Standard Methods for Examination of Water & Wastewater_9 potx

If the only hardness ions presentare calcium and magnesium, [C T,hard]eq= [Ca2+]eq+ [Mg2+]eq 10.1 The number of equivalents of one substance in a given chemical reaction is equal to the

Water Softening

Softening is the term given to the process of removing ions that interfere with theuse of soap These ions are called hardness ions due to the presence of multivalentcations, mostly calcium and magnesium In natural waters, other ions that may bepresent to cause hardness but not in significant amounts are iron (Fe2+), manganese(Mn2+), strontium (Sr2+), and aluminum (Al3+)

In the process of cleansing using soap, lather is formed causing the surfacetension of water to decrease This decrease in surface tension makes water moleculespartially lose their mutual attraction toward each other, allowing them to wet ‘‘foreign”solids, thereby, suspending the solids in water As the water is rinsed out, the solidsare removed from the soiled material In the presence of hardness ions, however,soap does not form the lather immediately but reacts with the ions, preventing theformation of lather and forming scum Lather will only form when all the hardnessions are consumed This means that hard waters are hard to lather Hard waters arethose waters that contain these hardness ions in excessive amounts Softening usingchemicals is discussed in this chapter Other topics related to softening are discussed

A very soft water has a slimy feel For example, rainwater, which is exceedinglysoft, is slimy when used with soap For this reason, hardness in water used fordomestic purposes is not completely removed Hardness is normally removed to thelevel of 75 to 120 mg/L as CaCO3

10.2 TYPES OF HARDNESS

Two basic types of hardness are associated with the ions causing hardness: ate and noncarbonate hardness When the hardness ions are associated with the ions in water, the type of hardness is called carbonate hardness; otherwise,the type of hardness is called noncarbonate hardness An example of carbonatehardness is Ca(HCO3)2, and an example of noncarbonate hardness is MgCl2.10

carbon-HCO3−

In practice, when one addresses hardness removal, it means the removal of thecalcium and magnesium ions associated with the two types of hardness Our dis-cussion of hardness removal therefore is divided into the categories of calcium andmagnesium Corresponding chemical reactions are developed If other specific hard-ness ions are also present and to be removed, such as iron, manganese, strontium,and aluminum, corresponding specific reactions have to be developed for them Theremoval of iron and manganese will be discussed in a separate chapter

In general, water is softened in three ways: chemical precipitation, ion exchange,and reverse osmosis Only the chemical precipitation method is discussed in this chapter

10.3 PLANT TYPES FOR HARDNESS REMOVAL

In practice, two types of plants are generally used for chemical precipitation hardnessremoval: One type uses a sludge blanket contact mechanism to facilitate the precip-itation reaction The second type consists of a flash mix, a flocculation basin, and

a sedimentation basin The former is called a solids-contact clarifier The latterarrangement of flash mix, flocculation, and sedimentation were discussed in previouschapters on unit operations

A solids-contact clarifier is shown in Figure 10.1 The chemicals are introducedinto the primary mixing and reaction zone Here, the fresh reactants are mixed bythe swirling action generated by the rotor impeller and also mixed with a return sludgethat are introduced under the hood from the clarification zone The purpose of thereturn sludge is to provide nuclei that are important for the initiation of the chemicalreaction The mixture then flows up through the sludge blanket where secondaryreaction and mixing occur The reaction products then overflow into the clarificationzone, where the clarified water is separated out by sedimentation of the reactionproduct solids The clarified water finally overflows into the effluent discharge Thesettled sludge from the clarification is drawn off through the sludge discharge pipe

FIGURE 10.1 Solids-contact clarifier (Courtesy of Infilco Degremont, Inc.)

Clarified water

Motor drive

Secondary mixing and Impeller

Chemical feed cat walk

Return sludge

Hood Clarified water

In the literature, hardness is frequently expressed in terms of CaCO3 Expressingconcentrations in terms of CaCO3 can be confusing For example, when the hardness

of water is 60 mg/L as CaCO3, what does this really mean? Related to this question

is a second question: Given the concentration of a hardness substance, how is thisconverted to the equivalent CaCO3 hardness concentration?

To obtain the mass of any substance, the number of equivalents of that substance

is multiplied by its equivalent mass Because the number of equivalents of all stances participating in a given chemical reaction are equal, what differentiates thevarious species in this chemical reaction are their respective equivalent masses Thus,

sub-to obtain the concentration of 60 mg/L as CaCO3, this equal number of equivalentsmust have been multiplied by the equivalent mass of calcium carbonate This sectionwill determine this equivalent mass This is actually 50 and converting the concentration

of any hardness substance to the equivalent CaCO3 hardness concentration is obtained

by multiplying the number of equivalents of the substance by 50

Let [C T,hard]eq represent the total concentration of hardness in equivalents, wherethe symbol [ ] is read as “the concentration of ” and the subscript eq means that theconcentration is expressed in terms of equivalents If the only hardness ions presentare calcium and magnesium,

[C T,hard]eq= [Ca2+]eq+ [Mg2+]eq (10.1)

The number of equivalents of one substance in a given chemical reaction is equal

to the same number of equivalents of any other substance in this reaction, so it isentirely correct to arbitrarily express the total concentration of hardness in terms ofonly one of the ions that participates in the chemical reaction The concentrations

of the other ions must then be subsumed in the concentration of this one ion beingchosen For example, if the total hardness is to be expressed in terms of the mag-nesium ion only, the previous equation will become

