superpro design v7

PHYSICAL AND CHEMICAL PROCESSES OF WATER POLLUTION CONTROL

PRECIPITATION


PRECIPITATION

PRECIPITATION

Practiced in water and wastewater treatment for

- Phosphorus removal

AlPO4(s), FePO4(s), Ca5(OH)(PO4)3(s)

- Hardness removal, “softening”

CaCO3(s), Mg(OH)2(s)

- Iron and Manganese removal

Fe(OH)3(s), MnO2(s)

- Removal of heavy metals, radionucleus

As(V), Ba, Cd, Hg, Zn, Ni, Cr(III)

M+n

+ nOH- ⇋ M(OH)n(s)

M+2

+ S-2

⇋ MS(s)

- Water conditioning (to control scale formation or corrosion)


PRECIPITATION

Removal of Hardness by Precipitation

Harness = ∑ divalent cations

= Ca2+

, Mg2+

, Fe2+

, Mn2+

, Sr2+

...

If it is too high, precipitation of soap, scaling on pipes, boilers,

cooling towers, heat exchangers

If it is too low, water is corrosive

Major ions


PRECIPITATION

Forms of Hardness

Ca-hardness due to Ca

Mg-hardness due to Mg

Carbonate harness – associated with CO3-2

and HCO3- alkalinity

Nancarbonate Hardness – associated with other anions

Lime-Soda Softening Process

Lime: Ca(OH)2, CaO → Ca++

+ OH-

Soda: Na2CO3 → Na+ + CO3

-2

Rxns:

Ca++

+ CO3-2

↔ CaCO3(s) Ksp = [Ca++

] [CO3-2

] = 5.2×10-9

at 20˚C

Mg++

+ 2OH- ↔ Mg(OH)2(s) Ksp = [Mg

++] [OH

-]2 = 1.8×10

-11 at 20˚C

TH (Total Hardness)

TH (Total Hardness)


PRECIPITATION

Strategy: Raise the pH by limen (HCO3- + OH

- → CO3

-2 + H2O)

So, CaCO3 will be precipitate, Mg(OH)2 will also precipitate

For precipitation of CaCO3, sufficient carbonate is needed. If iti is not enough, add

Na2CO3 as a source of CO3-2

(Na2CO3 is three times expensive than lime).

Do not rely on precipitation of Ca(OH)2, Ksp = 5.5×10-6

or MgCO3, Ksp = 5.6×10-5

Rxns with Lime:

1) H2CO3* + Ca(OH)2 → CaCO3(s) + 2H2O

Ca(OH)2 → Ca++

+ 2OH-

H2CO3* + OH

- → HCO3

- + H2O

HCO3- + OH

- → CO3

-2 + H2O

Ca++

+ CO3-2

↔ CaCO3(s) (pH > 6.3)


PRECIPITATION

2) Ca(OH)2 → Ca++

+ 2OH-

2HCO3- + 2OH

- → 2CO3

-2 + 2H2O

Ca++

+ (Ca++

)orig. + 2CO3-2

→ 2CaCO3(s)

(Ca++

)orig. + Ca(OH)2 + 2HCO3- → 2CaCO3 + 2H2O

1 mole 1 mole 2 eq.

3) Ca(OH)2 → Ca++

+ 2OH-

Mg++

+ 2OH- ↔ Mg(OH)2(s)

Ca++

+ HCO3- + Ca(OH)2 → 2CaCO3 + 2H2O

Mg++

+ 2Ca(OH)2 + 2HCO3- → Mg(OH)2 + 2CaCO3 + 2H2O

1 mole 2 mole 2 eq


PRECIPITATION

4) If there is not enough CO3, add Na2CO3

Na2CO3 → 2Na+ + CO3

-2

Ca+2

+ CO3-2

→ CaCO3

Ca+2

+ Na2CO3 → CaCO3 + 2Na+

1 mole 1 mole

5) Ca(OH)2 → Ca++

+ 2OH-

Mg++

+ 2OH- ↔ Mg(OH)2(s)

