|When a dissolved solvent is added to a solution, the
solute is spread out through the large volume and the numebr of moes
per unit volume decreases. The solution is said to be
diluted. Such dilutions are a natural part
of chemistry in the lab. If two solutions of different solutes
moixed, the total volume increases and becomes occupied by both
solutes. As a result the concentration of both solutes is
less. At other times dilution is deliberate and essential.
Stock solutions of common chemicals are often prepared in large volumes
in preset concentrations. The chief
supply of HCl in the lab may be 1.00 M HCl. In your experiment,
may requir for a much less conencrtated solution, so the concentrated
must be diluted first.
When carrying out a dilution with an acid. Always add acid to water. Never add water to the acid.
When acid and water mix a great deal of heat is released. Acid
to water gives a large heat sink for the heat to dissipate in. If
add water to acid, the heat generated will cause the water to boil, and
splatter you with the hot acid solution.
|All the calculations that you need to do are based on a
simple fact. As a solution is diluted, the number of moles of
solute doesn't change; the
solute simply spreads out through a larger volume.
|We will use the following symbols: V for volume and M for
molarity. Therefore Vc and Mc are the
concentrated solutions volume
and molarity. Vd and Md are then the diluted
volume and molarity.
Vc Mc = Vd Md
|Problem: How can we prepare 100 of 0.040 K2Cr2O7
from 0.200 M K2Cr2O7?
|First assemble the data into a table:
Vc = ? Vd = 100 mL
Mc = 0.200 M Md = 0.040 M
Next, use the formula from above. Vc X
M = 100 mL X 0.040 M
Place 20.0 mL of 0.200 M K2Cr2O7
into a 100 mL volumetric flask and then add water until the final
|Practise Problem: Describe how to make 500 mL of
0.20 M NaOH from 0.50 M NaOH.
|How many millilitres of water would have to be added to
100 mL of 0.40 M HCl to give a solution with a concentration of 0.10 M?
|First assemble the
data. Vc =
100 mL Vd
Mc = 0.40 M Md = 0.10 M
Use the equation: 100 mL X 0.40 M = Vd
X 0.10 M
The final volume must be 400 mL. But since we started
with 100 mL we will add 300 mL of water until the volume reaches 400 mL.
|Practise Problem: How many millilitres of water would have
to be added
to 300 mL of 0.5 M NaOH to give a solution with a conentration of 0.2 M?