Background Chemistry and Fluid Mechanics doc

For oxidation–reduction reactions, the equivalent mass is defined as the mass of substance per mole of electrons involved Snoeyink and Jenkins, 1980.. The mass of any substance participa

Background Chemistry and Fluid Mechanics

As mentioned, this chapter discusses the background knowledge needed in order tounderstand the subsequent chapters of this book The student must have alreadygained this knowledge, but it is presented here as a refresher Again, the backgroundknowledge to be reviewed includes chemistry and fluid mechanics Before they arediscussed, however, units used in calculations need to be addressed, first This isimportant because confusion may arise if there is no technique used to decipher theunits used in a calculation

UNITS USED IN CALCULATIONS

In calculations, several factors may be involved in a term and it is important to keeptract of the units of each of the factors in order to ascertain the final unit of the term.For example, consider converting 88 kilograms to micrograms To make this con-version, several factors are present in the term for the calculation Suppose we makethe conversion as follows:

This is how it is done Focus on the right-hand side of the equation of the firstcalculation It is known that the unit of 88 is kg, and we want it converted to µg.

Remember that conversions follow a sequence of units For example, to convert

88 kg = 88 1000( ) 1000( ) 1000( ) = 88,000,000,000 µg

88 kg 88 kg 1000g

kg -

g -

mg -

26 Physical–Chemical Treatment of Water and Wastewater

kg to µg, the sequence might be any one of the following:

(3)(4)

In Sequence (3), the conversion follows the detailed steps: first, from kg to g,then from g to mg and, finally, from mg to µg Looking back to Equation (1), this

is the sequence followed in the conversion The first 1000 then refers to the g; thesecond 1000 refers to the mg; and the last 1000 refers to the µg Note that in thisscheme, the value of a given unit is exactly the equivalent of the previous unit Forexample, the value 1000 for the g unit is exactly the equivalent of the previous unit,which is the kg Also, the value 1000 for the mg unit is exactly the equivalent of itsprevious unit, which is the g, and so on with the µg

In Sequence (4), the method of conversion is a short cut This is done if thenumber of micrograms in a kilogram is known Of course, we know that there are 106micrograms in a kilogram Thus, using this sequence to convert 88 kg to micrograms,

we proceed as follows:

Example 1 Convert 88,000,000,000 µg to kg using the detailed-step and theshort-cut methods

Solution: Detailed step:

Note that the first (10−3) refers to the mg; the second (10−3) refers to the g; and thelast (10−3) refers to the kg There is no need to write the units specifically

1 60

 

  1 60

 

  1 24

 

  - 24,462 m3/d Ans

Background Chemistry and Fluid Mechanics 27

Now, let us turn to a more elaborate problem of substituting into an equation

Of course, to use the equation, all the units of its parameters must be known

Accordingly, when the substitution is done, these units must be satisfied For

exam-ple, take the following equation:

(5)

It is impossible to use the above equation if the units of the factors are not

known Thus, any equation, whether empirically or analytically derived, must always

have its units known In the above equation, the following are the units of the

factors:

I amperes, A

[C o] gram equivalents per liter, geq/m3

Q o cubic meters per second, m3/s

η dimensionless

m dimensionless

Now, with all the units known, it is easy to substitute the values into the equation;

the proper unit of the answer will simply fall into place

The values can be substituted into the equation in two ways: direct substitution

and indirect substitution Direct substitution means substituting the values directly

into the equation and making the conversion into proper units while already

substi-tuted Indirect substitution, on the other hand, means converting the values into their

proper units outside the equation before inserting them into the equation These

methods will be elaborated in the next example

Example 3 In the equation I = , the following values for the

factors are given: [C o] = 4000 mg/L of NaCl; Q o = 378.51 m3/d; η= 0.77, m= 400,

and M= 0.90 Calculate the value of I by indirect substitution and by direct

substi-tution

Solution: Indirect substitution:

In indirect substitution, all terms must be in their proper units before making the

28 Physical–Chemical Treatment of Water and Wastewater

And, now, substituting, therefore,

Direct substitution:

Note that the conversions into proper units are done inside the equation, and

that the conversions for [C o ] and Q o are inside pairs of braces {} Once accustomed

to viewing these conversions, you may not need these braces anymore

One last method of ascertaining units in a calculation is the use of consistent

units If a system of units is used consistently, then it is not necessary to keep track

of the units in a given calculation The proper unit of the answer will automatically

fall into place

The system of units is based upon the general dimensions of space, mass, and

time Space may be in terms of displacement or volume and mass may be in terms

of absolute mass or relative mass An example of absolute mass is the gram, and an

example of relative mass is the mole The mole is a relative mass, because it expresses

the ratio of the absolute mass to the molecular mass of the substance When the

word mass is used without qualification, absolute mass is intended

The following are examples of systems of units: meter-kilogram-second (mks),

meter-gram-second (mgs), liter-gram-second (lgs), centimeter-gram-second (cgs),

liter-grammoles-second (lgmols), meter-kilogrammoles-second (mkmols),

centi-meter-grammoles-second (cgmols), etc Any equation that is derived analytically

