Solution to Exercises for Experiment 19 10.00 mL of 2.00 x 10

10.00 mL of 2.00 x 10-3 M Fe(NO3)3 (Fe3+)
+
0.200 M HNO3 (H+)
4.00 mL of 2.00 x 10-3 M HSCN
Total Volume when
reaction starts (i.e.,
when all reagents are
mixed) is 20,00 mL.
6.00 mL H2O
[Fe3+]I : 10.00 mL • 2.00 x 10-3 M = 20.00 mL • C2 ⇒ C2 = 1.00 x 10-3 M
C2 = initial reactant
concentration
[HSCN]I : 4.00 mL • 2.00 x 10-3 M = 20.00 mL • C2 ⇒ C2 = 4.00 x 10-4 M
When reactants are
mixed their
concentration is
diluted.
[H+]I : 10.00 mL • 0.200 M = 20.00 mL • C2 ⇒ C2 = 0.100 M
1.
Fe3+
+
HSCN
⇌
FeSCN2+
+
H+
I (M)
1.00 x 10-3
4.00 x 10-4
0
0.100
C (M)
- 0.53 x 10-4
- 0.53 x 10-4
+ 0.53x 10-4
+ 0.53 x 10-4
E (M)
9.47 x 10-4
3.47 x 10-4
0.53 x 10-4
0.100
Common information used for
Questions 1 and 2
Solution to Exercises for Experiment 19
The change in
concentration (x)
[FeSCN 2 + ] [H + ]
[0.53 x 10 −4 ] [0.100]
Kc =
=
= 16.1
[Fe3+ ] [HSCN ]
[9.47 x 10 −4 ] [3.47 x 10 −4 ]
2.
I (M)
Fe3+
1.00 x 10-3
C (M)
- 0.265 x 10-4
E (M)
9.74 x 10-4
Kc =
+
2 HSCN
4.00 x 10-4
- 2(0.265 x 10-4)
3.47x 10-4
⇌
Fe(SCN)2+
0
+ 0.265 x 10-4
0.265 x 10-4
[Fe(SCN )+2 ] [H + ]2
[0.265 x 10 −4 ] [0.100]2
=
= 2.26 x 10 3
3+
2
−4
−4 2
[Fe ] [HSCN ]
[9.74 x 10 ] [3.47 x 10 ]
+
2 H+
0.100
+ 2(0.265 x 10-4)
0.100
The change in
concentration (x)