Ca++

+ Na2CO3 → CaCO3(s) + 2Na+

Mg++

+ Ca(OH)2 + Na2CO3 → Mg(OH)2 + CaCO3 + 2Na+

1 mole 1 mole 1 mole


REVIEW

CT, Alkalinity, pH Relationship

CT (mole/L) = Total Inorganic Carbon = [H2CO3*] + [HCO3

-] + [CO3

-2]

Alkalinity (eq/L) = [HCO3-] + 2[CO3

-2] + [OH

-] - [H

+]

1

2 3 1 1 20 2

T

1

3 21

T 1

12

32

T 2 1 2

[H CO ]1

C [H ] [H ]

[HCO ] [H ]1

C [H ]

[CO ] [H ] [H ]1

C

*K K K

K

K

K K K

where

3

1

2 3

H HCOK

H CO

and

2

3

2

3

H COK

HCO

Alkalinity = (α1 +2α2) CT - [H+] + [OH

-]

PRECIPITATION


COAGULATION / FLOCCULATION

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

pH

pC

[H2CO3] [HCO3-] [CO3

2-]

[H+] [OH-]

CT= 10-3M


PRECIPITATION

pC-pH Diagrams

a) Mg(OH)2 Mg++

+ 2OH- Ksp = 10

-10.7

2

w

sp][H

K]Mg[K

log [Mg++

] + 2logKw – 2log[H+] = -10.7 log[Mg

++] = 17.3 – 2pH Line A

-28

b) Ca(OH)2 Ca++

+ 2OH- Ksp = 10

-5.3 log[Ca

++] = 22.7 – 2pH Line B

c) CaCO3 Ca++

+ CO3-2

Ksp = 10-8.3

When pH > pKA2, CO3-2

is predominant.

Ksp = [Ca++

] [CO3-2

] and [Ca++

] = [CO3-2

]

[Ca++

] = (Ksp)1/2

= 10-4.15

= 7.1×10-5

mole/L = 3 mg/L

Lowest possible Ca++

concentration achivable after softening with lime.


PRECIPITATION

When pKA1 < pH < pKA2

2/1

A2

spK

][HKCa

log[Ca++

] = 2

1log

A2

sp

K

K-

2

1pH

When pH < pKA1

log[Ca++

] =2

1log

A2A1

sp

KK

K- pH Line C


PRECIPITATION

Process Variations

1) Straight line addition

- No Mg removal if Mg-H < 40 mg/L as CaCO3

- Enough alkalinity for removal of Ca++

2) Lime-soda ash treatment

- Mg-H < 40 mg/L as CaCO3

- Not eough alkalinity for removal of Ca++

4.3 6.3 10.3 pH

log C

CaCO3

4.15 or [Ca++

] = 7×10-5

M about pH 10.3


PRECIPITATION

3) Excess lime-soda-ash treatment

- Mg-H is high requires excess lime to increase pH to 12-13 so that reaction

kinetics is fast (supersaturate solution) (Excess ≈ 1.25 meq/L)

- Usually not enough alkalinity, add Na2CO3

.

8.7 12 pH

Mg(OH)2

(Mg)ini.

(Mg)fin.

log C


PRECIPITATION

Recarbonation:

Addition of CO2 into water to lower the pH and also remove additional CaCO3 by

increasing CT. The use of strong acid for lowering the pH is called neutralization.

Two-stage recarbonation:

1) Add CO2, increase CT, ppt. More CaCO3

2) Lower the pH to desired level

1) Add CO2 → H2CO3*, titrate excess OH

-

H2CO3* + OH

- → HCO3

- + H2O

HCO3- + OH

- → CO3

-2 + H2O

Ca++

+ CO3-2

→ CaCO3

pH ≈ 10-10.5 and [Ca++

] at min., all CT is in CO3

2) H2CO3* + CO3

-2 → 2 HCO3

-

Desired pH determines CO2 requirement.


PRECIPITATION

Add Ca(OH)2 and Na2CO3 separately, otherwise

Ca(OH)2 + Na2CO3 → CaCO3 + 2 NaOH

Floccula.