does not need to have its units specified, because the units will automatically

conform to the general dimension of space, mass, and time In other words, the

units are automatically specified by the system of units chosen For example, if the

mks system of units is chosen, then the measurement of distance is in units of

meters, the measurement of mass is in units of kilograms, and the measurement of

time is in seconds Also, if the lgs system of units is used, then the volume is in

liters, the mass is in grams, and the time is in seconds To repeat, if consistent

units are used, it is not necessary to keep track of the units of the various

factors, because these units will automatically fall into place by virtue of the

choice of the system of units The use of a consistent system of units is illustrated

in the next example

I 96,494 C [ ]Q o o η m/2( )

M

- 96,494 68( ) 0.0044( ) 0.77( ) 400/2( )

0.90 -

3

58.45 -(1000)

24 60 ( ) 60 ( ) -

0.90 -

=

122.84 A Ans

=

Example 4 The formula used to calculate the amount of acid needed to lowerthe pH of water is

Calculate the amount of acid needed using the lgmols system of units

Solution: Of course, to intelligently use the above equation, all the factorsshould be explained We do not need to do it here, however, because we only need

to make substitutions Because the lgmols units is to be used, volume is in liters,mass is in gram moles, and time is in seconds Therefore, the corresponding con-centration is in gram moles per liter (gmols/L) Another unit of measurement ofconcentration is also used in this equation, and this is equivalents per liter For thelgmols system, this will be gram equivalents per liter (geq/L)

Now, values for the factors need to be given These are shown below and notethat no units are given Because the lgmols system is used, they are understood to

be either geq/L or gmols/L Again, it is not necessary to keep track of the units; theyare understood from the system of units used

The literature shows confused definitions of equivalents and equivalent masses and

no universal definition exists They are defined based on specific situations and arenever unified For example, in water chemistry, three methods of defining equivalentmass are used: equivalent mass based on ionic charge, equivalent mass based on

acid–base reactions, and equivalent mass based on oxidation–reduction reactions(Snoeyink and Jenkins, 1980) This section will unify the definition of these terms

by utilizing the concept of reference species; but, before the definition is unified,

the aforementioned three methods will be discussed first The result of the discussion,then, will form the basis of the unification

Equivalent mass based on ionic charge In this method, the equivalent mass

is defined as (Snoeyink and Jenkins, 1980):

(6)

For example, consider the reaction,

Calculate the equivalent mass of Fe(HCO3)2

When this species ionizes, the Fe will form a charge of plus 2 and the bicarbonateion will form a charge of minus 1 but, because it has a subscript of 2, the total ioniccharge is minus 2 Thus, from the previous formula, the equivalent mass isFe(HCO3)2/2, where Fe(HCO3)2 must be evaluated from the respective atomicmasses The positive ionic charge for calcium hydroxide is 2 The negative ioniccharge of OH− is 1; but, because the subscript is 2, the total negative ionic chargefor calcium hydroxide is also 2 Thus, if the equivalent mass of Ca(OH)2 were to

be found, it would be Ca(OH)2/2 It will be mentioned later that Ca(OH)2/2 is notcompatible with Fe(HCO3)2/2 and, therefore, Equation (6) is not of universal appli-cation, because it ought to apply to all species participating in a chemical reaction.Instead, it only applies to Fe(HCO3)2 but not to Ca(OH)2, as will be shown later

Equivalent mass based on acid–base reactions In this method, the equivalent

mass is defined as (Snoeyink and Jenkins, 1980):

(7)

where n is the number of hydrogen or hydroxyl ions that react in a molecule

For example, consider the reaction,

Now, calculate the equivalent mass of H3PO4 It can be observed that H3PO4 converts

to Thus, two hydrogen ions are in the molecule of H3PO4 that react and theequivalent mass of H3PO4 is H3PO4/2, using the previous equation The number ofhydroxyl ions that react in NaOH is one; thus, using the previous equation, theequivalent mass of NaOH is NaOH/1

Equivalent mass molecular weight

ionic charge -

=

Fe HCO( 3)2+2Ca OH( )2→Fe OH( )2+2CaCO3+2H2O

Equivalent mass molecular weight

Equivalent mass based on oxidation–reduction reactions For oxidation–

reduction reactions, the equivalent mass is defined as the mass of substance per mole

of electrons involved (Snoeyink and Jenkins, 1980)

For example, consider the reaction,

The ferrous is oxidized to the ferric form from an oxidation state of +2 to +3 Thedifference between 2 and 3 is 1, and because there are 4 atoms of Fe, the amount ofelectrons involved is 1 × 4 = 4 The equivalent mass of Fe(OH)2 is then 4Fe(OH)2/4.Note that the coefficient 4 has been included in the calculation This is so, because

in order to get the total number of electrons involved, the coefficient must be included.The electrons involved are not only for the electrons in a molecule but for the electrons

in all the molecules of the balanced chemical reaction, and, therefore must accountfor the coefficient of the term For oxygen, the number of moles electrons involvedwill also be found to be 4; thus, the equivalent mass of oxygen is O2/4