In this equation, the part of the total hardness contained in calcium is subsumed

concentration in terms of magnesium but that the calcium ion concentration issubsumed in it and expressed in terms of magnesium equivalents Moreover, since

an equivalent of one is equal to the equivalent of another, it is really immaterialunder what substance the total equivalents of hardness is expressed Then, in asimilar manner, the total concentration of hardness may also be expressed in terms

of the calcium ion alone as follows

In this equation, the part of the total hardness contained in magnesium is now

means that it is a concentration in terms of calcium but the magnesium ion

is subsumed in it and expressed in terms of calcium equivalents If other hardness

ions are present as well, then the concentrations of these ions will also be subsumed

in the one particular ion chosen to express the hardness

When any ion participates in a chemical reaction, it will react to the satisfaction

of its ionic charge For example, when the calcium and magnesium ions participate

in a softening reaction, they will react to the satisfaction of their ionic charges of

two Thus, if the chemical reaction is written out, the number of reference species

for Ca2+and Mg2+ will be found to be two and the respective equivalent masses are

then Ca/2 and Mg/2 In terms of molar concentrations, [Ca2+]eq is then equal to

[Ca2+](Ca/Ca/2) = 2[Ca2+] By analogy, = 2 , where [Ca2+]′ is now

an equivalent molar concentration in terms of the calcium ion that subsumes all

concentrations Also, [Mg2+]eq= [Mg2+](Mg/Mg/2) = 2[Mg2+] and, again, by analogy,

= 2[Mg2+]′ [Mg2+]′ is the equivalent molar concentration in terms of themagnesium ion that also subsumes all concentrations Considering Eqs (10.2) and

(10.3) simultaneously and expressing the molar concentrations of the calcium and

magnesium hardness in terms of calcium only,

The term 2[Ca2+]′ can be converted to [CaCO3]′ This is done as follows: In

CaCO3, one mole of [Ca2+] is equal to one mole of [CaCO3] Hence, 2 =

2[CaCO3]′ and Equation (10.4) becomes

[CaCO3]′ is an equivalent molar concentration expressed in terms of moles of

CaCO3 Thus, to get the equivalent calcium carbonate mass concentration, it must

be multiplied by CaCO3= 100 Therefore,

2+

[ ]eq′2 - Ca

2+

[ ]eq′2 -

CaCO3[ ]′ 100( ) [C T,hard]eq

2 -

Ca2+

[ ]eq′2

Thus, the concentration of hardness expressed in terms of the mass of CaCO3

is equal to the number of equivalents of hardness ([C T,hard]eq, , or )divided 2 times the molecular mass of calcium carbonate Or, simply put, theconcentration of hardness expressed in terms of the mass of CaCO3 is equal to the

number of equivalents of hardness ([C T,hard]eq, , or ) times 50.Note that Equation (10.7) merely states the concentration of hardness in terms

of the mass of CaCO3 Although the symbol for calcium carbonate is being used, itdoes not state anything about the actual concentration of the CaCO3 species present;

it is even possible that no species of calcium carbonate exists, but MgCO3 or anyother species where the concentrations are simply being expressed in terms of CaCO3

To apply Equation (10.7), suppose that the concentration of Mg2+ in a sample

of water is given as 60 mg/L as CaCO3 The meq/L of Mg2+ is then equal to [Mg2+]eq=

= 60/50 = 1.2 and the concentration of magnesium in mg/L is 1.2(Mg/2) =1.2(24.3/2) = 14.58 mg/L

Take the following reaction and the example of the magnesium ion:

In this reaction, the equivalent mass of CaCO3 is 2(CaCO3)/2 = 100 Again, in anygiven chemical reaction, the number of equivalents of all the participating speciesare equal Thus, if the number of equivalents of the magnesium ion is 1.2 meq/L,the number of equivalents of CaCO3 also must be 1.2 meq/L Thus, from this reaction,the calcium carbonate concentration corresponding to the 1.2 meq/L of Mg2+ would be1.2(100) = 120 mg/L as CaCO3 Or, because the species is really calcium carbonate,

it is simply 120 mg/L CaCO3––no more ‘‘as.”

How is the concentration of 120 mg/L CaCO3 related to the concentration of

60 mg/L as CaCO3 for the magnesium ion? The 60 mg/L is not a concentration ofcalcium carbonate but a concentration of the magnesium expressed as CaCO3 Thesetwo are very different In the “120,” there is really calcium carbonate present, while

in the “60,” there is none

Again, expressing the magnesium concentration as 60 mg/L CaCO3 does notmean that there are 60 mg/L of the CaCO3 but that there are 60 mg/L of the ion ofmagnesium expressed as CaCO3 Expressing the concentration of one substance interms of another is a normalization This is analogous to expressing other currencies

in terms of the dollar Go to the Philippines and you can make purchases with thedollar, because there is a normalization (conversion) between the dollar and the peso

Example 10.1 The concentration of total hardness in a given raw water is found

to be 300 mg/L as CaCO3 Calculate the concentration in milligram equivalents perliter

