TABEL
PENGAMATAN
[H2C2O4] (kira-kira)
|
[H2C2O4] sebenarnya
|
[NaOH] standar
|
Volume H2C2O4
|
Volume NaOH
|
c
|
0,3
|
0,3
|
0,1
|
10 mL
|
55 mL
|
2,75 M
|
0,2
|
0,2
|
0,1
|
10 mL
|
4,4 mL
|
0,22 M
|
0,1
|
0,1
|
0,1
|
10 mL
|
21,4 mL
|
10,7 M
|
0,05
|
0,05
|
0,1
|
10 mL
|
5,4 mL
|
0,27 M
|
0,01
|
0,01
|
0,1
|
10 mL
|
2,5 mL
|
0,125 M
|
0,005
|
0,005
|
0,1
|
10 mL
|
1,3 mL
|
0,065 M
|
c = [H2C2O4] sesudah dicampur
dengan arang aktif
PERHITUNGAN
No
|
x
|
c
|
-log x
|
-log c
|
1/x
|
1/c
|
1
|
13,475.10-2
|
2,75 M
|
0,87
|
-0,44
|
7,42
|
0,36
|
2
|
8,8.10-5
|
0,22 M
|
4,05
|
0,65
|
11,363. 10-3
|
4,54
|
3
|
22,684.10-2
|
10,7 M
|
0,64
|
-1,03
|
4,408
|
0,09
|
4
|
11,88.10-4
|
0,27 M
|
2,92
|
0,56
|
8,41. 10-2
|
3,7
|
5
|
2,875.10-4
|
0,125 M
|
3,54
|
0,903
|
3,478. 10-3
|
8
|
6
|
0,6.10-4
|
0,065 M
|
4,22
|
1,19
|
16,67.10-3
|
15,3
|
a) Perhitungan untuk memperoleh nilai c
Persamaan reaksi:
Rumus : (M V) H2C2O4
= ½(M V) NaOH
1.
M1
. 10mL = ½ . 55mL . 0,1M
M1 = 2,75 M
2.
M1
. 10mL = ½ . 4,4mL . 0,1M
M1 = 0,22 M
3.
M1
. 10mL = ½ . 21,4mL . 0,1M
M1 = 10,7 M
4.
M1
. 10mL = ½ . 5,4mL . 0,1M
M1 = 0,27 M
5.
M1
. 10mL = ½ . 2,5mL . 0,1M
M1 = 0,125 M
6.
M1
. 10mL = ½ . 1,3mL . 0,1M
M1 = 0,065 M
b)
Perhitungan
untuk nilai x (mol zat terlarut yang terserap oleh adsorban)
Rumus : x = | [
H2C2O4 ] awal - [H2C2O4
] akhir ] |
x V NaOH
1.
x
= | [ 0,3-2,75] |M x 55.10-3Liter
=
13,475.10-2mol
2.
x
= | [ 0,2 – 0,22 ] |M x 4,4.10-3
Liter = 8,8.10-5mol
3.
x
= | [ 0,1 – 10,7 ] |M x 21,4.10-3
Liter = 22,684.10-2mol
4.
x =
| [ 0,05 – 0,27 ] |M x 5,4.10-3
Liter = 11,88.10-4mol
5.
x
= | [ 0,01 – 0,125 ] |M x 2,5.10-3
Liter = 2,875.10-4mol
6.
x =
| [ 0,005 – 0,065 ] |M x 1,3.10-3
Liter = 0,6.10-4mol
Berdasarkan persamaan empiris Isoterm Freundlich :
x/m = kc1/n
log x – log m = log k + 1/n log c
|
Keterangan :
x = [ H2C2O4
awal - H2C2O4 akhir
] x V NaOH
m = massa arang
aktif
k = tetapan
c = [ H2C2O4 ] sesudah dicampur arang aktif x V
naOH
n = slope/ kemiringan kurva
Berdasarkan rumus di atas dapat ditentukan nilai n dari kurva di atas,
k = 3,0035 ( intersep dari kurva)
m = 2,5 gram @ labu
erlenmeyer
log m = 0,39
maka,
1.
n = 0,26
2.
n = -0,13
3.
n = 0,68
4.
n = -0,14
5.
n = -0,204
6.
n = -0,23
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