VII. Calculations:
Determination
(1)Clorox (2)HughesDensity of commercial bleaching solution, g⋅mL–1 1.0041 1 .0945
(1)
.
density volumemass mL.
g
10 gmL10 041
== =1 0041 : -^1(2)density
.
volumemass
mLg
10 gmL10 945
== =1.0945 : -^1Volume of commercial bleaching solution diluted to 100 mL, mL 4
10 mL=_e 25 mLio 100 xmL mL mLmL x
10 100 x mLmL xmL25
25
= _i ==^1004Number of moles of S 2 O 32 −ion required for titration, mol(1)25 9. mL Na S O 2 2310001 LmL 0 00259. 0 0256. moles Na S OL of sol n^2 ’^2311 mol Na S Omol S O^22233J^2 -
LKJL` f KKN
PONPj p OO...
L
mol Na S O
mLLmL
101
10001
1(^223) ##25 9 =0 00259 25.9 mL = 0.00259 L 0.00259 0.00418
(2) 41 85. mL Na S O 2 2310001 LmL 0 004185. 0 04185. moles Na S OL of sol n^2 ’^2311 mol Na S Omol S O 2
23
J 2 32 -
L
K
^ c c KK
N
P
O
h m m OO
...
L
moles Na S O
mLLmL
101
10001
1(^223) ##41 85 =0 00418541.85 mL = 0.04185 L
Number of moles of I 2 produced in the titration mixture, mol
(1) ..moles S O ion
mol S O ion
0 00259 mol I
2
(^12) 0 001295
(^23)
(^23)
2 - 2 - =
J
L
KK N P j OO 0.001295 0.0020925 (2) ..moles S O ion moles S O mol I ion 0 004185 2 (^12) 0 0020925 (^23) (^23) 2 - 2 - = J L KK
N
P
j OO
Number of moles of OCl–ion in diluted bleaching solution titrated, mol
(1) 0 001295..moles I 2 1 mol OCl ion 1 mol I 2 =0 001295
- ` je o 0.001295 0.0020925
(2) 0 0020925..moles I 2 1 mol OCl ion 1 mol I 2 =0 0020925- ` je o
Mass of NaOCl present in diluted bleaching solution titrated, g
(1).
.
moles OCl mol OCl ion.
mol NaOCl
mol NaOClg NaOCl
0 001295 1 g
1
174 44
- =0 09305
J
L
- =0 09305
KJ
L` KN
PON
Pj O 0.09305 0.15577(2).
.
moles OCl mol OCl ion.
mol NaOCl
mol NaOClg NaOCl
0 0020925 1 g
1
174 44
` -jf - pe o=0 15577Percent NaOCl in commercial bleaching solution, %
(1).
.
g %. %g
10 0410 09305
f p_ 100 i= 09267 0.9267 1.423(2).
.
g %.%g
10 0410 15577
f p_ 100 i=1 423Laboratory Experiments