10.5 SOFTENING OF CALCIUM HARDNESS

Calcium hardness may either be carbonate or noncarbonate The solubility productconstant of CaCO3 is K sp= [Ca2+][ ] = 5(10−9) at 25°C A low value of the K sp

means that the substance has a low solubility; a value of 5(10−9) is very low Because

of this very low solubility, calcium hardness is removed through precipitation ofCaCO3 Because there are two types of calcium hardness, there corresponds twogeneral methods of removing it When calcium is associated with the bicarbonateion, the hardness metal ion can be easily removed by providing the hydroxide radical.The H+ of the bicarbonate becomes neutralized by the OH− provided forming waterand the ion necessary to precipitate calcium carbonate The softening reaction

is as follows:

(10.8)The precipitate, CaCO3, is indicated by a downward pointing arrow,↓ As shown,the two bicarbonate species in the reactant side are converted to the two carbonatespecies in the product side The carbonate ion shown unpaired will pair with whatevercation the OH− was with in the reactant side of the reaction

Once all the bicarbonate ions have been destroyed by the provision of OH−, theyconvert to the carbonate ion, as indicated If the OH− source contains the calcium ion,the carbonates all precipitate as CaCO3 according to the following softening reaction:

as calcium carbonate The usual source of the carbonate ion is soda ash

10.6 SOFTENING OF MAGNESIUM HARDNESS

As in the case of calcium hardness, magnesium can also be present in the form of

carbonate and noncarbonate hardness The K sp of Mg(OH)2 is a low value of 9(10−12).Thus, the hardness is removed in the form of Mg(OH)2 To remove the carbonatehardness of magnesium, a source of the OH− ion is therefore added to precipitatethe Mg(OH)2 as shown in the following softening chemical reaction:

(10.10)

C T,hard

30050 - 6.0 meq/L Ans

The carbonate ions in the product side will pair with whatever cation the OH− waswith in the reactant side of the reaction If this cation is calcium, in the form ofCa(OH)2, then the product will again be the calcium carbonate precipitate.

In natural waters, the form of noncarbonate hardness normally encountered isthe one associated with the sulfate anion; although, occasionally, large quantities ofthe chloride and nitrate anions may also be found The softening reactions for theremoval of the noncarbonate hardness of magnesium associated with the possibleanions are as follows:

(10.11)(10.12)(10.13)

As shown in all the previous reactions, the removal of the magnesium hardness,both carbonate and noncarbonate, can use just one chemical This chemical is normallylime, CaO

10.7 LIME–SODA PROCESS

As shown by the various chemical reactions above, the chemicals soda ash and limemay be used for the removal of hardness caused by calcium and magnesium Thus,

the lime–soda process is used This process, as mentioned, uses lime (CaO) and

soda ash (Na2CO3) As the name of the process implies, two possible sets of chemicalreactions are involved: the reactions of lime and the reactions of soda ash Tounderstand more fully what really is happening in the process, it is important todiscuss these chemical reactions Let us begin by discussing the lime reactions.CaO first reacts with water to form slaked lime, before reacting with the bicar-bonate The slaking reaction is

(10.14)After slaking, the bicarbonates are neutralized according to the following reac-tions:

(10.15)(10.16)

Note that in Equation (10.16) two types of solids are produced: Mg(OH)2 andCaCO3 and that the added calcium ion from the lime that would have produced anadded hardness to the water has been removed as CaCO3 Although the hardnessions have been precipitated out, the resulting solids, however, pose a problem ofdisposal in water softening plants

MgSO4+2OH− Mg OH( )2↓ SO4

2−

+

→MgCl2+2OH−→Mg OH( )2↓ 2Cl+

Mg NO( 3)2+2OH−→Mg OH( )2↓ 2NO+ 3

CaO+HOH→Ca OH( )2

Ca HCO( 3)2+Ca OH( )2→2CaCO3↓ 2HOH+

Mg HCO( 3)2+2Ca OH( )2→Mg OH( )2↓ 2CaCO+ 3↓ 2 HOH+

Magnesium, whether in the form of the carbonate or noncarbonate hardness, isalways removed in the form of the hydroxide Thus, to remove the total magnesiumhardness, more lime is added to satisfy the overall stoichiometric requirements forboth the carbonates and noncarbonates (Later, we will also discuss the requirement

of adding more lime to raise the pH.) The pertinent softening reactions for theremoval of the noncarbonate hardness of magnesium follow

(10.17)(10.18)(10.19)

As shown from the previous reactions, there is really no net mole removal ofhardness that results from the addition of lime For every mole of magnesium hardnessremoved [MgSO4, MgCl2, or Mg(NO3)2], there is a corresponding mole of by-productnoncarbonate calcium hardness produced [CaSO4, CaCl2, or Ca(NO3)2] Despite thefact that these reactions are useless, they still must be considered, because they willalways transpire if the noncarbonate hardness is removed using lime For this reason,the implementation of the lime-soda process should be such that as much magnesium

as possible is left unremoved and rely only on the removal of calcium to meet thedesired treated water hardness If the desired hardness level is not met by the removal

of only the calcium ions, then removal of the magnesium may be initiated This willentail the use of lime followed by the possible addition of soda ash to remove theresulting noncarbonate hardness of calcium As shown by Eqs (10.17), (10.18), and(10.19), soda ash is needed to remove the by-product calcium hardness formed fromthe use of the lime

