Water StabilizationStabilization

Department of Civil, Water Resource and Environmental Department of Civil, Water Resource and Environmental EngineeringEngineering

King Fahd UniversityKing Fahd University

Water Stabilization

As in water softening, when the concentrations of CaCO3 and Mg(OH)2 exceed their solubilities, the

solids may continue to precipitate.

This condition can cause scale to form, a solid that deposits due to precipitation of ions in solution.

To prevent scale formation, the water must be stabilized.

Because of the universal presence of carbon dioxide, any water body is affected by the reaction products of carbon dioxide and water.

The species produced from this reaction form the carbonate system equilibria. As discussed later, the stability or instability of water can be gaged using these equilibria.

H2O

CO2

If the pH is high, stabilization may be accomplished using one of several acids or using CO2, a process called recarbonation.

If the pH is low, stabilization may be accomplished using lime or some other bases.

1. Carbonate Equilibria

The carbonate equilibria is a function of :

the ionic strength of water,

activity coefficient,

and the effective concentrations of the ionic species.

Carbonate Equilibria:

• Calcium is one of the major cations that can form scales as a result of the instability of water.

• Calcium plays an important role in the carbonate equilibria.

• We will therefore express the carbonate equilibria in terms of the interaction of the calcium ion and the carbonate species which are the reaction products of carbon dioxide and water.

Carbonate Equilibria:

• Since the equilibria occur in water, the dissociation of the water molecule must also be involved.

• Using calcium as the cation, the equilibrium

equations of the equilibria along with the respective equilibrium constants at 25 ºC are as follows (Rich, 1963):

Carbonate Equilibria:

*32

335.6 }}{{10

COH

HCOHK l

3

2333.10

2

}}{{10

HCO

COHK

}}{{10 14 OHHK w

}}{{)10(8.4 23

29, 3

COCaK CaCOsp

Carbonate Equilibria:

• The Ks are the values of the respective equilibrium constants.

• Ksp.CaC03 is the equilibrium constant for the solubility of CaC03.

• The pair of braces, { }, are read as "the activity

of,"

Carbonate Equilibria:

• the equilibrium constants are calculated using the activity.

• Activity: is a measure of the effectiveness of a given species in its participation in a reaction.

• It is an effective or active concentration and has units of concentrations.

Carbonate Equilibria:

{sp} = γ [sp]• Where sp represents any species involved in

the equilibria.

• The pair of brackets, [], is read as "the concentration of,“.

• γ is the activity coefficient.

1.1 Ionic Strength:• As the particle ionizes, the number of particles

increases. • the activity coefficient is a function of the

number of particles in solution. • The number of particles is characterized by

the ionic strength μ.

1.Carbonate Equilibria:

2][2

1ii zsp

1.1 Ionic Strength:• i is the index for the particular species and z

is its charge. • The concentrations are in gmmols/L In terms

of the ionic strength. • the activity coefficient is given by the DeBye-

Huckellaw as follows (Snoeyink and Jenkins, 1980; Rich, 1963):

1.1 Ionic Strength:

)(14.11

)(5.0 2

10

iz

LangelierLangelier estimated estimated as as (TDS (TDS 2.5 2.5 10 10-5-5))

)(14.11

)(5.0log

2

iz

RussellRussell estimated estimated as as ) ) specific conductance specific conductance 1.6 1.6 1010-5-5 ((

Example 1

• The pH of a solution is 7. Calculate the hydrogen ion concentration?

Solution:

pH = -log10{H+}

7 = -log10{H+}

{H+} = 10-7 gmmols/L Ans.

Example 2

• The concentration of carbonic acid was analyzed to be 0.2 gmmols/L. If the pH of the solution is 7, what is the concentration of the bicarbonate ion if the temperature is 25°C?

• Solution:

.][/)10(93.8}{

1000/2.0

}{10

}{

}}{{10

34

3

37

*32

335.6

AnsHCOLgmolHCO

HCO

COH

HCOHK l

Example 3

• A sample of water has the following composition: CO2 = 22.0 mg/L, Ca2+ = 80 mg/L, Mg2+ = I2.0 mg/L, Na+ = 46.0 mg/L, HCO3

- = 152.5 mg/L, and SO42- = 216 mg/L.

