Electroplating Presentation

8/10/2019 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


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


Needs 9.48 ml Ca(OH)2

For precipitating Cr3+



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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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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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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



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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



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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


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 :)

Documents

Electroplating Presentation