As noted before, the calcium ion is removed in the form of CaCO3 This is thereason for the use of the second chemical known as soda ash for the removal of thenoncarbonate hardness of calcium Thus, another set of chemical reactions involvingsoda ash and calcium would have to written The pertinent softening reactions are

as follows:

(10.20)(10.21)(10.22)

Some of the calcium hardness of these reactions would be coming from the by-productnoncarbonate hardness of calcium that results if lime were added to remove thenoncarbonate hardness of magnesium

It is worth repeating that soda ash is used for two purposes only: to remove theoriginal calcium noncarbonate hardness in the raw water and to remove the by-productcalcium noncarbonate hardness that results from the precipitation of the noncarbonate

MgSO4+Ca OH( 2)→Mg OH( )2↓ CaSO+ 4MgCl2+Ca OH( )2→Mg OH( )2↓ CaCl+ 2

Mg NO( 3)2+Ca OH( )2→Mg OH( )2↓ Ca NO+ ( 3)2

CaSO4+Na2CO3→CaCO3↓ Na+ 2SO4CaCl2+Na2CO3→CaCO3↓ 2NaCl+

Ca NO( 3)2+Na2CO3→CaCO3↓ 2NaNO+ 3

hardness of magnesium It is also important to remember that by using lime thecarbonate hardness of magnesium does not produce any calcium noncarbonate hard-ness It is only the noncarbonate magnesium that is capable of producing the by-product calcium noncarbonate hardness when lime is used

10.7.1 C ALCULATION OF S TOICHIOMETRIC L IME R EQUIRED

Because the raw water is exposed to the atmosphere, there will always be some CO2dissolved in it As will be shown later, carbon dioxide also consumes lime We will,however, ignore this requirement for the moment, and discuss it in the latter part of thischapter Ignoring carbon dioxide, the amount of lime needed comes from the require-ment to remove the carbonate hardness of calcium and the requirements to remove boththe carbonate and the noncarbonate hardness of magnesium We will first calculate theamount of lime required for the removal of the carbonate hardness of calcium

to be removed From Equation (10.15), taking the positive oxidation state, the number

of reference species is 2 moles of positive charges Thus, the equivalent mass ofcalcium bicarbonate is /2 and the number of equivalents of the

num-ber of equivalents of CaO (or slaked lime) used to neutralize the bicarbonate

From Eqs (10.14) and (10.15), the equivalent mass of CaO is CaO/2 Thus, the

(CaO/2) Let be the total calcium bicarbonate in the raw

(10.23)

From Equation (10.16), again taking the positive oxidation state, the number of erence species is 2 moles of positive charges Thus, the equivalent mass of magnesium

of equivalents of CaO (or slaked lime) used to neutralize the magnesium bicarbonate From Eqs (10.14) and (10.16), the equivalent mass of CaO is 2CaO/2 Thus,

The removal of the noncarbonate hardness of magnesium using lime as theprecipitant produces a corresponding amount of noncarbonate hardness of calcium

as a by-product This by-product requires the use of soda ash for its removal Thus,the amount of lime needed for the removal of this magnesium noncarbonate hardnesswill be discussed in conjunction with the determination of the stoichiometric sodaash required to remove the by-product calcium noncarbonate hardness that results

10.7.2 K EY TO U NDERSTANDING S UBSCRIPTS

also be using other mass variables such as M solidsNonCarb, and and fractional

variables such as fMgCa and fCa As we continue to develop equations, it would befound that ascertaining the meaning of the subscripts of the variables would becomedifficult Thus, a technique must be developed to aid in remembering

First, for the mass variables, M refers to the mass of the type of species involved The subscript of M is a string of words In this string, T may or may not be present.

If the M of a given species can be both a partial mass and a total mass, the T would

be present as the first one in the subscipt to represent the total mass Then if it isnot present, the mass is a partial mass Partial here means a part of the total For

be both a partial mass and a total mass Thus, the first one refers to the total massand the second refers to the partial mass of the type of species involved

The next part of the string subscript is CaHCO3 This refers to the type ofspecies involved This is not really the correct formula for calcium bicarbonate, butbecause it is used as a subscript, it can be easily taken to mean calcium bicarbonate

bonate and the partial mass of calcium bicarbonate, respectively Calcium bonate, , is said to be the type of species involved

bicar-If M can only have a total mass, the T is not used For example, the total solids,

M solids, can come from the solids produced from carbonate hardness, from

the solids produced from noncarbonate hardness, M solidsNonCarb, and from the calciumcarbonate solids produced from carbon dioxide, In this case, since the

designation for the components of total solids use different subscripts, the M of total solids can only have the total mass and therefore M solids is used instead of M Tsolids

No confusion occurs, however, if M Tsolids is used, because this would definitely refer

to the total

Not counting T, a second portion of the string subscript (the third portion, if T

is present) may also be present This will then refer to the reason for the existence

of the type of mass involved For example, consider , mentioned ously In this symbol, CO2 is the reason for the existence of the mass of CaCO3.Thus, is the mass of calcium carbonate produced from carbon dioxide.Another example where the second portion of a string subscript is the reasonfor existence of the type of mass involved is In this symbol,

previ-is the reason for the exprevi-istence of CaO In the pertinent chemical reaction, calcium

M T CaHCO3 MCaHCO3 MCaOMgHCO3

oxide is not produced from calcium bicarbonate, but, rather, calcium oxide is usedbecause of the bicarbonate Nevertheless, the bicarbonate is still the reason for theexistence of calcium oxide.