Recarbo. Recarbo. Filter

Ca(OH) 2 Na2CO3

∞ ∞

CO2 CO2

Mg(OH) 2

CaCO3

ppt.

pH = 11-12

Floccula.

pH ≈ 8

pH = 10-10.5


PRECIPITATION

Split Treatment:

Advantages: - less chemical required

- recarbonation may be eliminated due to lower pH of bypass

Limitations: - less removal of hardness

- applied only to clean waters which are usable w/o

coagulation (by-pass portion)

Floccula.

Lime

∞

sludge

by-pass

may be followed by

Na2CO3 treatment


PRECIPITATION

Example:

Con. (mg/L) Mwt mmol/L meq/L

H2CO3* 16 62.0 0.25 0.50

Ca++

100 40.0 2.50 5.00

Mg++

40 24.3 1.65 3.30

HCO3-

183 61.0 3.00 3.00

CO3-2

0 60.0 0.00 0.00

pH = ?

*]COH[

]HCO][H[10K

32

-

33.6

A1

M1017.4)10(3.00

)10(0.2510]H[ 8

3-

-33.6

pH = 7.4

Ca-H = 5.0 meq/L

Mg-H = 3.3 meq/L

TH = 8.3 mg/L


PRECIPITATION

Alk. = [HCO3-] + 2[CO3

-2] + [OH

-] – [H

+]

Alk. ≈ [HCO3-] = 3.0 meq/L

Carbonate Hardness = 3.0 meq/L

Non-carbonate Hardness = 8.3 – 3.0 = 5.3 meq/L (associated with SO4-2

, Cl-, NO3

- etc.)

Apply excess lime-soda ash softening:

Chemical Requirement:

For Rxn 1: H2CO3* + Ca(OH)2 →

0.5 meq/L 0.5 meq/L

Ca(OH)2 req’d. = 0.25 mmole/L

For Rxn 2: (Ca++

)orig. + Ca(OH)2 + 2HCO3- →

3.0 meq/L of Ca hardness is carbonate hardness

Ca(OH)2 req’d. = 2

0.3= 1.5 mmole/L


PRECIPITATION

For Rxn 3: Mg++

+ 2Ca(OH)2 + 2HCO3- →

No Mg hardness associated with carbonate

For Rxn 4: Ca++

+ Na2CO3 →

Non-carbonate Hardness = 5.0 – 3.0 = 2.0 meq/L

Na2CO3 req’d. = 1 mmole/L

For Rxn 5: Mg++

+ Ca(OH)2 + Na2CO3 →

Mg hardness is all non-carbonate hardness = 3.3 meq/L

= 1.65 mmole/L

Ca(OH)2 req’d. = 1.65 mmole/L

Na2CO3 req’d. = 1.65 mmole/L

Excess lime ≈ 1.25 meq/L = 0.625 mmole/L

∑ Lime = 0.25 + 1.5 + 1.65 + 0.625 = 4.025 mmole/L = 298 mg/L

∑ Na2CO3 = 1.0 + 1.65 = 2.65 mmole/L = 281 mg/L

Sludge produced?


PRECIPITATION

Scale Formation and Protection Agents

= Corrosion =

CaCO3 Ca+2

+ CO3-2

Ksp = 10-8.3

CO3-2

+ H+ HCO3

- KA = 10

10.3

CaCO3 + H+ Ca

+2 + HCO3

- Keq = 100

Example:

pH = 7.3

Ca++

= 215 mg/L = 5.5×10-3

M

Alk. = 500 mg/L as CaCO3 = 0.01 eq/L = [HCO3-] = 0.01 M

91.0C

]HCO[

T

-

31

CT = 0.011 M


PRECIPITATION

1) 107710

(0.01))10(5.4

]H[

]HCO][Ca[3.7

-3-

3

Q 1100

1077

K eq

Q

; oversaturation

2) Lingelier’s Saturation Index:

Emprical relation: SI = pH – pHs

Actaul pH Equlibrium pH

]H[

]HCO][Ca[ K

-

3

eq

eq

-

3

K

]HCO][Ca[ ]H[

7--3

S 105.4100

(0.01))10(5.4]H[

pHs = 6.3

SI = 7.3 – 6.3 = 1.0 ; oversaturation

when SI = 0 – equilibrium

when SI < 0 – undersaturation

* Needs correction for T, SO4-2

, NH4+ etc.


