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The general formula utilized for
determining the normality of a solution is:
Normality X Equivalent Weight X
Volume = Grams
Equivalents/liter X Grams/equivalent X
Liters = Grams needed
The EQUIVALENT WEIGHT is the amount of
solute needed to be the equivalent of one mole of hydrogen ions. Therefore, the equivalent weight is dependent
on the valence of the solute. For
solutes with a valence of one (i.e. NaCl) the
molecular weight and equivalent weight are the same. When the valence of the
solute is more than one (i.e. H3PO4, valence = 3), then the equivalent weight
is equal to the molecular weight divided by the valence.
NOTE: This formula assumes that the
solute has a percent assay=100% and that it is a solid. Remember, when actually making the solution,
a small amount of solvent is used to dissolve the solute. Sufficient solvent is then added to bring the
final volume to the specified amount. In
other words, a sufficient quantity of solvent is added (q.s.)
to yield the final volume, after the solute is first dissolved into a volume of
solvent.
The general formula calculates how
many grams of a pure solid that are required to make a
specific volume of a specified normality.
In some situations it will be necessary to rearrange the formula.
To determine normality:
Normality = Grams/(Equivalent Weight X Volume)
Equivalents/liter = Grams/[(grams/equivalent) X Liters]
It is important to keep track of the
units used to make sure that the units on each side of the equals sign are the
same. In the formula above, on the right
side grams cancels grams and 1/(1/equivalent) is the
same as equivalent. Therefore, the right
side of the equation cancels to equivalents/liter which is the same as the left
side of the equation.
Liquids and impure solutes:
There are several ways to calculate
normality for liquids and impure solids.
The easiest way is to first determine how much of a pure solid would be
necessary to make a specified solution.
Then determine how much of the impure solid or liquid is required to
give the equivalent amount of pure solid.
Percent assay < 100%: After determining how much of the pure solute
is required, divide that number by the DECIMAL EQUIVALENT of the percent assay
(85% = 0.85). The amount of impure
solute required will ALWAYS be more than the amount of pure solute.
Liquids: Most liquids are not pure. Therefore when calculating the amount of
solute needed, two corrections are required.
It is necessary to determine how many grams of pure solute are present
in a milliliter of the solute. First
determine the density of the solute in grams/ml. This is given in the problem or in the real
world, written on the side of the bottle.
Multiply the density by the DECIMAL EQUIVALENT of the percent
assay. This yields the number of grams
of pure solute present in one ml of solute.
Divide this number into the number of grams of pure solute required to
determine the number of ml of solute that are necessary. Again, enough solvent is added to q.s. the solution to the required volume.
EXAMPLE: One hundred grams of a pure solute are needed
to create a specified solution. The
available solute is a liquid with a density of 1.2g/mland a percent assay =
70%.
1.2g/ml X 0.70 = 0.84 grams
pure solute/ml
100g/(0.84g/ml) = 119.05 ml
Therefore, 119.05 ml of the existing
solute are needed to yield 100 grams of pure solute.
To determine normality:
Multiply the volume used by the
density to find grams added
Grams added times decimal equivalent of
the percent assay = grams pure solute
Equivalents/liter = Grams/[(grams/equivalent) X Liters]
HYDRATES can be considered impure
solids. Divide the molecular weight of
the anhydrous solute by the formula weight of the hydrate to determine the
"percent assay". Finish the
calculation as outlined above. Remember
the molecular weight of water is 18g/mole.