Green Chemistry and the Ten Commandments

M = mass of solute (4.11.3)
(molar mass of solute) × (number of liters of solution)
A solution of known concentration that is added to a reaction mixture during the
procedure of titration is a standard solution. One of the most common of these is a
standard base solution of sodium hydroxide, NaOH. Typically, the concentration of
sodium hydroxide in such a standard solution is 0.100 mol/L. Suppose that it is desired
to make exactly 2 liters of a solution of 0.100 mol/L sodium hydroxide. What mass of
NaOH, molar mass 40.0 g/mol, is dissolved in this solution? To do this calculation, use
Equation 4.11.3 rearranged to solve for mass of solute:

Mass NaOH = M^ ×^ (molar mass NaOH)^ × (liters NaOH)^ (4.11.4)

Mass NaOH = 0.100 mol/L^ × 40.0 g/mol^ × 2.00 L = 8.00 g NaOH (4.11.5)

A common titration procedure is to use a standard solution of base to titrate an
unknown solution of acid or to use standard acid to determine base. As an example
consider a sample of water used to scrub exhaust gas from a hospital incinerator. The
water is acidic because of the presence of hydrochloric acid produced by the scrubbing
of HCl gas from the incinerator stack gas where the HCl was produced in the burning of
polyvinyl chloride in the incinerator. Suppose that a sample of 100 mL of the scrubber
water was taken and that the volume of a 0.125 mol/L standard NaOH consumed was
11.7 mL. What was the molar concentration of HCl in the stack gas scrubber water? To
solve this problem it is necessary to know that the reaction between NaOH and HCl is,

NaOH + HCl → NaCl + H 2 O (4.11.6)

a neutralization reaction in which water and a salt, NaCl are produced. Examination of
the reaction shows that 1 mole of HCl reacts for each mole of NaOH. Equation 4.11.1
applies to both the standard NaOH solution and the HCl solution being titrated leading
to the following equations:

MHCl =

molesHCl

and MNaOH =

molesNaOH

LitersHCl LitersNaOH

When exactly enough NaOH has been added to react with all the HCl present, the
reaction is complete with no excess of either HCl or NaOH. In a titration this end point
is normally shown by the change of color of a dye called an indicator dissolved in the
solution being titrated. At the end point molesHCl = molesNaOH and the two equations
above can be solved to give,

MHCl × LitersHCl =^ MNaOH × LitersNaOH (4.11.7)

which can be used to give the molar concentration of HCl:

100 Green Chemistry, 2nd ed