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 5.00-ml sample was diluted to one liter, then 20 mil of the diluted acid were titrated with 0.10-molar 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 H₂SO₄, 1.50 meq; but since each mole of H₂SO₄, 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