The portions of a string subscript can be easily identified, since they ously stand out For example, portions CaCO3 and CO2 conspicuously stand out in

conspicu- CaCO3 and CO2 are said to be portions of the string subscript of

In some situations, a string subscript may appear to have several portions

con-spicuously standing out For example, consider M solidsNonCarb This symbol seems to

have three portions standing out; however, solidsNonCarb does not refer to three different entities but to only one and this is the noncarbonate solids Thus, M solidsNonCarb

does not have three subscript portions

As far as the masses are concerned, consider this last example, Thissymbol indicates that the reason for the existence of the CaCO3 solid is Mg, however,

a next symbol Ca also appears The appearance of Ca simply indicates that Mg, asthe reason for the existence of CaCO3, bears a relationship to calcium This rela-tionship can only happen when the noncarbonate hardness of magnesium is involved,for this will produce the corresponding noncarbonate hardness of calcium––therelationship An important thing to remember, as far as our method of subscripting

is concerned, is that whenever Mg and Ca are used as subscripts, they refer to the

noncarbonate form of their respective hardness For example, M T Ca refers to the

total noncarbonate calcium hardness and MCa refers to a partial noncarbonate calcium

Mg is the reason for the existence of the calcium carbonate This magnesium, ofcourse, produced the noncarbonate Ca (the relationship) for the eventual production

of the calcium carbonate In other words, to repeat, whenever Ca and Mg are used

as subscripts, they are intended to refer only to their respective noncarbonate form

of hardness

Now, consider the fractional variables For example, consider fMgCa and fCa As

in the mass variables, the first subscript of the fractional variable refers to the type

of fraction of the mass Thus, in fMgCa, the type of fraction is the fraction of Mg;and, considering the fact that when Mg and Ca are used as subscripts, they refer to

the noncarbonate forms of hardness, fMgCa stands for the fraction of the noncarbonate

form of magnesium hardness, with Ca reminding about the relationship fMgCa could

just simply be written as fMg and they would be the same, but Ca is there, again,just as a reminder that it is also involved in the chemistry of the reaction By the

convention we have just discussed, fCa stands for the fraction of the noncarbonateform of calcium hardness As a further example, to what does refer? The

is the fraction of the calcium carbonate (bicarbonate) hardness

10.7.3 C ALCULATION OF S TOICHIOMETRIC S ODA A SH R EQUIRED

The amount of soda ash needed comes from the requirement to remove the bonate hardness of calcium In addition, if the noncarbonate hardness of magnesiumwas precipitated using lime, additional soda ash will also be required to remove the

by-product noncarbonate hardness of calcium that results from the magnesiumprecipitation (if additional hardness removal is desired)

Let MCa be the mass of the calcium species to be removed from the noncarbonatehardness of calcium From Eqs (10.20), (10.21), or (10.22), taking the positiveoxidation state, the number of reference species is 2 moles of positive charges foreach of these reactions Thus, the equivalent mass of calcium is Ca/2 and the number

of equivalents of the MCa mass of calcium is MCa/(Ca/2) This is also the samenumber of equivalents of Na2CO3 needed to precipitate the noncarbonate hardness

of calcium to CaCO3 Again, from Eqs (10.20), (10.21), or (10.22), the equivalentmass of Na2CO3 is Na2CO3/2 Thus, the mass of soda ash, M sod AshCa, needed toprecipitate the calcium species from the noncarbonate hardness of calcium is

MCa/(Ca/2)(Na2CO3/2) Let M T Ca be the total calcium in the noncarbonate hardness

of calcium in the raw water and fCa be its fractional removal Thus, MCa= fCaM T Ca and

(10.25)

Note that in M sodAshCa, Ca is the reason for the use of soda ash

To calculate the soda ash requirement for the calcium noncarbonate hardness

by-product, let MMgCa represent the mass of the magnesium species that precipitatesand results in the production of the additional calcium hardness cation, where the

Ca, again, is written as a reminder Refer to Eqs (10.17), (10.18), and (10.19) tosee how the calcium hardness is produced from the precipitation of noncarbonatemagnesium The number of equivalents of the calcium hardness produced is equal

to the number of equivalents of the noncarbonate hardness of magnesium precipitated

[which is equal to MMgCa/(Mg/2), where Mg/2 is the equivalent mass of Mg asobtained from Eqs (10.17), (10.18), and (10.19)] Because this is calcium hardness,

we already have the method of determining the amount of soda ash needed to remove

it as shown in Equation (10.25) Letting this amount be M sodAshMgCa , M sodAshMgCa is

equal to MMgCa/(Mg/2)(Na2CO3/2) Let M T Mg be the total magnesium in the

noncar-bonate hardness of magnesium and fMgCa be its fractional removal Thus, MMgCa =

fMgCaM T Mg, with this and Equation (10.25),

(10.26)