Now, we are going to unify this equivalence using the concept of the referencespecies The positive or negative charges, the hydrogen or hydroxyl ions, and themoles of electrons used in the above methods of calculation are actually referencesspecies They are used as references in calculating the equivalent mass Note thatthe hydrogen ion is actually a positive charge and the hydroxyl ion is actually anegative charge From the results of the previous three methods of calculatingequivalent mass, we can make the following generalizations:

1 The mass of any substance participating in a reaction per unit of thenumber of reference species is called the equivalent mass of the substance,and, it follows that

2 The mass of the substance divided by this equivalent mass is the number

of equivalents of the substance

The expression molecular weight/ionic charge is actually mass of the substance

per unit of the reference species, where ionic charge is the reference species The

expression molecular weight n is also actually mass of the substance per unit of the

reference species In the case of the method based on the oxidation–reductionreaction, no equation was developed; however, the ratios used in the example areratios of the masses of the respective substances to the reference species, where 4,the number of electrons, is the number of reference species

From the discussion above, the reference species can only be one of two bilities: the electrons involved in an oxidation–reduction reaction and the positive(or, alternatively, the negative) charges in all the other reactions These species(electrons and the positive or negative charges) express the combining capacity orvalence of the substance The various examples that follow will embody the concept

possi-of the reference species

Again, take the following reactions, which were used for the illustrations above:

4Fe(OH)2+ O2+ 2H2O → 4Fe(OH)3Fe(HCO2)3+ 2Ca(OH)2→ Fe(OH)2+ 2CaCO3+ 2H2O

4Fe OH( )2+O2+2H2O→4Fe OH( )3

In the first reaction, the ferrous form is oxidized to the ferric form from an oxidationstate of +2 to +3 Thus, in this reaction, electrons are involved, making them thereference species The difference between 2 and 3 is 1, and since there are 4 atoms of

Fe, the amount of electrons involved is 1 × 4 = 4 For the oxygen molecule, its atomhas been reduced from 0 to −2 per atom Since there are 2 oxygen atoms in the molecule,the total number of electrons involved is also equal to 4 (i.e., 2 × 2 = 4) In both thesecases, the number of electrons involved is 4 This number is called the number of

reference species, combining capacity, or valence (Number of reference species will

be used in this book.) Thus, to obtain the equivalent masses of all the participatingsubstances in the reaction, each term must be divided by 4: 4Fe(OH)2/4, O2/4, 2H2O/4,and 4Fe(OH)3/4 We had the same results for Fe(OH)2 and O2 obtained before

If the total number of electrons involved in the case of the oxygen atom weredifferent, a problem would have arisen Thus, if this situation occurs, take theconvention of using the smaller of the number of electrons involved as the number

of reference species For example if the number of electrons involved in the case ofoxygen were 2, then all the participating substances in the chemical reaction wouldhave been divided by 2 rather than 4 For any given chemical reaction or series ofrelated chemical reactions, however, whatever value of the reference species ischosen, the answer will still be the same, provided this number is used consistently.This situation of two competing values to choose from is illustrated in the secondreaction to be addressed below Also, take note that the reference species is to betaken from the reactants only, not from the products This is so, because the reactantsare the ones responsible for the initiation of the interaction and, thus, the initiation

of the equivalence of the species in the chemical reaction

In the second reaction, no electrons are involved In this case, take the conventionthat if no electrons are involved, either consider the positive or, alternatively, thenegative oxidation states as the reference species For this reaction, initially considerthe positive oxidation state Since the ferrous iron has a charge of +2, ferrousbicarbonate has 2 for its number of reference species Alternatively, consider thenegative charge of bicarbonate The charge of the bicarbonate ion is −1 and becausetwo bicarbonates are in ferrous bicarbonate, the number of reference species is,again, 1 × 2 = 2 From these analyses, we adopt 2 as the number of reference speciesfor the reaction, subject to a possible modification as shown in the paragraph below.(Notice that this is the number of reference species for the whole reaction, not onlyfor the individual term in the reaction In other words, all terms and each term in achemical reaction must use the same number for the reference species.)