PRECIPITATION

3) How much CaCO3 will ppt.?

Quantity and kinetics

Stability index (Ryznar) = 2pHs – pH

St. Index < 6 scaling

7 > St. Index > 6 no difficulties

St. Index > 7 corrosive

St. Index = 2×(6.3) – 7.3 = 5.3 (scaling tendency)

(See Figure 13)

Water A (75ºC) Water B (75ºC)

pH = 6.5 pH = 10.5

pHs = 6.0 pHs = 10.0

SI = + 0.5 SI = + 0.5

St. Index = 5.5 (scale forming) St. Index = 9.5 (corrosive)


PRECIPITATION

To eliminate oversaturation:

1) Acidification:

Original Conditions:

[Ca++

] = 5.4×10-3

M

[HCO3-] = 0.01 M pH = 7.3

From *]COH[

]HCO][H[10K

32

-

33.6

A1

, *]COH[ 32 = 1×10-3

M

Assume 35% neutralization by H2SO4

[HCO3-] = 0.01 – 0.35 × (0.01) = 6.5×10

-3 M

*]COH[ 32 = 0.35 × (0.01) + 1×10-3

= 4.5×10-3

M

From (1) [H+] = 3.5×10

-7 ; pH = 6.5

]H[

]HCO][Ca[ K

-

3

eq

= 100

From (2) [HCO3-] = 6.4×10

-3 M

H2SO4 req’d. = 0.35 (0.01) 2

1×98×10

3 = 170 mg/L


PRECIPITATION

2) Softening by Ca(OH)2:

Ca++

+ Ca(OH)2 + 2HCO3- → 2CaCO3 + 2H2O

Bring pH up to about 10.3.

[Ca++

]0 = 5.4×10-3

M

[HCO3-]0 = 0.01

1 mole Ca isremoval by 2 eq HCO3-

X 0.01 eq

X = 5×10-3

M Ca will be removed

Ca(OH)2 required = 5×10-3

M = 280 mg/L as CaCO3 [Ca++

]f = 4×10-4

M

[CO3-2

]f = 4104

spK= 1.3×10

-5 M

Check: 5-

2

-10.3-5-

3 103.1)(10)10(1.34

]HCO[

AK

M CT = 2.6×10-5

M

Bring pH down to about 7 by H2SO4

[Ca++

]f = 4×10-4

M 93 % removal

[HCO3-]f = 2.6×10

-5 M 99.7 % removal


PRECIPITATION

PHOSPHORUS REMOVAL

The Types of P

1 – Orthophosphate: H3PO4, H2PO4-, HPO4

-2, PO4

-3

pKA1 = 2.1, pKA2 = 7.2, pKA3 = 12.3

2 – Condensed (Polyphosphates): Inorganic, contains more than one P per molecule, used

in corrosion control and indetergents as complexation

agents. (Ex. Pyrophosphate H4P2O7 and its ionization

produsts) They hydrolize to orthophosphates.

3 – Oraganic Phosphorus: ATP, ADP, etc.

They also hydrolize to orthophosphates.

Typical Municipal Waste: 10-15 mg/L as P (30% from human waste, 70% from detergents)

P beeing a nutrient, causes algal blooms (eutrophication).

106 CO2 + 16 NO3- + HPO4

-2 + 18 H

+ + 122 H2O C106H263O110N16P + 154O2

light


PRECIPITATION

In most lakes, P is limiting nutrient (P > 0.01 mg/L) and stimulates algal growth.

80% removal of P or a residual of 1 mg/L as P is required in effluents discharged to

streams which are tributaries of Great Lakes.