What is the ionic strength of the sample?• Solution:

2][2

1ii zsp

IonMg/LMol. Massgmmols/L

Ca2+8040.10.001995

Mg2+12.024.30.0004938

Na+46.0230.002

HCO3-152.5610.0025

SO42-21696.10.0022

μ = 1/2[0.001995(22) + 0.0004938(22) + 0.002(1) + 0.0025(1) + 0.0022(22)]

= 0.023 Ans.

Example 4

• In Example 3, calculate the activity coefficient and the activity in mg/L of the bicarbonate ion.

• Solution:

)(14.11

)(5.0 2

10

iz

86.010 )023.0(14.11

)023.0)(1(5.0

}HCO3- = {0.86(0.0025) = 0.00215 mg/L Ans.

}sp = {γ [sp]

1.2. Equilibrium Constant As a Function Of Temperature:

• The equilibrium constants given previously were at 25°C.

• To find the values of the equilibrium constants at other temperatures, the Van't Hoff equation is needed;

2

ln

RTdt

Kd

• Where:• T: is the absolute temperature. • ΔHº: is the standard enthalpy change, where the

standard enthalpy change has been adopted as the change at 25°C at one atmosphere of pressure.

• R: is the universal gas constant.

• The value of R depends upon the unit used for the other variables. Table 11.1 gives its various values and units

• Table 11.2 shows values of interest in water stabilization. It is normally reported as enthalpy changes.

• The enthalpy change is practically constant with temperature; thus ΔHº may be replaced by ΔHº298 Doing this and integrating the Van't Hoff equation from KT1 to KT2 for the equilibrium constant K and from T1 to T2 for the temperature,

)](exp[ 1221

29812 TT

TRTKK TT

“This equation expresses the equilibrium constant as a function of temperature”.

1.3. ΔH298o, For Pertinent Chemical

Reactions Of the Carbonate Eq.:According to Hess's low, if the chemical reaction can be written in steps, the enthalpy changes can

be obtained as the sum of the steps;

• The values of the ΔH298o’s are obtained from

the previous Table 11.2.

• The values in the table indicate ΔH298o of

formation having negative values.• If the reaction is not a formation but a breakup

such as;

22)( 2

1OHHOH l , The sign is positive.

Example 5

Hand-off

• Any questions?

2. Criteria For Water Stability at Normal Conditions:

• In the preceding discussions, a criterion for stability was established using the equilibrium constant Ksp.

• At normal conditions, as especially in the water works industry, specialized forms of water stability criteria have been developed.

• These are saturation pH, Langelier index, and the precipitation potential of a given water.

2.1. Saturation pH and Langelier Index:

• Because pH is easily determined the determination of saturation pH is convenient method of determining the stability of water.

• If the condition is at equilibrium no precipitate or scale will form.

• If the pH of the sample is determined , this can be compared with the equilibrium pH to see if the water stable or not.

• Therefore, we now proceed to derive the equilibrium pH. “Equilibrium pH is also called saturation pH”.

• In natural systems, the value of the pH is strongly influenced by the carbonate equilibria reactions.

• The CO32- species of these reactions well pair

with a cation, thus the equilibrium reactions into a dead end by forming a precipitate.

• For example, the complete carbonate equilibria reactions as follows:

wKOHHHOH

13*32 KHCOHCOH

• c is the charge of the cation that pairs with CO3

2- forming the precipitate Cation2(CO3)c(s).

• We call the formation of this precipitate as the dead end of the carbonate equilibria, since the carbonate species in solution are removed by the precipitation.

223)()(3 KCOHHCO aqaq

spc

sc KcCOCationCOCation 23)(32 2)(

• In order to find the dead end cation , several cations can possibly pair with the carbonate.

• The pairing will be governed by the value of Ksp.• The cation with smallest Ksp value is the one that

can form a dead end for the carbonate equilibria reactions.

• however, of all the possible cations, Ca2+ is the one that is found in great abundance in the nature compared to the rest.