In the case of the calcium hydroxide, since calcium has a charge of +2 and thecoefficient of the term is 2, the number of reference species is 4 Thus, we have nowtwo possible values for the same reaction In this situation, there are two alternatives:the 2 or the 4 as the number of reference species As mentioned previously, eithercan be used provided, when one is chosen, all subsequent calculations are based onthe one particular choice; however, adopt the convention wherein the number ofreference species to be chosen should be the smallest value Thus, the number ofreference species in the second reaction is 2, not 4—and all the equivalent masses

of the participating substances are obtained by dividing each balanced term of thereaction by 2: Fe(HCO3)2/2, 2Ca(OH)2/2, Fe(OH)2/2, 2CaCO3/2 and 2H2O/2

Note that Ca(OH)2 has now an equivalent mass of 2Ca(OH)2/2 which is differentfrom Ca(OH)2/2 obtained before This means that the definition of equivalent mass

in Equation (6) is not accurate, because it does not apply to Ca(OH)2 The equivalentmass of Ca(OH)2/2 is not compatible with Fe(HCO3)2/2, Fe(OH)2/2, 2CaCO3/2, or2H2O/2 Compatibility means that the species in the chemical reaction can all berelated to each other in a calculation; but, because the equivalent mass of Ca(OH)2

is now made incompatible, it could no longer be related to the other species in thereaction in any chemical calculation In contrast, the method of reference speciesthat is developed here applies in a unified fashion to all species in the chemicalreaction: Fe(HCO3)2, Ca(OH)2, Fe(OH)2, CaCO3, and H2O and the resulting equiv-alent masses are therefore compatible to each other This is so, because all the speciesare using the same number of reference species

Take the two reactions of phosphoric acid with sodium hydroxide that follow.These reactions will illustrate that the equivalent mass of a given substance dependsupon the chemical reaction in which the substance is involved

Consider the positive electric charge and the first reaction Because Na+ of NaOH(remember that only the reactants are to be considered in choosing the referencespecies) has a charge of +1 and the coefficient of the term is 2, two positive charges(2 × 1 = 2) are involved In the case of H3PO4, the equation shows that the acidbreaks up into and other substances with one H still “clinging” to the PO4

on the right-hand side of the equation This indicates that two H+’s are involved inthe breakup Because the charge of H+ is +1, two positive charges are accordinglyinvolved In both the cases of Na+ and H+, the reference species are the two positivecharges and the number of reference species is 2 Therefore, in the first reaction,the equivalent mass of a participating substance is obtained by dividing the term(including the coefficient) by 2 Thus, for the acid, the equivalent mass is H3PO4/2;for the base, the equivalent mass is 2NaOH/2, etc

In the second reaction, again, basing on the positive electric charges and forming similar analysis, the number of reference species would be found to be +3.Thus, in this reaction, for the acid, the equivalent mass is H3PO4/3; for the base, theequivalent mass is 3NaOH/3, etc., indicating differences in equivalent masses withthe first reaction for the same substances of H3PO4 and NaOH Thus, the equivalentmass of any substance depends upon the chemical reaction in which it participates

per-In the previous discussions, the unit of the number of reference species was notestablished A convenient unit would be the mole (i.e., mole of electrons or mole ofpositive or negative charges) The mole can be a milligram-mole, gram-mole, etc Themass unit of measurement of the equivalent mass would then correspond to the type

of mole used for the reference species For example, if the mole used is the gram-mole,the mass of the equivalent mass would be expressed in grams of the substance pergram-mole of the reference species; and, if the mole used is the milligram-mole,

H3PO4+2NaOH→2Na++HPO42−+2H2O

H3PO4+3NaOH→3Na++PO43−+3H2O

HPO42−

the equivalent mass would be expressed in milligrams of the substance per mole of the reference species and so on

milligram-Because the reference species is used as the standard of reference, its unit, themole, can be said to have a unit of one equivalent From this, the equivalent mass

of a participating substance may be expressed as the mass of the substance perequivalent of the reference species; but, because the substance is equivalent to thereference species, the expression “per equivalent of the reference species” is thesame as the expression “per equivalent of the substance.” Thus, the equivalent mass

of a substance may also be expressed as the mass of the substance per equivalent

of the substance

Each term of a balanced chemical reaction, represents the mass of a participatingsubstance Thus, the general formula for finding the equivalent mass of a substanceis

Example 5 Water containing 2.5 moles of calcium bicarbonate and 1.5 moles

of calcium sulfate is softened using lime and soda ash How many grams of calcium

carbonate solids are produced (a) using the method of equivalent masses and (b)

using the balanced chemical reaction? Pertinent reactions are as follows:

Solution:

number of reference species = 2

number of equivalents of substance -

=

term in balanced reactionnumber of moles of reference species -

=

Ca HCO( 3)2+Ca OH( )2→2CaCO3↓ 2H+ 2OCaSO4+Na2CO3→CaCO3+Na2SO4

a

( ) Ca HCO( 3)2+Ca OH( )2→2CaCO3↓ 2H+ 2O

Therefore, eq mass of CaCO3

2CaCO32 - 100

eq mass of Ca HCO( 3)2

Ca HCO( 3)22 - 40+2 1[ +12+3 16( )]

2 -162

2 - 81

qeq of Ca HCO( 3)2

2.5 162( )81 - 5 geq of CaCO3

g of CaCO3 solids = 5 100( ) = 500 g Ans

CaSO4+Na2CO3→CaCO3+Na2SO4

number of reference species = 2

Several methods are used to express concentrations in water and wastewater and it

is appropriate to present some of them here They are molarity, molality, mole fraction,mass concentration, equivalents concentration, and normality

Molarity Molarity is the number of gram moles of solute per liter of solution,

where from the general knowledge of chemistry, gram moles is the mass in gramsdivided by the molecular mass of the substance in question Take note that thestatement says “per liter of solution.” This means the solute and the solvent are taken

together as a mixture in the liter of solution The symbol used for molarity is M.