Primary clarification removes 5-10% of P

Primary + secondary treatment removes 10-20% of P

Removal of P by Chemical Precipitation

Salts of Al (III), Fe (III) and Ca

Al2(SO4)3·XH2O Al+3

, Al(OH)+2

, Al(OH)2+ + Al(OH)4

- etc., Al(OH)3(s)

Al+3

+ PO4-3

Al(PO4)(s)

So, Al(III) required is more than 1:1 ratio and same for Fe(III)

Fe+3

+ PO4-3

Fe(PO4)(s)

Need site-specific experiments to determine chemical requirment, which depends upon pH,

alkalinity, initial PT and desired % P removal.


PRECIPITATION


PRECIPITATION

Fig. 6-4 Iron leakage during phosphorus removal


PRECIPITATION

Typical Al or Fe

requirement per mole P % Removal

1.5 75

1.7 85

2.3 95

Al+3

+ PO4-3

AlPO4 Ksp = [Al+3

][PO4-3

]

[Al+3

] ↓ as pH ↑ Ksp = [Al+3

](α3 PT)

α3 ↑ as pH ↑ PT = ][Al

K

3

3

sp

So, there is a minimum point for PT at about pH = 6-7

For FePO4, min PT at pH = 5-6 (See Fig. 9.1, a & b)

Use Fe(III) salts, if alkalinity is low.

When salts are added, the pH drops (remember hydrolysis reactions and alkalinity

consumption). This helps the precipitation, if the pH is above the min. solubility point.

Otherwise, low pH is detrimental. (See Fig. 2-4)

See Fig. 4-3

(wt ratio)


PRECIPITATION

Fig. 9.1 Conc. of ferric and aluminum phosphate in

Equilibrium with phosphorus (a) Fe(III) phosphate,

(b) Al(III) phosphate

Fig. 11 Lime dosage required to raise the

pH to 11 as a function of raw waste

water alkalinity


PRECIPITATION

Fig. 2-4 Alum coagulation of sewage Fig. 2-2 Lime coagulation of sewage


PRECIPITATION

Sodium Aluminate

NaAlO2 → Na+ + AlO2

- + H2O → Al

+3 + Al(OH)2

+ + Al(OH)2

+ etc. + OH

-

Aluminate

OH- is released in the reaction. So, aluminate behaves as a base, and pH↑. If water is acidic,

use NaAlO2 not Al2(SO4)3 or use a mixture to adjust to the optimum pH. (See Fig. 7.5)

Higher Al(III) requirement with aluminate than alum.

Addition of Ca++

Ca(OH)2, CaO : Both Ca++

and OH- source

Ca++

+ CO3-2

CaCO3(s) and

5Ca++

+ OH- + 3PO4

-3 Ca5(OH)(PO4)3(s)

Hydroxyapatile

Ca requirement for P-removal is a function of carbonate alkalinity. Avoid if alkalinity is too high

(CaCO3 sludge)

High pH is required (pH = 10-11)

Neutralize or recarbonate (CO2) to bring the pH down after precipitation. (See Fig 2-2 and 11)


PRECIPITATION

P Removal in Various Stages

During primary treatment:

Increase BOD and SS removal

Least efficient use of salts

Fe and Al sludges are difficult to dewater

During secondary treatment:

Lime cannot be used because of high pH

Sludge to digester increases in volume

Overdose of Al and Fe salts may cause

low pH toxicity

Aeration

(1) (2) (3)

Primary

Clarifier

Secondary

Clarifier

Tertiary

Clarifier

During tertiary treatment:

Highest capital cost

Highest metal leakage in the effluent

Most efficient metal use

Lowest P in effluent

Lime recovery possible

Sludge CaO + CO2

Separate disposal of chemical precipitation

Heat

Recalcination


PRECIPITATION

Example:

A biolocically treated municipal waste contains 10 mg/L PT as P. The pH is 7.5, Ca++

concentraion

is 50 mg/L and alkalinity is 100 mg/L as CaCO3.

a) What are the major species of P?

b) When 250 mg/L Ca(OH)2 is added to the waste, the pH increases to 11. How much P will

precipitate at this pH?

c) What is the final alkalinity after treatment?

