electroplating presentation
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IN ELECTROPLATING WASTE WATER
Luh Laksmi Dharayanti Satria 15311010
Milna Kurniawati 15311027
Budi Khairunisa Solekha 15311021
Kaysha Floren Devinadiar 15311035
THE REMOVAL OF
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ELECTROPLATING INDUSTRY
Disintegrated
Contributes massive quantity of
hexavalent chromium
to water body
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Protein synthesis
Insulin synthesis
Fatty acid formation
toxic
mutagenic
teratogenic
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Knowing the dosesof FeSO4 and Ca(OH)2 that needed to remove Cr6+
Comparing
the total need of FeSO4and Ca(OH)2 from stoichiometry andexperiment.
Knowing the total sludge produced after the coagulation-flocculation
process
Compare the total sludge produced by stoichiometric and by
eperimental work.
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METHODS
ION EXCHANGE
ADSORPTION
PRECIPITION
Cr ions exchanged by other ions in the resin
Cr ions adsorbed by adsorbent (i.e. activated carbon)
Cr ions reacted with Ca(OH)2and forming sediment
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PROCEDURE BY PRECIPITATING
Cr6+ waste water H2SO4until pH < 3
Strongly acidic Cr6+
waste waterFeSO4.7H2O
Cr3+ and Fe3+ in
waste waterCa(OH)2
1.
2.
3.
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[STOICHIOMETRY]
Suppose,Cr6+ in water will mostly be in the form of Cr2O7
2-
Cr2O72-+ 14H++ 6e-2Cr3++ 7H2O
Fe2+Fe3++ e------------------------------------------------------------------------------------------------
Cr2O72-+ 6Fe2++ 14H+2Cr3++ 6Fe3++ 7H2O
1 mol Cr
2
O
7
2-
needs 6 mol Fe2+
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1 mol Cr
2
O
7
2- needs6 mol 6Fe2+
How much Fe will be needed for 1 mol of Cr6+
?
Fraction Cr6+ =
CrO
= 5
16=
1
16= 0.4815 mol
1 mol Cr3+needs =6
.815= 12.5 mol Fe2+.
So,
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1 mol Cr3+needs12.5 mol Fe
2+
.
We have,
100 ppm = 100 mg/L FeSO4 solution
1
5= 1.92 mmol Cr3+
For 1 L of waste,
there will be : And it needs :
1.92 x 12.5 = 24.04 mmol Fe2+
24.04 56
1 /= 13.46 ml FeSO4.7H2O
So,
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In short,
We need 13.46 ml FeSO4.7H2Oto treat 1 L waste
water
For reducing Cr6+into Cr3+
Precipitated by adding Ca(OH)2
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Cr3+ + 3OH- Cr(OH)31.92 mmol 5.76 mmol 1.92 mmol
Fe3+ + 3OH- Fe(OH)324.04 mmol 72.12 mmol 24.04 mmol +
---------------------------------------------------------------------------------
77.88 mmol
77.88 mmol x7
1= 53.9312 ml Ca(OH)
2
We have,
100 ppm = 100 mg/L FeSO4 solution
So, lime needed will be :
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From over all stoichiometry :
1 L Cr6+ waste water (100 ppm)
Needs 13.46 ml FeSO4.7H2O
For reducing Cr6+into Cr3+
Needs 53.9312 ml Ca(OH)2
For precipitating Cr3+
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[THEORETICAL PRACTICE]
Wisnjuprapto, 2013
1 L Cr6+ waste water (100 ppm)
Needs 16.03 ml FeSO4.7H2O
For reducing Cr6+into Cr3+
Needs 9.48 ml Ca(OH)2
For precipitating Cr3+
[THEORETICAL PRACTICE]
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Variation Dose ( )
For Volume 1L For Volume 500 ml
FeSO
4
.7H
2
O Ca(OH)
2
FeSO
4
.7H
2
O Ca(OH)
2
1 50 6.73 26.9656 3.37 13.48
2 75 10.095 40.4484 5.05 20.22
3 100 13.46 53.9312 6.73 26.97
4 125 16.825 67.414 8.41 33.71
5 150 20.19 80.8968 10.10 40.45
6 250 33.65 134.828 16.83 67.41
[STOICHIOMETRY]
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[THEORETICAL PRACTICE]
Variatiom Dose ( )
For Volume 1L For Volume 500 ml
FeSO
4
.7H
2
O Ca(OH)
2
FeSO
4
.7H
2
O Ca(OH)
2
1 50 8.015 4.74 4.01 2.37
2 75 12.0225 7.11 6.01 3.56
3 100 16.03 9.48 8.02 4.74
4 125 20.0375 11.85 10.02 5.93
5 150 24.045 14.22 12.02 7.11
6 250 40.075 23.7 20.04 11.85
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initial temperature
initial pH of wastewater
final pH of wastewater
6.95
25.3oC
2.97
final temperature 25.3oC
