Page 323 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:12 PM Page 306

306 Modern Analytical Chemistry

which can be solved for the grams of sulfanilamide
0
M b ´ V b ´ FW sulfanilamide (.1251 M)(0.04813 L)(168.18 g/mol)
=
2 2
0
= .5063 g sulfanilamide
Thus, the purity of the preparation is
g sulfanilamide 0 5063 g
.
´100 = ´100 =98 58.% w/w sulfanilamide
g sample 0.5136 g

This approach is easily extended to back titrations, as shown in the following
example.

EXAMPLE .4
9
The amount of protein in a sample of cheese is determined by a Kjeldahl
analysis for nitrogen. After digesting a 0.9814-g sample of cheese, the nitrogen
+
is oxidized to NH 4 , converted to NH 3 with NaOH, and distilled into a
collection flask containing 50.00 mL of 0.1047 M HCl. The excess HCl is then
back titrated with 0.1183 M NaOH, requiring 22.84 mL to reach the
bromothymol blue end point. Report the %w/w protein in the cheese given
that there is 6.38 g of protein for every gram of nitrogen in most dairy
products.
SOLUTION

In this procedure, the HCl reacts with two different bases; thus
Moles HCl = moles HCl reacting with NH 3 + moles HCl reacting with NaOH
Conservation of protons requires that

Moles HCl reacting with NH 3 = moles NH 3
Moles HCl reacting with NaOH = moles NaOH

A conservation of mass on nitrogen gives the following equation.
Moles NH 3 = moles N
Combining all four equations gives a final stoichiometric equation of

Moles HCl = moles N + moles NaOH
Making appropriate substitutions for the moles of HCl, N, and NaOH gives

g N
M a ´ V a = + M b ´
V b
AW N
which we solve for the grams of nitrogen.

g N=(M a ´V a – M b ´V b ) ´AW N

(0.1047 M ´0.05000 L – 0.1183 M ´0.02284 L) ´14.01 g/mol = 0.03549 g N