The mass of magnesium species from the noncarbonate hardness of magnesiumthat precipitates also needs additional lime for its precipitation as magnesiumhydroxide Note that this is the same mass of magnesium that produced the additional

by-product calcium hardness As mentioned above, this mass is MMgCa Again, from

Eqs (10.17), (10.18), and (10.19), the number of equivalents of the MMgCa mass of

magnesium is MMgCa/(Mg/2) This is the same number of equivalents of the limeneeded to precipitate Mg(OH)2 From the preceding equations, the equivalent mass

of CaO is CaO/2 Let the mass of lime needed to precipitate the Mg(OH)2 resulting

M sodAsh Ca fCaM T Ca

Ca/2 - Na( 2CO3/2) 2.65 fCaM TCa

M sod Ash MgCa fMgCaM T Mg

Mg/2 - Na( 2CO3/2) 4.36 fMgCaM T Mg

from MMgCa mass of magnesium be MCaOMgCa Thus, MCaOMgCa is MMgCa/(Mg/2)(CaO/2),

but because MMgCa= fMgCaM T Mg,

(10.27)

The solids produced in water softening plants, if not put to use, pose a disposalproblem Conceptually, because of their basic nature, they can be used in absorptiontowers that use alkaline solutions to scrub acidic gas effluents These solids, as far

as the lime-soda process is concerned, come from the solids produced in the removal

of (1) the carbonate hardness and (2) the noncarbonate hardness

Let us first derived the equations for solids production from the removal of thecarbonate hardness The sources of these solids are the CaCO3 produced from theremoval of the carbonate hardness of calcium [Equation (10.15)], the CaCO3 pro-duced from the removal of the carbonate hardness of magnesium [Equation (10.16)],and the Mg(OH)2 produced also from the removal of the carbonate hardness ofmagnesium [Equation (10.16)]

Thus, using Equation (10.15), the corresponding production of CaCO3 is

(10.28)

is the calcium carbonate produced Note that the second subscript,CaHCO3, is the reason for the existence of the carbonate

From Equation (10.16), the corresponding production of CaCO3 is

(10.29)

is the calcium carbonate produced Note that, similarly with MCaCO3CaHCO3,the second subscript, MgHCO3, is the reason for the existence of the carbonate.Also, from the same Equation (10.16), the amount of Mg(OH)2 solids produced is

(10.30)

MCaOMgCa fMgCaM T Mg

Mg/2 - CaO( /2) 2.30 fMgCaM T Mg

is the amount of the Mg(OH)2 solids produced Also, note the reasonfor its existence, which is MgHCO3.

Combining Eqs (10.28), (10.29), and (10.30), the total mass of solids, M soildsCarb,produced from the removal of the carbonate hardness using the lime–soda ashprocess is therefore

Equations (10.17), (10.18), and (10.19) show the production of the Mg(OH)2

solids From these equations, the amount of hydroxide solids, MMgOHMgCa, produced

from the MMgCa mass of magnesium, is

MMgCa= fMgCaM T Mg mass of magnesium is obtained as follows: From Eqs (10.17),(10.18), and (10.19), one mole of Mg produces one mole of Ca From Eqs (10.20),(10.21), and (10.22), one mole of Ca also produces the same one mole of CaCO3.Thus,

70 g Assuming the latter takes precedence, the bicarbonate species will now be posed of one gram-equivalent of Mg(HCO3)2 and 1.5 gram-equivalent of Ca(HCO3)2.The corresponding amount of lime needed to remove the bicarbonates is 1.0(2CaO/2) + 1.5(CaO/2) = 98 g These two simple calculations produce two very differentresults, so knowledge of the order of precedence of the reactions is very important.The order of precedence can be judged from the solubility of the precipitates, namely:CaCO3 and Mg(OH)2.

com-The K sp for CaCO3 is 5(10−9) at 25°C, while that for Mg(OH)2 is 9(10−12), also

at 25°C Consider calcium carbonate, first Before the solid is formed, the product

of the concentrations of the calcium ions and the carbonate ions must be, at least,equal to 5(10−9) Let x be the concentration of the calcium ion; thus, x will also be

the concentration of the carbonate ion Therefore,

x2= 5(10−9) ⇒ x = 0.000071 gmole/L

Now, consider the magnesium hydroxide Again, before the solid is formed, theproduct of the concentrations of the magnesium ions and the hydroxide ions must

be, at least, equal to 9(10−12) In magnesium hydroxide, there are two hydroxide ions

to one of the magnesium ion Thus, letting y be the concentration of the magnesium ions, the concentration of the hydroxide ions is 2y Therefore,

y(y2) = 9(10−12) ⇒ y = 0.00021 gmole/L

Now, comparing the two answers, CaCO3 is less soluble than Mg(OH)2, because

it only needs 0.000071 gmole/L to precipitate it compared to 0.00021 gmole/L forthe case of Mg(OH)2 Therefore, calcium carbonate will precipitate first and Equation(10.15) takes precedence over Equation (10.16) In this example, the answer would

be 70 g instead of 98 g

For the removal of the hardness of calcium and magnesium, the effect of carbondioxide on the amounts of sludge produced and chemicals used were not considered;but because the type of raw water treated is exposed to the atmosphere, CO2 alwaysdissolves in it Therefore, in addition to the chemical reactions that have beenportrayed, the precipitant chemicals must also react with carbon dioxide This means