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[STOICHIOMETRY]
Variation
Dose Sediment
FeSO
4
.7H
2
O Ca(OH)
2
Empty Filter
Weight (gr)
Filter +
Sediment (gr)
Total Sediment
(gr)
1 3.365 13.4828 1.11972 1.3245 0.20478
2
5.0475 20.2242 2.32126 2.4261 0.10484
3 6.73 26.9656 1.1859 1.4328 0.2469
4 8.4125 33.707 1.169 1.4422 0.2732
5 10.095 40.4484 1.1547 1.5028 0.3481
6 16.825 67.414 1.1432 1.6523 0.5091
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[THEORETICAL PRACTICE]
Variation
Dose Total Sludge Produced
FeSO
4
.7H
2
O Ca(OH)
2
Empty Filter
Weight (gr)
Filter +
Sediment
(gr)
Total Sediment
(gr)
1 4.0075 2.37 1.1287 1.2675 0.1388
2 6.01125 3.555 1.1294 1.3086 0.1792
3 8.015 4.74 1.1865 1.4177 0.2312
4 10.01875 5.925 1.1419 1.3829 0.241
5 12.0225 7.11 1.1185 1.376 0.2575
6 20.0375 11.85 2.2772 3.1233 0.8461
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0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 1 2 3 4 5 6 7
S
m
e
g
Variation
Comparation Graph
of Sediment (gr) VS Variation in Both Versions
B
A
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[STOICHIOMETRY]
1
2 3
4 5 6
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1
2 3
4 5 6
[THEORETICAL PRACTICE]
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6 6
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Cr3+ and Fe3+ in
waste water
3.
Na(OH)2
OH-will form ionic bond with Cr3+
and Fe3+
Fe(OH)3 and Ca(OH)3
These new complexes collide each
other, forming bigger complex,
weight more, forming bigger flock
Precipitated
Negative pressure vacuum used to
separate sediment and liquid part
Weight in to examine total sediment
performed
AAS is needed for a further a do
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Theoretically,
In the need of 16.03 ml FeSO4.7H2Oand9.48 ml Ca(OH)2
Will produce :
0,38 ppm Fe(OH)3dari setiap 1ppm FeSO4.7H2O yang digunakan
1,84 ppm CaSO4dari setiap 1ppm Ca(OH)2yang digunakan
1,98 ppm Cr(OH)3dari setiap kandungan 1 ppm Cr+6
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0
0.1
0.2
0.3
0.4
0.5
0.6
0 50 100 150 200 250 300
S
m
e
g
Dose ( )
Comparation Graph
of Sediment (gr) VS Variation
A
theory
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0
0.1
0.2
0.3
0.4
0.50.6
0.7
0.8
0.9
0 1 2 3 4 5 6 7
S
m
e
g
Variation
B
theory
Comparation Graph
of Sediment (gr) VS Variation
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0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 1 2 3 4 5 6 7
S
m
e
g
Variation
Comparation Graph
of Sediment (gr) VS Variation
B
A
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For the theoritical one, the total doses of FeSO4and Ca(OH)2are: 20.0375 ml
and 11.85 ml (version B). For the stoichiometic one, the total doses of FeSO4and
Ca(OH)2are 16.825 ml and 67.414 (version A)1
The theoritical one needs less coagulant and flocculant than the stoichiometric
one
For the theoritical one, the total sludge produced is 0.84610 gr (version B). For
the stoichiometic one, the total sludge produced is 0.5091 gr (version A)
The theoritical one produced more total sludge than the theoritical one4
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[Bellaria Ekaputri]
Kalau kurang waktu untukjar testdan filtrasi dalam 1 hari, kenapa tidak
dikerjain sampai sekarang-sekarang, sampai h-beberapa sebelum presentasi
aja?
1
Tidak bisa, soalnya semakin lama hasil jartestnya dibiarkan, semakin baguskarakteristik endapannya, perlakuan satu variabel dan yang lainnya jadi
berbeda, idealnya pengukuran endapan harus dilakukan di satu waktu.
[Kevin Barlian]Maaf saya kurang memperhatikan, sebenarnya variabel percobaannya apa?
2
Variabel dosis koagulan, mulai dari 50%, 75%, 100%, 125%, 150%, dan 250% dosis
optimal koagulan menurut perhitungan dan teori.
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[Kakak S2]
Lain kali, analisis penyebab kesalahan praktikum tidak usah diisampaikan, itu
cukup sebagai analisis untuk kalian sendiri saja3
Oke kak (y)
[Piyarat Kittiwat]
Thank you for the English presentation!4 Yourewelcome Palm :)
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