M 1 ) = 16 M


• V 2 = 8.00 L


• M 2 = 0.50 M

Unknown

• volume of stock HNO 3 (V 1 ) =? L

The unknown in the equation is V 1 , the necessary volume of the concentrated stock solution.

Step 2: Solve.

V 1 =

M 2 ×V 2

27.13 M

=

0 .50 M× 8 .00 L

16 M

= 0 .25 L=250 mL

Step 3: Think about your result.

250 mL of the stock HNO 3 solution needs to be diluted with water to a final volume of 8.00 L. The dilution from 16
M to 0.5 M is a factor of 32.