Scenario:A student diluted 50.00 mL of a commercial bleach to 250.00 mL in a volumetric
flask and then titrated a 20. mL aliquot. The titration required 35.50 mL of 0.10000 M Na 2 S 2 O 3
solution. The price of the gallon jug of bleach was $1.00. The density of the bleach was
1.15 g ⋅mL–1.
Analysis:
- Calculate the number of moles of S 2 O 3 2–ion required for titration.
.'
'
'
'
.
.
mL Na S O sol n
mL Na S O sol n
L sol n
L sol n
mol Na S O
mol Na S O
mol S O
mol S O
1
35 50
1000
1
1
0 10000
1
1
3 550 10
22 3
22 3
22 3
22 3
2 3 3
(^23)
=
- -- 2
2
- Calculate the number of moles of I 2 produced in the titration mixture.
.
. mol S O
mol S O
mol I 1 775 10 mol I
1
3 550 10
2
1 3
2
(^32)
3
(^23)
## (^2) = -
--
2
2
- Calculate the number of moles of OCl–ion present in the diluted bleaching solution that
was titrated.
.
. mol I
mol I
mol OCl 1 775 10 mol OCl
1
1 775 10
1
(^21)
2
3
#^3
##=
- Calculate the mass of NaOCl present in the diluted bleaching solution titrated.
. mol OCl.
mol OCl
mol NaOCl
mol NaOCl
g NaOCl
1
1 775 10
1
1
1
74 44
#^3
##
= 0.1321 g NaOCl
- Determine the volume of commercial bleach present in the diluted bleaching solution
titrated (aliquot).
.
.
mL.
mL
mL commercial bleach
1
20 00
250 00
diluted bleach sol’n 50 00
# diluted bleach sol’n
= 4.00 mL commercial bleach
- Calculate the mass of commercial bleach titrated.
.
.
mL commercial bleach.
mL commercial bleach
g commerical bleach
1
400
100
115
# = 4.60 g commercial bleach
- Determine the percent NaOCl in the commercial bleach.
%.
.
whole %.%
part
g commercial bleach
g NaOCl
(^100460)
0 1321
##== 100 2 87
- Calculate the mass of 1.00 gallon of the commercial bleach. 1 U.S. gallon = 3785 mL.
..
..
gal comm bleach ..
gal
mL
mL comm bleach
g comm bleach
1
100
1
3785
100
115
= ## = 4.35 × 103 g
Part III: AP Chemistry Laboratory Experiments