For example, consider calcium carbonate This substance has a mass density equals 2.6 g/cc Suppose, 35 mg is dissolved in a liter of water, find the correspondingmolarity

The mass of 35 mg is 0.035 g Calcium carbonate has a molecular weight of

100 g/mol; thus, 0.035 g is 0.035/100 = 0.00035 gmol The volume corresponding

to 0.35 g is 0.035/2.6 = 0.0135 cc = 0.0000135 L and the total volume of themixture then is 1.0000135 L Therefore, by the definition of molarity, the corre-sponding molarity of 35 mg dissolved in one liter of water is 0.00035/1.0000135

= 0.00035 M.

Molality Molarity is the number of gram moles of solute per 1000 g of solvent.

Take note of the drastic difference between this definition of molality and thedifinition of molarity The solvent is now “separate” from the solute The symbol

used for molality is m.

Therefore, eq mass of CaCO3 CaCO3

2 - 50

eq mass of CaSO4 CaSO4

2 - 40+32+4 16( )

2 - 136

2 - 68

qeq of CaSO4 1.5 136( )

68 - 3 geq of CaCO3

g of CaCO3 solids = 3 50( ) = 150 g Ans

Total grams of CaCO3 = 500+150 = 650 Ans b

( ) Ca HCO( 3)2+Ca OH( )2→2CaCO3↓ 2H+ 2O

g of CaCO3 solids 2CaCO3

Ca HCO( 3)2 - 2.5( ) 162( ) 500

CaSO4+Na2CO3→CaCO3+Na2SO4

g of CaCO3 solids CaCO3

CaSO4 - 1.5( ) 136( ) 150

Total grams of CaCO3 = 500+150 = 650 Ans

ρCaCO3

Now, consider the calcium carbonate example above, again, and find the sponding molality The only other calculation we need to do is to find the number

corre-of grams corre-of the liter corre-of water To do this, an assumption corre-of the water temperaturemust be made Assuming it is 5°C, its mass density is 1.0 g/cc and the mass of oneliter is then 1000 g Thus, the 0.00035 gmol of calcium carbonate is dissolved in

1000 gm of water This is the very definition of molality and the corresponding

molality is therefore 0.00035m Note that there is really no practical difference

between molarity and molality in this instance Note that the mass density of waterdoes not vary much from the 5°C to 100°C

Mole fraction Mole fraction is a method of expressing the molar fractional

part of a certain species relative to the total number of moles of species in the

mixture Letting n i be the number of moles of a particular species i, the mole fraction

of this species x i is

(8)

N is the total number of species in the mixture.

Example 6 The results of an analysis in a sample of water are shown in thetable below Calculate the mole fractions of the respective species

Solution: The calculations are shown in the table, which should be explanatory

self-Mass concentration Generally, two methods are used to express mass

concen-tration: mass of solute per unit volume of the mixture (m/v basis) and mass of the

Ca(HCO 3 ) 2 150 40.1(2) + 2{1 + 12 + 3(16)} = 202.2 0.74a 0.32bMg(HCO 3 ) 2 12.0 24.3(2) + 2{1 + 12 + 3(16)} = 170.6 0.07 0.03

solute per unit mass of the mixture (m/m basis) In environmental engineering, themost common expression in the m/v basis is the mg/L The most common in the m/mbasis is the ppm, which means parts per million In other words, in a ppm, there isone mass of the solute in 106 mass of the mixture.

One ppm for a solute dissolved in water can be shown to be equal to one mg/L.This is shown as follows: 1 ppm = (1 mg)/(106

mg) = (1 mg)/(103

g) The massdensity of water at 5°C is 1.0 g/cc Therefore,

The mass density of water decreases from 1.0 g/cc at 5°C to 0.96 g/cc at 100°C.Thus, for practical purposes,

Molar concentration In concept, molar concentrations can also be expressed

on the m/v basis and the m/m basis; however, the most prevalent practice in ronmental engineering is the m/v basis Molar concentration, then, is the number ofmoles of the solute per unit volume of the mixture There are several types of moles:milligram-moles, gram-moles, tonne-moles, and so on corresponding to the unit ofmass used In chemistry, the gram moles is almost exclusively used When the type

envi-of moles is not specified, it is understood to be gram moles So, normally, molarconcentration is expressed in gram moles per liter (gmmols/L)

An important application of molar concentration is in a molar mass balance.For example, in the removal of phosphorus using alum, the following series ofreactions occurs:

An indicator of the efficiency of removal is that the concentration of phosphorus

in solution must be the minimum To find this phosphorus in solution, it is necessary

to perform a phosphate (PO4) molar mass balance in solution using the abovereactions AlPO4(s) is a solid; thus, not in solution and will not participate in thebalance The other species, however, are in solution and contain the PO4 radical

respective formulas; thus, the formulas are said to contain the PO4 radical Because theycontain this PO4 radical, they will participate in the molar mass balance Let

1 ppm = 1 mg/L

Al3++PO43−
AlPO4 s( )↓AlPO4 s( )↓
Al3++PO43−

represent the total molar concentration of the phospate radical contained in all thespecies Then, the molar mass balance on PO4 becomes

The symbol needs to be explained further, since this mode of ing is used in the unit processes part of this book First, [ ] is read as “the concentration

subscript-of.” The symbol sp stands for species and its first subscript, PO4, stands for the type

of species, which is the PO4 radical The second subscript, Al, stands for the “reason”for the existence of the type of species In other words, the use of alum (the Al) isthe reason for the existence of the phosphate radical, the type of species

Example 7 In removing the phosphorus using a ferric salt, the following series

Example 8 State the symbol in words

Solution: is the total molar concentration of the radical as aresult of using a ferric salt Ans

Example 9 The complexation reaction of the calcium ion with the carbonatespecies, OH−, and are given below:

spPO

4 Al[ ]=[PO43−]+[HPO42−]+[H2PO4−]+[H3PO4]

[spPO

4 Al]

Fe3+ PO4

3−
FePO4↓+

[spPO4FeIII]

3−

SO4 2−

CaCO3o
Ca2++CO32−

CaHCO3+
Ca2++HCO3−CaOH+
Ca2++OH−CaSO4o
Ca2++SO42−

Write the molar mass balance for the total molar concentration of Ca.

Solution: Let [CaT] represent the total molar concentration of Ca Therefore,

Equivalent concentration and normality This method of expressing

concen-trations is analogous to molar concenconcen-trations, with the solute expressed in terms ofequivalents One problem that is often encountered is the conversion of a molar

concentration to equivalent concentration Let [C] be the molar concentration of any

substance, where the symbol [ ] is read as “the concentration of.” Convert this toequivalent concentration

To do the conversion, first convert the molar concentration to mass concentration

by multiplying it by the molecular mass (MM) Thus, [C] (MM) is the corresponding

mass concentration By definition, the number of equivalents is equal to the massdivided by the equivalent mass (eq mass) Therefore, the equivalent concentration,

[C]eq, is

The concentration expressed as geq/L is the normality The symbol for normality is N.

Example 10 The concentration of Ca(HCO3)2 is 0.74 gmol/L Convert thisconcentration to geq/L

Solution:

No of reference species = 2

Another problem often encountered in practice is the conversion of equivalentconcentration to molar concentration Let us illustrate this situation using the car-bonate system The carbonate system is composed of the species ,and H+ In addition, Ca2+ may also be a part of this system Note that OH and H+are always part of the system, because a water solution will always contain thesespecies To perform the conversion, the respective equivalent masses of the speciesshould first be found In order to find the number of reference species, the pertinent

2 -

C

[ ]eq

0.74 202.2( )202.2 2 - - 1.48 geq/L

CO32−, HCO3−, OH

chemical reactions must all be referred to a common end point For the carbonatesystem, this end point is the methyl orange end point when H+ is added to the system

to form H2CO3

The reaction to the end point of is

Because the number of reference species from this reaction is 1, the equivalent mass

of is For , the reaction to the end point is

This reaction gives the number of reference species equal to 2; thus, the equivalentmass of is CO3/2 The reaction of H+

and OH− to the end point gives theirrespective equivalent masses as H/1 and OH/1 The reaction of Ca2+

in the carbonatesystem is

This gives the equivalent of Ca2+ as Ca/2 The conversion from equivalent to molarconcentrations is illustrated in the following example

Example 11 An ionic charge balance in terms of equivalents for a carbonatesystem is shown as follows:

Convert the balance in terms of molar concentrations

−

HCO31 -; (OH)eq mass

OH1 -

H+

1 ; (Ca2+)eq mass Ca

2 -

- HCO3

−[ ] HCO( 3)HCO 3

1 - - OH

−[ ] OH( )OH 1 - -

1 - Ca

2+

Ca 2 - -+

=

2 CO[ 32−]+[HCO3−]+[OH] = [ ] 2 CaH+ + [ 2+] Ans

A CTIVITY AND A CTIVE C ONCENTRATION

In simple language, activity is a measure of the effectiveness of a given species in

a chemical reaction It is an effective or active concentration and is proportional to

concentration It has the units of concentration Since activity bears a relationship

to concentration, its value may be obtained using the value of the correspondingconcentration This relationship is expressed as follows:

(9)

where sp represents any species involved in the equilibria such as Ca2+, , and so on, in the case of the carbonate equilibria The pair of braces, { }, is read asthe “activity of ” and the pair of brackets, [ ], is read as “the concentration of;” γ iscalled activity coefficient

The activity coefficient expresses the effect of ions on the reactive ability of aspecies As the species become crowded, the reactive ability or effectiveness of thespecies to react per unit individual of the species is diminished As they becomeless concentrated, the effectiveness per unit individual is improved Thus, at verydilute solutions, the activity coefficient approaches unity; at a more concentratedsolution, the activity coefficient departs from unity

Because of the action of the charges upon each other, the activity of the ionizedparticles is smaller than those of the unionized particles Ionized particles tend tomaintain “relationship” between counterparts, slowing down somewhat their inter-actions with other particles Thus, Ca2+ and , which are ionization products ofCaCO3, have activity coefficients less than unity; they tend to maintain relationshipwith each other rather than with other particles Thus, their activity with respect toother particles is diminished On the other hand, a solid that is not ionized such asCaCO3 before ionization, has an activity coefficient of unity; a liquid or solvent such

as water (not ionized) has an activity coefficient of unity Gases that are not ciated have activity coefficients of unity (Because the activity coefficients are unity,all molar concentrations of these unionized species have a unit of activity.) In otherwords, unionized particles are free to interact with any other particles, thus havingthe magnitude of the highest possible value of the activity coefficient They are notrestricted to maintain any counterpart interaction, because they do not have any

Let a reactant be represented by the solid molecule A a B b As this reactant is mixedwith water, it dissolves into its constituent solute ions The equilibrium dissolutionreaction is

(10)

The ratio is called the reaction quotient Note that in the definition of

reaction quotients, the activities of the reactants and products are raised to their

respective coefficients At equilibrium, the reaction quotient becomes the equilibriumconstant Thus, the equilibrium constant is

(11)

Note the pair of braces denoting activity Because A a B b is a solid, its activity is

unity For this reason, the product K{A a B b } is a constant and is designated as K sp

K sp is called the solubility product constant of the equilibrium dissolution reaction.

Equation (11) now transforms to

(12)

At equilibrium, neither reactants nor products increase or decrease with time Thus,

K sp’s, being constants, can be used as indicators whether or not a given solid willform or dissolve in solution

For example, CaCO3 has a K sp of 4.8(10−9) at 25°C This value decreases to2.84(10−9) at 60°C The equilibrium reaction for this solid is

Assuming the reaction is at equilibrium at 25°C, the ions at the right side of thereaction will neither cause CaCO3 to form nor dissolve At 25°C, by the definition

of the solubility product constant and, since the reaction is assumed to be in equilibrium,the product of the activities will equal 4.8(10−9) As the temperature increases from

25°C to 60°C, the product of the activities decreases from 4.8(10−9) to 2.84(10−9).This means that there are fewer particles of the ions existing than required to maintainequilibrium at this higher temperature As a consequence, some of the particles willcombine to form a precipitate, the CaCO3 solid

Table 1 shows values of K sp’s of solids that are of importance in ascertainingwhether or not a certain water sample will form or dissolve the respective solids

The subscripts s and aq refers to “solid” and “aqueous,” respectively In writing the

chemical reactions for the discussions in this book, these subscripts will not beindicated unless necessary for clarity

Of all the K sp’s discussed previously, the one involving the carbonate systemequilibria is of utmost importance in water and wastewater treatment This is becausecarbon dioxide in the atmosphere affects any water body As carbon dissolves inwater, the carbonate system species and are formed Cationswill then interact with these species and, along with the H+ and OH− that alwaysexist in water solution, complete the equilibrium of the system is a mixture

of CO2 in water and H2CO3 (note the absence of the asterisk in H2CO3) The CO2

in water is written as CO2(aq) and is carbonic acid

Another important application of the equilibrium constant K in general, [Equation

(11)], is in the coagulation treatment of water using alum, Al2(SO4)3·14 H2O (The

“14” actually varies from 13 to 18.) In coagulating a raw water using alum, a number

of complex reactions are formed by the Al3+ ion These reactions are as follows:

(13)

TABLE 1

Solubility Product Constants of Respective Solids at 25 °C

1.8(10−10) Chloride analysis 1.9(10−33) Coagulation

10−10 Sulfate analysis 4.8(10−9) Hardness removal, scales 7.9(10−6) Hardness removal 2.4(10−5) Flue gas desulfurization

10−25 Phosphate removal 5.0(10−6) Phosphate removal 3.9(10−11) Fluoridation 6.7(10−31) Heavy metal removal 5.6(10−20) Heavy metal removal 1.1(10−36) Coagulation, iron removal,

corrosion 7.9(10−15) Coagulation, iron removal,

corrosion

10−5 Hardness removal, scales 1.5(10−11) Hardness removal, scales 4.5(10−14) Manganese removal 1.6(10−14) Heavy metal removal 4.5(10−17) Heavy metal removal

From A P Sincero and G A Sincero (1996) Enviromental Engineering: A Design

Approach Prentice Hall, Upper Saddle River, NJ, 42.

and apply at 25°C Note that the subscript c is used for the equilibrium

constants of the complexes; it stands for “complex.” Al(OH)3(s) is not a complex;thus, does not contain the subscript c.

