NEUTRALIZATION
When an acid and a base are mixed together the result is a salt and water
forming. If you add equal amounts of H+ (from the acid) and OH- (from the
base) you will also end up with a neutral solution. The general equation for
the reaction of an acid with a base is shown below.
Acid + Base ------------------------> Salt + Water
The water that forms is the result of the H+ combining with the OH-. The
salt that forms is the combination of the other ions that were attached to the
H+ and the OH-. A couple of examples are shown below.
HNO3 + NaOH ----------------------> H2O + NaNO3
H2SO4 + Ca(OH)2 --------------------> 2H2O + CaSO4
3HCl + Al(OH)3 -------------------> 3H2O + AlCl3
Notice that some acids and bases can provide more than one H+ or OH-.
When performing a neutralization reaction, you have to make the H+ = OHin order to achieve a neutral solution, so we have to take into consideration
the number of protons in the acid and the number of hydroxide ions in the
base. Different acids and bases provide different equivalents of H+ and OH.
equivalent - an equivalent is the amount of an acid or a base that yields one
mole of H+ (acids)
or one mole of OH- (bases).
An acid that has one proton, like HCl is said to be monoprotic. If you have
one mole of HCl it will yield one mole of H+.
Therefore, one mole of HCl has one equivalent of H+. An acid that has two
protons, like H2SO4, is said to be diprotic. If you have one mole of H2SO4 it
will yield two moles of H+. Therefore, one mole of H2SO4 can provide 2
equivalents of H+.
The same idea holds true for bases. If you have one mole of Ca(OH)2, it will
provide 2 equivalents of OH-,
since there are 2 OH- in the formula.
Sample Problem 1
How many equivalents are in 37.5 grams of HNO3?
Solution
Since HNO3 has one H+ in the formula, one mole of nitric acid will yield one
equivalent. We need to determine the number of moles in 37.5 grams of
HNO3 before we can determine the number of equivalents.
37.5 g HNO3 x 1 mole
---------- = .595 moles HNO3 x 1 equivalent
63 g
-------------- = .595
equivalents
1 mole
Sample Problem 2
How many equivalents are in 87.3 grams of Ca(OH)2?
87.5 g Ca(OH)2 x 1 mole
---------- = 1.18 moles Ca(OH)2 x 2 equivalents
74.1 g
---------------- =
2.36 equivalents
1 mole
We can also relate equivalents to the concentration of an acid or a base. The
concentration of solutions are normally in the units of mole/liter (M), but
with acids and bases the common unit used is normality. Normality takes
into consideration the number of H+ or OH- in the acid or base. The formula
for normality is shown below.
equivalents
normality (N) = ------------liter
The molarity (M) of an acid can be converted into normality by multiplying
the molarity times the number of H+ in the acid. See the examples below.
1.5M HCl x 1 H+ = 1.5N HCl
same normality as molarity)
(an acid with 1 H+ will have the
1.5M H2SO4 x 2 H+ = 3.0N H2SO4
1.5M H3PO4 x 3 H+ = 4.5N H3PO4
The normality shows that even though the molarity of each acid was the
same, the H3PO4 can neutralize more base because of its added H+. The
molarity of a base can also be converted into normality by multiplying the
molarity of the base times the number of OH- in the base. See the examples
below.
1.5M NaOH x 1 OH- = 1.5N NaOH (a base with 1 OH- will have the
same normality as molarity)
1.5M Ca(OH)2 x 2 OH- = 3.0N Ca(OH)2
1.5M Al(OH)3 x 3 OH- = 4.5N Al(OH)3
The concentration of an acid or base can be determined by using the idea of
neutralization and normality. When the volume of acid times the normality
of acid equals the volume of base times the normality of base, a neutral
solution will result. This can be shown with the following equation.
NA x VA = NB x VB
Sample Problem 1
How many mL of 3.50M HCl would be needed to neutralize 175 mL of
1.25M NaOH?
Solution
The acid and base in this problem have 1 H+ and 1 OH-, so their normality
will be equal to their molarity. The normality and volumes can be plugged
into the equation and the volume of HCl needed can be calculated.
3.50N HCl x VA = 1.25N NaOH x 175 mL
1.25N x 175 mL
--------------------- =
3.50N
62.5 mL
Sample Problem 2
What is the molarity of H2SO4 if 345 mL of this acid will neutralize 267 mL
of .375M NaOH?
Solution
Convert the molarity of base into normality.
.375M NaOH x 1 OH= .375N NaOH
Plug the numbers into the equation and solve for the normality of the H2SO4.
NA x 345 mL = .375N NaOH x 267 mL
.375N x 267 mL
NA = ---------------------- = .290N H2SO4
345 mL
Now convert the normality of the H2SO4 back into molarity by dividing by
the number of H+ in the acid.
.290N H2SO4
------------------ = .145M H2SO4
2 H+