Solution. The base solution contains 0.01 meq per
ml, since KOH yields only one OH^{} ion per molecule. The
83 mil of base solution contain
83 mil X 0.01 meq per ml = 0.83 meq of base
Since the molecular weight of acetic acid is 60.05 g per mole,
or 60.05 mg per mmole, the amount of acetic acid in the sample titrated
is
60.05 mg per mmole X 0.83 mmole = 44.8 mg of acetic acid
Example. In an analysis of industrial sulfuric acid, a 5.00ml
sample was diluted to one liter, then 20 mil of the diluted acid
were titrated with 0.10molar NaOH. What is the concentration of
the original acid, if 15.0 ml of NaOH solution are required to neutralize
the acid?
Solution. The number of milliequivalents of NaOH
used is
15.0 ml x 0.10 meq per ml = 1.50 meq of NaOH
This much NaOH will neutralize the same number of meq of H_{2}SO_{4},
1.50 meq; but since each mole of H_{2}SO_{4}, contributes
two moles of protons, 1.50 meq are obtained from only 0.75
mmole of sulfuric acid. The diluted sample had 0.75 mmole in 20
nil, or 0.0375 mmole per ml.
The original undiluted sample was more concentrated by a factor
of 5/1000, and hence had a sulfuric acid concentration of
0.0375 moles per liter x (1000/5) = 7.5 moles per liter
