lampiran isoterm adsorpsi
DESCRIPTION
lampiran perhitunganTRANSCRIPT
LAMPIRANA. CONTOH PERHITUNGAN1. Konsentrasi asam asetat setelah adsorpsi
Sampel 1
CH3COOH=NaOHn M1 V1=n M2 V21 . M1 . 25=1 . 0,1 . 10,8M1=0,0432 M
Sampel 2
CH3COOH=NaOHn M1 V1=n M2 V21. M1 . 25=1 . 0,1 . 5,1M1=0,0204 M
Sampel 3
CH3COOH=NaOHn M1 V1=n M2 V21. M1 . 25=1 . 0,1 . 2M1=8 x 10-3 M
Sampel 4
CH3COOH=NaOHn M1 V1=n M2 V21. M1 . 25=1 . 0,1 . 0,9M1=3,6 x 10-3 M
Sampel 5
CH3COOH=NaOHn M1 V1=n M2 V21. M1 . 25=1 . 0,1 . 0,2M1=1,2 x 10-3 M
2. Konsentrasi asam asetat yang teradsorpsi (C) C awal C akhir
Sampel 1
C= C awal C akhir= 0,05 0,0432= 6,8 x 10-3 M
Sampel 2
C= C awal C akhir= 0,025 0,0204= 4,6 x 10-3 M
Sampel 3
C= C awal C akhir= 0,0125 8 x 10-3= 4,5 x 10-3 M
Sampel 4
C= C awal C akhir= 0,00625 3,6 x 10-3= 2,65 x 10-3 M
Sampel 5
C= C awal C akhir= 0,003 1,2 x 10-3= 1,8 x 10-3 M
3. Massa Zat yang Teradsorpsi (X)
Sampel 1
X1= C x BM CH3COOH x V filtrat/ 1000= 6,8 x 10 -3 x 60 x 96/1000= 0,039 g
Sampel 2
X2= C x BM CH3COOH x V filtrat/ 1000= 4,8 x 10 -3 x 60 x 96/1000= 0,0264 g Sampel 3
X3= C x BM CH3COOH x V filtrat/ 1000= 4,5 x 10 -3 x 60 x 94,5/1000= 0,0255 g
Sampel 4
X4= C x BM CH3COOH x V filtrat/ 1000= 2,65 x 10 -3 x 60 x 95/1000= 0,015 g
Sampel 5
X5= C x BM CH3COOH x V filtrat/ 1000= 1,8 x 10 -3 x 60 x 96,5/1000= 0,0104 g
4. Menghitung X/m
Sampel 10,039 g/1 g= 0,039
Sampel 20,0264 g/1 g = 0,0264
Sampel 30,0255 g/1 g = 0,0255
Sampel 40,015 g/1 g = 0,015
Sampel 50,0104 g/1 g = 0,0104
5. Menghitung log X/m
Sampel 1Log 0,039 = -1,408
Sampel 2Log 0,0264 = -1,5784
Sampel 3Log 0,0255 = -1,59346
Sampel 4Log 0,015 = -1,82391
Sampel 5Log 0,0104 = -1,98297
6. Menghitung log C
Sampel 1Log C = log 6,8 x 10 -3 = -2,16749 Sampel 2log 4,6 x 10-3 = -2,33724
Sampel 3Log 4,5 x 10-3 = -2,34679
Sampel 4Log 2,65 x 10-3 = -2,57675
Sampel 5Log 1,8 x 10-3 = -2,74473
x/mlog x/mclog c
0,039-1,408946,80E-03-2,16749
0,0264-1,57844,60E-03-2,33724
0,0255-1,593464,50E-03-2,34679
0,015-1,823912,65E-03-2,57675
0,0104-1,982971,80E-03-2,74473
Log X/m = log k + 1/n log Cy = 0,997 x + 0,749log k = 0,749k= 5,611/n = 0,977 n = 1,00