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151Chapter 9 Titrimetric Methods
.6 6 23
0.01052 L L
0.00987 mol K Cr O
mol K Cr O
mol Fe 10 mol Fe4
2 2 7
2 2 7
2
2
# #
#=
+
- +
 which means the original sample contained
. .6 23 3 1210 mol Fe
2 mol Fe
1 mol U 10 mol U4 42
2
4
4
# # #=
- +
+
+
- +
 he concentration of uranium in the original sample, therefore is
.
.
3 12
23 6
0.315 g sample
10 mol U
mol U
238.08 g U
100 %w/w U
4 4
4
4
4
# #
# =
- +
+
+
+
52. In this back-titration, iron, in the form of Fe2+, reacts with the ana-
lyte, chromium, and then with the titrant, K2Cr2O7. In its reaction 
with the analyte, iron loses one electron as it is oxidized from Fe2+ to 
Fe3+ and each chromium gains three electrons as it is reduced from 
Cr O2 7
2- to Cr3+; thus, six moles of Fe2+ reacts with each mole of 
Cr O2 7
2- . In its reaction with the titrant, iron loses one electron as it is 
oxidized from Fe2+ to Fe3+ and each chromium gains three electrons 
as it is reduced from Cr O2 7
2- to Cr3+; thus, the stoichiometry of the 
titration reaction requires that each mole of Fe2+ reacts with six moles 
of Cr O2 7
2- . he total moles of Fe2+ added to the original sample is
92
28
0.500 g Fe(NH ) (SO ) 6H O
3 .12 g Fe(NH ) (SO ) 6H O
1 mol Fe 1. 10 mol Fe
4 2 4 2 2
4 2 4 2 2
2
3 2
: #
:
#=
+
- +
 Of this iron, the moles that react with the titrant, K2Cr2O7 are
.6
829
389
4 27
0.01 L L
0.00 mol K Cr O
mol K Cr O
mol Fe 10 mol Fe4
2 2 7
2 2 7
2
2
# #
#=
+
- +
 which leaves 1.28×10–3 – 4.27×10–4 = 8.53×10–4 moles of Fe2+ 
to react with chromium in the original sample; thus, the mass of 
chromium in the original sample is
.8 53
49
10 mol Fe
6 mol Fe
1 mol Cr O
mol Cr O
2 mol Cr
mol Cr
51.996 g Cr
0.01 g Cr
4 2
2
2 7
2
2 7
2
# #
# # =
- +
+
-
-
 and the thickness of chromium is
thickness area
volume
30.0 cm
0.0149 g 7.20 g
1
mL
1 cm
6.90 10 cm
mL
2
3
5
# #
#
= =
=
-