Chemistry Reference
In-Depth Information
The formula we use for these calculations is shown here:
V
a
M
a
= V
b
M
b
V
a
= the volume of the acid
M
a
= the molar concentration of the H
+
ions in the acid
V
b
= the volume of the base
M
b
= the molar concentration of the OH
-
ions in the base
As you can imagine, we can use this formula to solve for any of the four
possible variables, provided that we know the other three values. Let's try
a simple example to start.
Example 1
A student carries out an acid-base titration in the laboratory and
finds that it takes 4.00 × 10
2
ml of a standard 1.0 M solution of NaOH
to neutralize 1.00 × 10
2
ml of an HCl solution with an unknown con-
centration. What would be the concentration of the HCl solution?
Let's begin by listing the information that we know and the original
formula. Please don't be confused by the use of scientific notation in the
problem. It is simply used to indicate a certain number of significant digits.
Given: V
b
= 4.00 × 10
2
ml; M
b
= 1.0 M; V
a
= 1.00 × 10
2
ml
Find: M
a
Original Formula: V
a
M
a
= V
b
M
b
We isolate for the unknown, M
a
, by dividing both sides of our original
equation by V
a
. Then, all we need to do is substitute, solve, and round to
the correct number of significant digits.
Given: V
b
= 4.00 × 10
2
ml; M
b
= 1.0 M; V
a
= 1.00 × 10
2
ml
Find:
M
a
V
b
M
b
V
a
(4.00 × 10
2
ml)(1.0 M)
(1.00 × 10
2
ml)
M
a
=
=
= 4.0 M
As you can see, titration problems are just a matter of simple algebra.
As much as I hate to make things more complicated, just when they seemed
so easy, I have to add one small detail. Hydrochloric acid is a monoprotic
acid, meaning that it theoretically yields one mole of hydrogen ions for
every mole of acid. When dealing with HCl, one mole of acid generates
one mole of H
+
. Similarly, one mole of NaOH yields one mole of OH
-
.
