Chemical Precipitation

Similar but NOT a Coagulation

Process

Removal of dissolved metal ions from solution by changing the solution composition, causing the metal ions to form insoluble metal complexes

Removal of dissolved metal ions from solution by changing the solution composition , causingthe metal ions to form insolublemetal complexes

Solu.on with soluble ions

+

Chemical precipitant

Insoluble complex

+

Treated water

Applications

Treatment of “hard” water – removal of Mg2+ and Ca2+

Phosphorus removal

Removal of metals from industrial wastewater

TheoriticalBackground

Solubility of Compounds

Compound Solubility, mg/L

CaCO3 18

CaCl2 745000

Ca(OH)2 770

CaSO4 1620

Mg(OH)2 40

Equilibrium Considera t i on

Precipitate will form when two solutionsare mixed if and only if the

concentrations of the metal ions is greater than the solubility limits of the

ions in solution for the particular solution composition

Applications

Basic Procedures

A. Chemical additionB. MixingC. FlocculationD. Sedimentation

How it WorksProcess Tank

Process Tank

S t i r r e r Unit 1. Ferric Addition 2. Lime addition

3. Alum addition4. Polymer addition5. After coagulation6. After sedimentation

Industrial Equipment

ApplicationsWater Softening

Hardness

Caused by Ca2+, Mg2+, Fe2+, Mn2+, Sr2+

and Al3+

Carbonate hardness – caused by anion CO -‐2, HCO -‐

3 3

Non-‐carbonate hardness – caused by anion SO4

-‐2, Cl-‐

Level of Acceptance Soft : < 60 mg/L as CaCO3

Moderately hard : 60 – 120 mg/L asCaCO3

Hard : 120 – 180 mg/L as CaCO3

Very hard : > 180 mg/L as CaCO3

Problems

Interfere with laundering –excessive soap consumption

Scale productionin hot-water heaters and pipes

Hair problem

SolutionWater Softening

Water Softening

To remove hardness by removing Ca2+ and Mg2+ via chemical percipitation

Use lime (CaO) or lime slurry (Ca(OH)2) and soda ash (Na2CO3)

Chemical Reactions(Carbonate Hardness)

Reaction of lime with existing CO2 in water

CO2 + Ca(OH)2 = CaCO3 + H2O (3.6-‐1)

Reaction of lime with bicarbonate

Ca(HCO3)2 + Ca(OH)2 = 2CaCO3 + 2H2O(3.6-‐2)

Chemical Reactions(Carbonate Hardness)

Conversion of Mg(HCO3)2 to MgCO3

Mg(HCO3)2 + Ca(OH)2 = CaCO3 + MgCO3

+ 2H2O

(3.6-‐3)

Precipitation of Mg2+

MgCO3 + Ca(OH)2 = Mg(OH)2 + CaCO3(3.6-‐4)

Chemical Reactions(Non-‐carbonate Hardness)

Precipitation of Mg2+

MgSO4 + Ca(OH)2 = Mg(OH)2 + CaSO4

MgCl2 + Ca(OH)2 = Mg(OH)2 + CaCl2

(3.6-‐5)

(3.6-‐6)

Precipitation of Ca2+ by soda ash (Na2CO3)

CaSO4 + Na2CO3 = CaCO3 + Na2SO4

CaCl2 + Na2CO3 = CaCO3 + 2NaCl

(3.6-‐7)

(3.6-‐8)

Lime-‐soda Ash Softening(Process Variations)

1. Excess lime softening

2. Selective calciumcarbonate removal

3. Split-treatment sof t ening

Excess Lime Softening

Removal of Ca2+ and Mg2+ hardness to practical limit of CaCO3 solubility by

stoichiometric addition of lime

Total hardness reduced to

40mg/L as CaCO3

30 mg/L of Ca2+

and

10 mg/L of Mg2+

3(as CaCO )

Precipitation of Mg2+ requires a

surplus of 35 mg/L of CaO above stoichiometric

requirement for

pH adjustment

After excess-‐lime treatment, water is highly alkaline and must be

neutralised.

This is carried out using CO2

(recarbonation)

How it Works

Example 1Water defined by the following analysis is to be sof t ened by excess lime treatment. Assume that the practical limit of hardness removal for CaCO3 is 30 mg/L and that of Mg(OH)2 is 10 mg/L as CaCO3.