HCO3−

that more lime than indicated in previous reactions would be needed The reactionthat portrays the consumption of lime by carbon dioxide is as follows:

(10.36)

As shown in Equation (10.36), solids of CaCO3 are produced Let be themass of carbon dioxide in the raw water Therefore, letting be the car-bonate solids produced from the dissolved carbon dioxide,

is practically dissolved when the pH is raised to 11

Also, from the figure, the lowest solubility of Mg2+ in equilibrium with Mg(OH)2occurs at about pH 10.4; raising the pH above this value does not affect the solubility,since the solubility is already zero at pH 10.4 At this value of pH, the solubility of

Ca2+ in equilibrium with CaCO3 is not affected very much Also, lowering the pH

to below 9.8 practically dissolves the Mg(OH)2 Thus, these two limiting conditionsleave only a very small window of pH 9.8 to 10.4 for an optimal concurrentprecipitation of Ca2+ and Mg2+ Above pH 10.4, the precipitation of CaCO3 suffersand below pH 9.8, the precipitation of Mg(OH) suffers

It has been found experimentally that adding lime to satisfy the stoichiometricamount for precipitating Mg(OH)2 does not raise the pH value to 10.4 Lime dissolves

in water according to the following reaction: CaO + HOH → Ca(OH)2
Ca2+ +2OH− Empirically, based on this reaction (equivalent mass = CaO/2), 1.0 milligram-equivalent per liter of water of excess alkalinity over the computed stoichiometricamount is needed to raise the pH to 10.4 This means that this amount must be added

if the pH is to be maintained at this level Letting M CaOExcess be the total amount ofexcess lime and be the volume of water treated in cubic meters,

(10.40)

FIGURE 10.2 Concentrations of calcium and magnesium ions in equilibrium with the

calcium carbonate and magnesium hydroxide solids, respectively (From Powell, S T (1954).

Water Conditioning for Industry McGraw-Hill, New York With permission.)

200 180 160 140 120 100 80 60 40 20

0

7 8 9 10 11 12 13 14

pH value

Concentration of magnesium ion in equilibrium with magnesium hydroxide

Concentration of calcium ion in equilibrium with calcium carbonate

V

M CaOExcess

1.0 CaO( /2) 1000( )V

1000 1000( ) - 0.028V kg

10.11 SUMMARY OF CHEMICAL REQUIREMENTS

AND SOLIDS PRODUCED

To summarize, let MCaO be the total lime requirement; M sodAsh be the total soda ash

requirement; and M solids be the total solids produced MCaO is equal to the amount of

mag-nesium (MCaOMgCa), the requirement to neutralize the dissolved carbon dioxide( ), and the requirement to raise the pH to 10.4 for the precipitation ofMg(OH)2 Thus,

(10.41)

Note: The unit of M CaOExcess is kilograms Thus, in order for the terms of this

equation to be dimensionally equivalent to M CaOExcess, they must all beexpressed in terms of kilograms

M sodAsh is equal to the amount of soda ash needed to precipitate the noncarbonatehardness of calcium and the amount of soda ash needed to precipitate the noncar-bonate hardness of calcium produced from the precipitation of the noncarbonatehardness of magnesium Thus,

(10.42)

M solids is equal to the solids produced from the removals of the carbonate and

the noncarbonate hardness (M solidsCarb and M solidsNonCarb, respectively) and the solidsproduced from the neutralization of the dissolved carbon dioxide using lime

M solids = M solidsCarb + M solidsNonCarb+ (10.43)

10.12 SLUDGE VOLUME PRODUCTION

The mass of solids precipitated enmeshes in an enormous amount of water As thesesolids are flocculated and settled, they retain extraordinary amounts of water resulting

in a huge volume of sludge to be disposed of The amount of the solids is negligiblecompared to the total amount of sludge In fact, sludges contains approximately 99%

water Let f solids be the fraction of solids in the sludge Then the mass of sludge M sludge

is M solids /f solids.If S sl is the specific gravity of the sludge, its volume in cubic meters is

(10.44)

The 1,000 in the denominator is the mass density of water in kg/m3 at the temperature

of 4°C used as the reference temperature in the definition of specific gravity Allunits in this equation must be in the mks (meter-kilogram-second) system

MCaOCaHCO3

MCaOMgHCO3

MCaOCO2

MCaO = MCaOCaHCO3+MCaOMgHCO3+MCaOMgCa+MCaOCO2+M CaOExcess

M sodAsh = M sodAsh Ca+M sodAsh MgCa

10.13 CHEMICAL SPECIES IN THE TREATED WATER

In addition to the H+ and OH− species, a great majority of ions present in the treatedwater are the cations Ca2+, Mg2+, and Na+ and the anions and Occa-sionally, the anions Cl− and may also be present when the hardness cations innatural waters are associated with these anions

The effluent water from the hardness removal reactor is basic and tends to depositscales in distribution pipes For this reason, this water should be stabilized Stabili-

zation is normally done using carbon dioxide, a process called recarbonation

Sta-bilization using carbon dioxide affects the concentration of the bicarbonate ion inthe treated water The concentrations of the , Cl− and ions are not affected,however, because they do not react with carbon dioxide Their concentrations remainthe same as when they were in the influent to the treatment plant The original cation

