Waste water treatment: Precipitation

Precipitation
Objective: To remove undesirable pollutants by a precipitation reaction.
M(aq) + N(aq)

MN(s)

Example removal of hardness by lime soda-ash and removal of iron and
manganese by oxidation.
The focus of this lecture is on removal of hardness by lime and soda-ash, but
before proceeding with the lime soda-ash method, we need to know how to
express concentration in units of meq/l and mg/l as CaCO 3?
meq Concentration (mg / L)
Concentration in
=
L
Eq. wt (mg / meq)
Eq. wt. =

Atomic or molecular wt
eq. #

Concentration of X (mg / L) × 50 ( mg CaCO3 / meq)
mg
of X as CaCO3 =
L
Eq. wt. of X ( mg / meq)


Symbol

At. or M. wt.

(mg/mmol)

Eq. #

Eq. wt.
(mg/meq)

Ca2+

40.1

2

20

Mg2+

24.3

2

12.2

K+

39.1

1

39.1

Na+

23

1

23

HCO3-

61

1

61

CO3 2-

60

2

30

Cl-

35.5

1

35.5

PO4 3-

95

3

31.7

SO42-

96

2

48

CaCO3

100

2

50


Lime Soda-Ash Softening
A precipitation method through which Ca is removed as

CaCO3 and Mg is removed as Mg(OH)2.
The first step is to remove CO2 from water:
CO2+H2O

H2CO3

Lime

H2CO3+Ca(OH)2

CaCO3 +2H2O

Now remove Ca or Mg that has alkalinity in the water:
Ca hardness

Lime

Ca(HCO3)2+Ca(OH)2

2CaCO3 +2H2O


Mg hardness

Lime

Mg(HCO3)2+2Ca(OH)2

2CaCO3 + Mg(OH)2+ 2H2O

Soda ash is needed if there is not enough alkalinity in the water:
Ca hardness

Soda ash

CaSO4+ Na2CO3
Mg hardness

Lime

CaCO3 + Na2SO4
Soda ash

MgSO4+ Ca(OH)2 +Na2CO3

CaCO3 + Mg(OH)2+ Na2SO4


Amount of lime and soda ash needed to remove hardness
1 meq/l of
meq/l needed of
lime

Soda ash

CO2

1

-

Ca(HCO3)2

1

-

Mg(HCO3)2

2

-

Ca with ions other than alkalinity

-

1

Mg with ions other than alkalinity

1

1

Excess

1.25

Practically, some Ca and Mg will be left in solution after treatment:

For Ca: 30 mg/l as CaCO3 =0.6 meq/l in equilibrium with CaCO3 is left
For Mg: 10 mg/l as CaCO3=0.2 meq/l in equilibrium with Mg(OH)2 is left


Example
A groundwater sample has the following characteristics. Find:
1.The total hardness
2.TDS
3.The amount of lime and soda ash needed to remove hardness
to practical limits.
4.The meq/l bar graph after lime soda addition.
Component

Concentration

pH (-)

6.3

CO2 (mg/l)

8.8

Ca2+ (mg/l)

40

Mg2+ (mg/l)

14.7

Na+ (mg/l)

13.7

HCO3- (mg/l)

164.7

SO42- (mg/l)

29

Cl- (mg/l)

17.8


Solution
1. Hardness: Convert the concentration of Ca and Mg to mg/l as CaCO3 using

Concentration of X (mg / L) × 50 (mg CaCO3 / meq)
mg
of X as CaCO3 =
L
Eq. wt. of X ( mg / meq)
Thus Ca= 100 mg/l as CaCO3, Mg=60 mg/l as CaCO3. Total Hardness is 160 mg/l as CaCO3
2. TDS= sum of all cations and anions in mg/l = 279.9 mg/l.
3. To determine the amount of lime and soda ash needed, we need first to convert
the concentration to meq/l, draw a meq/l bar graph and then use the table

Component

Concentration
(mg/l)

Equivalent weight
(mg/meq)

Concentration
(meq/l)

CO2

8.8

22

0.4

Ca2+

40

20

2.0

Mg2+

14.7

12.2

1.21

Na+

13.7

23

0.6

HCO3-

164.7

61

2.7

SO42-

29

48

0.6

17.8

35

0.51

Cl-


The meq/l bar graph of water before lime/soda-ash treatment

The amount of lime and soda ash needed to remove hardness to practical limits are
determined as shown in the table below

meq/l of
CO2 =0.4
Ca(HCO3)2=2.0
Mg(HCO3)2=0.7
CaSO4=0.0
MgSO4=0.51
Excess
Total

meq/l needed of
lime
Soda ash
0.4
2.0
1.4
0
0.51

0.51
1.25
5.56
0.51


The meq/l bar graph of water after lime/soda-ash treatment but before recarbonation

Note, the pH of the treated water is high (about 11), so CO2 is added (recarbonation)
to reduce the pH.