CO2

Mg2+

SO42-‐

Ca2+

Na+

Cl-‐

= 40.0 mg/L

= 13.7 mg/L

= 17.9 mg/L

Alk (HCO3-‐) =

= 8.8 mg/L

= 14.7 mg/L

= 29.0 mg/L

135 mg/L as CaCO3

1. Sketch a meq/L bar graph and list the hypothetical combinationsof chemical compounds in solution.

2. Calculate the sof t ening chemicals required expressing lime dosage asCaO and soda ash as Na2CO3.

3. Draw a bar graph for the sof t ened water before and afterrecarbonation. Assume that half the alkalinity in the sof t ened water is inthe bicarbonate form.

Selective CaCO3 RemovalTo soften water low in Mg2+

hardness (< 40 mg/L as CaCO3)

i.e. NO Mg2+ removal

No excess lime for Mg2+ removal

One step recarbonation

How it Works

Consider selective calcium carbonate removal sof t eningof a raw water with a bar graph as drawn in the figurebelow. Calculate the required lime dosage as CaO, and sketch the sof t ened water bar graph after recarbonation.

2.0 2.6 2.9

2.2 2.7 2.9

Example 2

Ca2+ Mg2+ Na+

HCO -3 SO4

2- Cl-

Split Treatment Softening

First Stage

First Stage

Mixing

Mixing point

AdvantageExcess lime in the first stage is utilised

in second stage

Reduce cost of lime and recarbonation

ApplicationsPhosphorus Removal

Involve precipitation with calcium, iron or aluminium.

Reaction with Ca2+ produce hyroxypalite, Ca5OH(PO4)3:

5Ca2+ + 7OH-‐+ 3H2PO4-‐ Ca5OH(PO4)3 +

6H2O

Precipitation is subjected to pH

ApplicationsHeavy Metal Removal

Industrial wastewaters eg. plating and polishing operations, mining, steel manufacturing, electronics manufacturing

Arsenic (As), barium (Ba), chromium (Cr), cadmium (Cd), lead (Pb), mercury (Hg), silver (Ag)

Types of Precipition

Hydroxide precipitation (OH-‐)

Sulphide precipitation (S-‐2)

Carbonate precipitation (CO32-‐)

Hydroxide Precipitation (OH-‐)

Add lime (CaO) or sodium hydroxide (NaOH) to precipitate heavy metals in the form of metalhydroxides

Cd2+ + Ca(OH)2 Cd (OH)2 + Ca2+

Will form floc and sludge in clarifier

Hydroxide Precipitation (OH-‐)

CaO in the form of slurry (Ca(OH)2) NaOH in the form of solution

NaOH is easier to handle but very corrosive.

Sulphide Precipitation (S2-‐)

Use of sulphide in the form of FeS, Na2S or NaHS

Better metal removal as sulphide salt has low solubility limit

Cu2+ + FeS CuS + Fe2+

Sulphide Precipitation (S2-‐)

Limitation: can produce H2S (gas) at low pH

2H+ + FeS H2S + Fe2+

At low pH, reaction to the right

pH > 8 for safe sulphide precipitation.

Carbonate Precipitation (CO3)2-‐

Use of carbonate in the form ofNa2CO3

Ni2+ + Na2CO3 NiCO3 + 2Na+

Equilibrium/ Solubility Limit

OH-‐/ S2-‐

precipitation

CO32-‐

precipitation

Example 3

One litre of an aqueous wastewater containing 10-‐4

moles/L Zn2+ is treated by hydroxide precipitation at pH=10 (Atomic weight of Zn = 65.4 g/mol)

Zn2+ + Ca(OH)2 Zn(OH)2 + Ca2+

1. Will a precipitate form?

2. What will the equilibrium composition of the solution be in the presence of the precipitate?

3. What is the % Zn removed from the solution?

4. What could be done to improve the treatment?

1. 10-‐4 mol/L Zn2+ x 65.4 g/mol x 103

mg/g = 6.54 mg/L

Refer to HO-‐ solubility graph (Fig. 3.3).

It will precipitate as the value is

greater than the equilibrium

concentration (red dashed line)

2. At pH 10, the equilibrium concentration of Zn2+ would be ~1mg/L

3. %Removal = (1 – Ce/Ci) x 100

= (1 – 1/6.54) x 100

= 84.7 %

4. Adjust pH to 9.3 or use sulphideprecipitation

6.54 mg/L

pH = 10

Solution

Read This!!Hammer and Hammer, Water and

Wastewater Technology, 2004, pg. 239 – 244