• Thus, although all the other cations have much more smaller Ksp’s than the calcium, they are of no use as dead ends if they do not exist (see Table 11.3).

• As will be shown later, the saturation pH may conveniently be expressed in terms of total alkalinity.

• The species which are the components of the carbonate equilibria , they also represent as components of the total alkalinity of the carbonate system equilibria.

• They may be added together to produce the value of the total alkalinity.

• A convenient common unit is the gram equivalent. Letting [A]geq represent the total alkalinity;

][][][][][ 233

HCOHCOOHA geqgeqgeq

• Equations to calculate the carbonate equilibria species:

}{

}{][][

H

KOHOHOH

OH

w

OHgeq

}{

}{}}{{][][

22

,

2

23

33

3

3

3

CaK

KH

K

COHHCOHCO

HCO

CaCOsp

HCOgeq

}{22

][][

2

,232

3

3

3

Ca

KCOCO

CO

CaCOsp

geq

H

HH

}{

][

• Then the total alkalinity equation becomes;

HCO

CaCOsp

HCO

CaCOsp

OH

wgeq

H

Ca

K

CaK

KH

H

KA

}{

}{2}{

}{

}{][

2

,

22

,

3

3

3

3

Let

,][ geqAA ,OH

wKB

,

1

}{ 22

,

3

3

HHCO

CaCOsp

CaK

KC

}{2 2

,

3

3

Ca

KDand

CO

CaCOsp

Solving for {H+},

}{2

4)()(}{

2

sH

C

CBADADH

thus, the saturation pH, pHs is

• The Langlier Index (or Saturation Index) (LI) is the difference between the actual pH and the saturation pH of the a solution, thus;

}{log10 ss HpH

spHpHLI

Example 6

Example 7

2.2. Determination of {Ca2+}:• The activity of the calcium ion is affected by

its complexation with anions.• Ca2+ forms complexes with the carbonate

species, OH- and SO42-.

• The complexation reactions are as follows: 2

32

3 COCaCaCO

32

3 HCOCaCaHCO OHCaCaOH 2

24

24 SOCaCaSO

Ca anion

= anion complex of calcium ion

= represent complex of calcium ion .

• Let the total concentration of the calcium species as determined by titration be [CaT].

• Thus, the concentration of the calcium ion [Ca2+

(aq)] is

• Table 11.4 shows the equilibrium constants of the previous complexes at 25 oC.

])[][][]([][][ 4332

)( CaSOCaOHCaHCOCaCOCaCa Taq

• By applying Hess’s law and using solubility product relations the above equation becomes:

cCaSO

SOCa

HCaOHcCaOHc

wCa

T

K

SO

HK

K

CaHCOCaCOCaCa

4

4][

][1

][][][][

24

332

Example 8

2.3. Total Alkalinity As Calcium Carbonate:

• The unit of concentration that we used for alkalinity is equivalent per unit volume.

• But alkalinity also can be expressed in terms of CaCO3.

• Expressing the alkalinity in terms of CaCO3 is a sort of equivalence.

• depending upon the chemical reaction it is involved with, CaCO3 can have more than one value for its equivalent mass.

• If the reaction is

CaCO3 has an equivalent mass of CaCO3/2, because the number of reference species is 2.

• To illustrate the use of this concept, assume 10-3

gmmol/L of the hydroxide ion and express this concentration in terms of CaCO3.

the pertinent reaction is

[OH-] = 10-3 gmmol/L = 10-3 geq/L

= 10-3(CaCO3/2) g/L as CaCO3

= 0.05 g/L as CaCO3 = 50 mg/L as CaCO3.

2323 2 CaCOHHCaCO

.HOHHOH

2.4. Precipitation Potential:Precipitation potential is another criterion for water stability, and application of this concept can help prevent situations like the one shown in this figure.

All solutions are electrically neutral and negative charges must balance the positive charges. Thus, the balance of charges, where concentration must be expressed in terms of equivalents, is

eqeqeqeqeq CaHOHHCOCO ][][][][][ 23

23

• Expressing in terms of moles,

• the amount of calcium carbonate that precipitates is simply the equivalent of the calcium ion that precipitates, Cappt.