The previous equations can be used to determine the conditions that will allowmaximum precipitation of the solid represented by Al(OH)3 The maximum precip-itation of Al(OH)3 will produce the utmost clarity of the treated water To allow forthis maximum precipitation, the concentrations of the complex ions Al(OH)2+,

, Al13(OH)34, Al , and and Al3+ must be held to aminimum This will involve finding the optimum pH of coagulation This optimum

pH may be determined as follows:

It is obvious that the complex ions contain the Al atom Thus, the technique is

to sum up their molar concentrations in terms of their Al atom content Once theyhave been summed up, they are then eliminated using the previous Κ equations,Eqs (13) to (18), with the objective of expressing the resulting equation in terms

of the constants and the hydrogen ion concentration Because the K constants are

constants, the equations would simply be expressed in terms of one variable, thehydrogen ion concentration, and the equation can then be easily differentiated toobtain the optimum pH of coagulation

7Al3+ 17H2O
Al7(OH)17

4+

17Η+++

Gleaning from Eqs (13) to (18), a molar mass balance on the aluminum atom

may be performed Let spAl represent all the species that contain the aluminum atomstanding in solution Thus, the concentration of all the species containing the alu-minum atom, is

(19)

Note that, because we are summing the species containing the aluminum atom,the coefficients of the terms of the previous equation contain the number of aluminumatoms in the respective species For example, contains 7 aluminum atoms;thus, its coefficient is 7 Similarly, contains 13 aluminum atoms; there-fore, its coefficient is 13 This explanation holds for the other species as well.Using Eqs (13) to (18) and the relations of molar concentration and activity,Equation (9), the following equations are obtained [for the purpose of eliminatingthe complex ions in Equation (19)]:

3

H+[ ]3

γAlK w

3 -

2

H+[ ]2

γAl OH ( )c K w

3 -

KAl

7 ( OH )17c K sp,Al OH( )

3 7

γH 4

H+[ ]4

γAl7( OH )17c K w

21 -

13

Al3+[ ]13

γAl13( OH )34cγH

34

H+[ ]34 -

KAl

13 ( OH )34c K sp,Al OH( )

3 13

γH 5

H+[ ]5

γAl13( OH )34c K w

39 -

are, respectively, theactivity coefficients of the aluminum ion and the hydrogen ion and the complexes

solubility product constant of the solid Al(OH)3(s) and K w is the ion product of water

KAl2(OH)2c K sp,Al OH2 ( )3γH

4

H+[ ]4

γAl2( OH )2c K w

6 -

=

γAl, γH, γAl(OH) c, γAl7(OH)17c, γAl13(OH)34c,γAl(OH)4c, γAl2(OH)2c

(OH)174+, (OH)345+, Al(OH)4− Al2(OH)24+ K sp,Al(OH)

The next part of the equation, , should be easy to derive

and will no longer be pursued Note that K w came from the ion product of waterthat reads:

expressed in terms of the K’s and the hydrogen ion concentration This equation

may then be differentiated and equated to zero to obtain the optimum pH We willnot, however, complete this differentiation and equate to zero in this chapter, butwill do this in the unit processes part of this book

(32)

Still another important application of the concept of K equilibrium constants is

the coprecipitation of FePO4 and Fe(OH)3 in the removal of phosphorus from water

As in the case of coagulation using alum, it is desired to have a final equation that

is expressed only in terms of the constants and the hydrogen ion Once this is done,the equation can then also be manipulated to obtain an optimum pH for the removal

of phosphorus

In phosphorus removal, the phosphorus must be in the phosphate form and,because the reaction occurs in water, the phosphate ion originates a series of reactionswith the hydrogen ion The series is as follows:

γAlK w

3 - KAl OH( )c K sp,Al OH( )3γH

2

H+[ ]2

γAl OH ( )c K w

3 -+

γAl7( OH )17c K w

21 -+

13KAl13(OH)34c K sp,Al OH13 ( )3γH

5

H+[ ]5

γAl13( OH )34c K w

39 -

γAl OH ( )4cγH[ ]H+ -+

2KAl2(OH)2c K sp,Al OH2 ( )3γH

4

H+[ ]4

γAl2( OH ) 2c K w

6 -+

- 10–12.3

(35)

From the previous equations, a molar mass balance on the phosphate radicalmay be performed Let represent the species in solution containing PO4species, using a ferric salt as the precipitant (Note the FeIII as the second subscript,indicating that a ferric salt is used.) Therefore,

, and [H3PO4] may be eliminated by expressing them in terms

of the K constants using Eqs (33) to (35) The results are:

KH

3 PO4 - γPO4γH