Assume closed system.

a) pKA1 = 2.1

pKA2 = 7.2 pH = 7.5 (H3PO4 and PO4-3

are negligible)

pKA3 = 12.3

α1 = 0.33 (H2PO4-) α2 = 0.66 (HPO4

-2)

PT = (10×10-3

) / (31 g/mole) = 3.22×10-4

M


PRECIPITATION

b) Check whether CaCO3 will precipitate

Ksp = [Ca++

][CO3-2

] = 10-8.3

[Ca++

]added = 250 L

mg

2Ca(OH) g 74

mole 1×10

3 = 3.38×10

-3 M

[Ca++

]orig. = 50 L

mg

40

1×10

3 = 1.25×10

-3 M

Total [Ca++

] = (3.38 + 1.25)×10-3

= 4.68×10-3

M

[CO3-2

] = ?

At pH 7.5

Alk. = [HCO3-] + 2[CO3

-2] + [OH

-] – [H

+]

Alk. = [HCO3-] = 100 mg/L × 0.50 = 2 meq/L

[HCO3-] = 2×10

-3 M

Since CO3-2

and H2CO3 are negligible at that pH.

CT = [HCO3-] = 2×10

-3 M

At pH 11

HCO3- + OH

- → CO3

-2 + H2O

[CO3-2

] = 2×10-3

M, and other carbonate

species are negligible.

(4.68×10-3

M) (2×10-3

M) > 10-8.3

CaCO3 will precipitate.


PRECIPITATION

Unknowns:

1) [Ca++

] 5) (Ca++

) in ppt. apalite

2) [CO3-2

] 6) (CO3-2

) in CaCO3

3) [PO4-3

] 7) (PO4) in apatite

4) (Ca++

) in ppt. CaCO3 8) [HPO4-2

]diss.

Equations:

1) [Ca++

] [CaCO3] = 10-8.3

2) [Ca++

]5 [OH

-] [PO4

-]3 = 10

-55.9

3) (Ca++

)T = [Ca++

] + (Ca++

)CaCO3 + (Ca++

)apatile = 4.63×10-3

M

4) (CO3-2

)T = [CO3-2

] + (CO3-2

)CaCO3 = 2×10-3

M

5) PT = [PO4-] + [HPO4

-2] + (PO4

-)apatile = 3.22×10

-4 M

6)

+

4 12 3

3 2

4

PO HK 10

HPO

.


PRECIPITATION

7) (Ca++

)CaCO3 = (CO3)CaCO3

8) 4 apatileapatile

1 1Ca PO

5 3

Reduced to

[Ca++

] 3

5/3

-16.09-9

101.2 ][Ca

10

][Ca

105

Trial and error procedure:

[Ca++

] = 2.10×10-3

M [HPO4-2

]diss. = 9.5×10-15

M

(Ca++

) in ppt. CaCO3 = 2.00×10-3

M (PO4) in apatite = 3.2×10-4

M (all)

(Ca++

) in ppt. apalite = 0.54×10-3

M [PO4-3

] = 4.7×10-16

M

(Ca)T

= 4.64×10-3

M [CO3-2

] = 2.4×10-6

M

(CO3-2

) in CaCO3 = 2.0×10-3

M

c) Alk. = 2[CO3-2

] + [OH-] = 2×(2.4×10

-6 M) + 10

-3 M = 10

-3 eq/L = 50 mg/L as CaCO3


Heavy Metal Removal

Chemical precipitation is one of the most common methods of heavy metal removal from wastewater.

Precipitating anions include S=, OH- and CO-3

These anions are pH dependent therefore solubility of ppt is also pH dependent

Complexing anions such NH3, CN and organics can hinder ppt.


Typical Solubility Graph of Metal Hydoxides


Typical Solubility graphs of Metal Sulfides


Reactions

+ 2+ -

(s)

+ +

(s)

+ - 0

(s) 2

+ - -

(s) 3

+

(s)

CdS + H = Cd +HS ; logK = -13.36

CdS + H = CdHS ; logK = 6.7

CdS + H + HS = Cd(HS) ; logK = -1.0

CdS + H + 2HS = Cd(HS) ; log K= 2.08

CdS + H + - -2

4

- +

2 (aq)

3HS = Cd(HS) ; log K= 3.53

H S = HS + H ; log K= -7

superpro design v7

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