• Because the number of moles of Cappt is equal to the number of moles of the carbonate solid CaCO3ppt that precipitates,

[CaCO3ppt] = [Cappt]

][2][][][][2 23

23

CaHOHHCOCO

• [CaCO3ppt] is the precipitation potential of calcium carbonate, Cappt, in turn, can be obtained from the original calcium, Ca2+

before, minus the calcium at equilibrium, Ca2+

after.

[Cappt] = [Ca2+before] - [Ca2+

after]

• To use the above equation [Ca2+before] must first be known.

• To determine [Ca2+after], the charge balance equation

derived previously will be used. [Ca2+after] is the [Ca2+] in

the charge balance equation, thus

][2][][][][2 23

23

afterCaHOHHCOCO

By substituting the carbonate system equilibria equations in the above equation and solving for , [Ca2+

after], produces

][4

][][2][8][][

][

2

,2

,

2

22

2

3

3

3

3

H

K

KHH

KH

KH

KH

Ca

CaHHCO

CaCOsp

CaCO

CaCOsp

HOH

w

HOH

w

after

• The [H+] in the previous equation is the saturation pH.

• Finally, the precipitation potential [CaCO3ppt] is:

]CaCO3ppt] = [Cappt] = [Ca2+before] - [Ca2+

after[

2.5. Determination of Percent Blocking Potential of Pipes:

• Let Volpipe be the volume of the pipe segment upon which the percent blocking potential is to be determined.

• The amount of volume precipitation potential in this volume after a time t is:

• = the mass density of the carbonate precipitate.• = the detention time of the pipe segment.

dCaCOpipeafterbefore ttVolCaCa3

/])[]([100 22

3CaCO

dt

td = Volpipe/Qpipe

• Qpipe = the rate of flow through the pipe.

• by substituting this expression for td, the percent blocking potential P block after time t is:

• = 2600 g/L.• Note that since the concentrations are expressed in gram

moles per liter, volumes and rates should be expressed in liters and liters per unit time, respectively.

)100(])[]([100

3

22

pipeCaCO

pipeafterbeforeblock Vol

tQCaCaP

3CaCO

Example 9

Example 10

• Assume that the pH of the treated water in Example 11.9 was raised to cause the precipitation of the carbonate solid. The water is distributed through a distribution main at a rate of 0.22 m3/s. Determine the length of time it takes clog a section of the distribution main 1 km in length, if the diameter is 0.42 m.

Solution:

Hand-off

• Any questions?

3. Recarbonation of Softened Water • After the softening process, the pH is so high

that reduction is necessary to prevent deposition of scales in distribution pipes.

• This can be accomplished inexpensively using carbon dioxide.

• We will therefore develop the method for determining the carbonic acid necessary to set the water to the equilibrium pH.

• In recarbonation, the available calcium ion in solution is prevented from precipitation.

• Therefore, it remains to determine at what pH will the equilibrium condition be, given this calcium concentration.

• This determination is, in fact, the basis of the Langelier saturation pH.

• Adding carbonic acid will increase the acidity of the solution after it has neutralized any existing alkalinity.

• Let the current pH be pHcur and the pH to which it is to be adjusted (the destination pH) be pHto.

• The hydrogen ion concentration corresponding to pHcur is 10-pHcur gram moles per liter and that corresponding to pHto is 10-pHto gram moles per liter.

• Assuming no alkalinity present, the total acid to be added is 10-pHto - 10-pHcur gram moles per liter.

Alkalinity is always present, however, so more acid must be added to counteract the natural alkalinity, [Acur]geq. Thus, the total acidity to be added, [Acadd]geq, is:

a

pHpH

geqcurgeqcadd

curto

AA

1010

][][

• Where Фa is the fractional dissociation of the hydrogen ion from the acid supplied.• For strong acids, Фa is unity; for weak acids, it may be calculated from equilibrium constants (Table 11.5).

Th anks

• Any questions?

Q & A

Thanks