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1400-CH09 9/9/99 2:13 PM Page 349

Chapter 9 Titrimetric Methods of Analysis 349

that is solved for the grams of NaOCl.

M 2- ´ V 2 - ´ FW NaOCl (. M)(0.00896 L)(74.44 g /mol)
0 09892
SO 3 SO 3 =
2
2
2 2
= . g NaOCl
0 03299

Thus, the %w/v NaOCl in the diluted sample is
Grams NaOCl 0 03299 g
.
´100 = ´100 =0 132% w/v NaOCl
.
mL sample 25.00 mL
Since the bleach was diluted by a factor of 40 (25 mL to 1000 mL), the
concentration of NaOCl in the bleach is 5.28% (w/v).

EXAMPLE .15
9
The amount of ascorbic acid, C 6 H 8 O 6 , in orange juice was determined by
oxidizing the ascorbic acid to dehydroascorbic acid, C 6 H 6 O 6 , with a known
–
–
excess of I 3 , and back titrating the excess I 3 with Na 2 S 2 O 3 . A 5.00-mL sample
–
of filtered orange juice was treated with 50.00 mL of excess 0.01023 M I 3 . After
the oxidation was complete, 13.82 mL of 0.07203 M Na 2 S 2 O 3 was needed to
reach the starch indicator end point. Report the concentration of ascorbic acid
in milligrams per 100 mL.
SOLUTION

–
–
Oxidizing ascorbic acid requires two electrons, and reducing I 3 to I also
requires two electrons. Thus
–
(Moles I 3 ) ascorbic acid = moles C 6 H 8 O 6
–
2–
For the back titration, the stoichiometric relationship between I 3 and S 2 O 3 is
(see Example 9.14)
– 2–
(Moles I 3 ) back titration = 0.5 ´moles S 2 O 3
–
The total moles of I 3 used in the analysis is the sum of that reacting with
2–
ascorbic acid and S 2 O 3
– – –
(Moles I 3 ) tot = (moles I 3 ) ascorbic acid + (moles I 3 ) back titration
or
– 2–
Moles I 3 = moles C 6 H 8 O 6 + 0.5 ´moles S 2 O 3
– 2–
Making appropriate substitutions for the moles of I 3 , C 6 H 8 O 6 , and S 2 O 3
g C 6 H O 6
8
M
M - ´ V - = +05 ´ 2 - V´ 2 -
.
FW C 6 H O 6
2
I 3 I 3 SO 3 SO 3
2
8
and solving for the grams of C 6 H 8 O 6 gives
FW C
05
(M - ´ V – . ´ M - ´ V ) ´ 6 H O 6 =
8
I 3 I 3 - SO 3 2 SO 3 2 -
2
2
13
[( .01023 M )( .0500 L ) – ( . )( .07203 M )( .01382 L )](176 . g/mol )
0
0
0
0
0
5
g
= . 0 00243 CH 8 O 6
6