Na+ from the influent raw water is also not affected for the same reason that it doesnot react with carbon dioxide Na+ is, however, introduced with the soda ash

indicated ions in milligram equivalents per liter in the influent to the reactor Also, let

, , and be the concentrations of the ions in the treated water.Thus,

The precipitation of CaCO3 and Mg(OH)2 is never complete At 0°C, the solubility

of CaCO3 is 15 mg/L and at 25°C, it is 14 mg/L These solubility values imply that

no matter how much precipitant is applied to the water, there will always remainsome calcium and carbonate ions which, in the aggregate, amounts to 14 to 15 mg/L

of CaCO3 dissolved in the treated water

For Mg(OH)2, its solubility at 0°C is 17 mg/L as CaCO3; at 18°C it is 15.5 mg/L

as CaCO3 As in the case of CaCO3, some ions of magnesium and, of course, thehydroxide will always remain dissolved in the treated effluent no matter how muchprecipitant is employed Operationally, let us adopt the following figures as the limit

of technology: CaCO3= 15 mg/L; Mg(OH)2= 16 mg/L as CaCO3 This brings atotal to 15 + 16 = 31 mg/L as CaCO3

10.13.2 C ONCENTRATION OF Ca 2++++

The concentration of the calcium ion in the treated water comes from the calciumbicarbonate not precipitated, the calcium from the noncarbonate hardness of cal-cium not precipitated, and the calcium that results from the precipitation of the

NO3−[ ]meq = [NO3−]meq,inf

noncarbonate hardness of magnesium Note that, as the treated water is recarbonated,the 15 mg/L of CaCO3 from the limit of technology converts to the bicarbonate form,

(10.45)

From this reaction, the number of milligram equivalents per liter of Ca2+ and are 15/(CaCO3/2) = 0.3, respectively

carbonate hardness of calcium was precipitated Let this mass be measured in

bicar-bonate not precipitated Thus,

(10.46)

is the volume of water treated in cubic meters

of calcium bicarbonate is Ca(HCO3)2/2 The calcium bicarbonate not precipitated

equivalents per liter of Ca2+ and , respectively

Now, also recall that MCa (= fCaM TCa) mass of calcium from the noncarbonatehardness of calcium precipitated As before, let this mass be measured in kilograms

(10.47)

produces [Ca2+]mgnot/(Ca/2) = 0.050[Ca2+]mgnot = 50.0 [MTCa (1 − fCa)/ ] milligramequivalents of Ca2+

Summing all the calcium ion concentrations, the total concentration [Ca2+]meq is

(10.48)

cubic meters [Ca2+]meq is then in milligram equivalents per liter

CaCO3+H2CO3(CO2+HOH→H2CO3)→Ca HCO( 3)2

M T CaHCO3(1–fCaHCO3) V 12.3 M[ T CaHCO3(1–fCaHCO3)/V]

-=

10.13.3 C ONCENTRATION OF Mg 2++++

The concentration of the magnesium ion in the treated water comes from the nesium from the magnesium bicarbonate not precipitated, and the magnesium fromthe noncarbonate hardness of magnesium not precipitated

mag-As the treated water is recarbonated, the 16 mg/L of Mg(OH)2 as CaCO3 fromthe limit of technology converts to the bicarbonate form,

(10.49)Equation (10.7) has been written as

(10.50)

Thus, the 16 mg/L of Mg(OH)2 as CaCO3 is equivalent to 16/50 = 0.32 meq/L of

The possibility exists that the 16 mg/L of Mg(OH)2 as CaCO3 will react in the

presence of excess bicarbonates to form magnesium carbonate; however, the K sp ofmagnesium carbonate is around 10−5 compared to that of the K sp of magnesiumhydroxide, which is 9(10−12) Thus, Mg(OH)2 will stay as Mg(OH)2 until recarbonation

the carbonate hardness of magnesium was precipitated Again, let all this mass bemeasured in kilograms Also, let [Mg(HCO3)2]mgnot, in mg/L, be the concentration

of magnesium bicarbonate not precipitated Thus,

(10.51)

is the volume of water treated in cubic meters

Mg(HCO3)2 ionizes to Mg(HCO3)2→ Mg2++ ; thus, the equivalent mass

of magnesium bicarbonate is Mg(HCO3)2/2 The magnesium bicarbonate not cipitated therefore produces:

pre-milligram equivalents of Mg2+ and , respectively

Now, also recall that MMgCa (= fMgCaM T Mg) mass of magnesium from the bonate hardness of magnesium precipitated As before, let this mass be measured

13.7 M T MgHCO3(1– fMgHCO3)

V

-=HCO3−

in kilograms Also, let [Mg2+]mgnot, in mg/L, be the concentration of magnesium notprecipitated Thus,

(10.52)Consider MgSO4 as representing the noncarbonate hardness of calcium species.This ionizes as MgSO4→ Mg2+ + ; thus, the equivalent mass of magnesium

in the noncarbonate hardness of magnesium is Mg/2 The magnesium not tated therefore produces:

precipi-milligram equivalents of Mg2+

Summing all the magnesium ion concentrations, the total concentration [Mg2+]meq

is

(10.53)

cubic meters [Mg2+]meq is then in milligram equivalents per liter

per liter, we obtain

V

+

-=

13.7 M T MgHCO3(1– fMgHCO3)

V

+