CIVE 350Water Chemistry

Pascal E. Saikaly

Department of Civil and Environmental Engineering

American University of Beirut

OutlineWater Chemistry

• Introduction• Physical properties of water• Chemical reactions

• Acid/Base, Precipitation/Dissolution, Complexation, Oxidation/Reduction

• Chemical units• Stoichiometry• Chemical equilibrium

• Precipitation reactions• Acid-Base reactions• The carbonate system• Solubility of gases in water

Water chemsitry• Almost every pollution problem that we face has a chemical basis (e.g. greenhouse effect, ozone depletion, toxic wastes, groundwater contamination, disinfection, softening, air pollution, and acid rain).

• A knowledge of the fundamentals of water chemistry is essential to the understanding of processes governing water quality and wastewater treatment.

Physical Properties of Water• Density: mass per unit volume (Kg/m3)

• Specific gravity

• The subscript denotes the density of water at 3.98oC, 1,000 Kg/m3.

• A measure of friction is viscosity (higher friction: harder to pump liquid)

• Dynamic viscosity or absolute viscosity, μ, has dimensions of mass per unit length per time, with units Pa . S.

• Kinematic viscosity (m2/s)

VMρ =

0ρρS =

ρμν =

Physical properties of water at 1 atm

Chemical Reactions• Important types of chemical reactions in environmental

engineering

• Acid/Base (e.g. dissociation of bicarbonate)HCO3

- H+ + CO32-

• Precipitation/dissolutionCaCO3 (s) Ca2+ + CO3

2-

• Ion-association or complexationFe3+ + OH- FeOH2+

• Redox reactions: Oxidation/Reduction (valence changes and transfer of electrons)Fe Fe2+ + 2e- (Oxidation, loss of electrons)2H+ + 2e- H2(g) (Reduction, gain of electrons)

Chemical Units• Methods of expressing concentration of a given material in

aqueous solution

• One mole of a substance contains: 6.023 X 1023 molecules.

• Number of moles = mass/molecular weight (g)/(g/mole)

• Molarity, M (mol/L): The number of moles of a specific species per liter of solution.

• 1-molar solution (1 M) is 1 mole/L• Molarity is related to mg/L by

mg/L = molarity X molecular weight X 103

mg/L = (moles/L) X (g/mole) X (103 mg/g)

• Molality, m (mol/Kg): The number of moles of a specific species per Kg of solvent.

Chemical units• Normality, N (equivalents/L): The number of equivalent weights of acid, base, or

redox-active species per liter of solution.

or

where, 1. For Acid/Base rxnn = number of protons (H+) or hydroxyl ions (OH-) that react in an acid-base reaction

2. For Redox rxnn = number e- transferred per mole of species

3. For precipitation rxnn = the valance of the element in question (see table next page)

4. For compoundsn = number of hydrogen ions that would be required to replace the cation

nweightMolecularweightEquivalent =

weightEquivalentliterpersubstanceofMassNormality = nMolarityNormality ×=

Typical valences of elements and compounds in water

Example 1Find the normality of the following solutionsA) 155 mg Ca3(PO4)2/liter given that Ca3(PO4)2

participates in the dissolution rxnCa3(PO4)2 3Ca2+ + 2PO4

3-

B) 49 mg H3PO4/L with respect the rxnH3PO4 2H+ + HPO4

2-

C) 45 mg CO32-/L with respect the rxn

CO32- + H2O HCO3

- + OH-

D) Equivalent weight of O2 for the reduction rxnO2 + 4H+ +4e- 2H2O

E) Equivalent weight of Ca2+

Example 1A) n = 6. Each Ca3(PO4)2 forms 6 + and 6 – charges. MW of

Ca3(PO4)2 is 310 g/mole. Equivalent weight = (310 g/mole)/(6 eg/mole) = 51.67 g/eg or 51.67 mg/meqNormality = (155 mg/L)/(51.67 mg/meq) = 3 meq/L

B) n = 2. Equivalent weight = (98 g/mole)/ (2 eq/mole) = 49 g/eqor 49 mg/meqNormality = (49 mg/L)/(49 mg/meq) = 1 meq/L

C) n =1. Equivalent weight = (60 g/mole)/(1 eq/mole) = 60 g/eq or 60 mg/meqNormality = (45 mg/L)/(60 mg/meq) = 0.75 meq/L

D) n = 4. Equivalent weight = (32 g/mole)/(4 eq/mole) = 8 g/eqE) n = 2. Equivalent weight = ( 40.08 g/mole)/(2 eq/mole) = 20.04

g/eq

Example 2Find the weight of sodium bicarbonate, NaHCO3, necessary to make a 1 M solution. Find the normality of the solution.

NaHCO3 Na+ + HCO3-

HCO3- H+ + CO3

2-

HCO3- + H2O H2CO3 + OH-

Example 2The molecular weight of NaHCO3 is 84. using the equationmg/L = molarity X molecular weight X 103

= (1 mole/L)(84 g/mole)(103 mg/g) = 84,000HCO3

- is able to give or accept only one proton; therefore n = 1. Equivalent weight = (84 g/mole)/(1 eq/mole) = 84 g/eq or 84 mg/meqNormality = (84,000 mg/L)/(84 mg/meq) = 1000 meq/L or 1 eq/L

Stoichiometry• Chemical reaction

• Qualitative information: Which chemicals are interacting to produce which end products

• Quantitative: Principle of conservation of mass

• Stoichiometry: The balancing of equations so that the same number of each kind of atom appears on each side of the equation and the subsequent calculations, which can be used to determine amounts of each compound involved.

• Atomic weight of an atom is the mass of the atom measured in atomic mass units.

• Molecular weight of a molecule is the simply the sum of the atomic weights of all the constituent atoms (e.g. CH4, atomic weight of C is 12 and atomic weight of H is 1. Therefore, the MW of methane = 12 + 4(1) = 16 g/mole).

StoichiometryVarious ways to express the oxidation of methane• CH4 + 2O2 CO2 + 2H2O

1 molecule + 2 molecules 1 molecule of + 2 molecules ofof methane of oxygen carbon dioxide water

or,1 mole + 2 mole 1 mole of + 2 moles ofof methane of oxygen carbon dioxide water

or , 16 g + 64 g 44 g of + 36 g ofof methane of oxygen carbon dioxide water

The mass is conserved in the last expression ( 80 g = 80 g)

Example 3Consider a 1.67 X 10-3 M glucose solution that is completely oxidized to CO2 and H2O. Find the mg/L of oxygen required to complete the reaction.

C6H12O6 + 6O2 6CO2 + 6H2O180 g 192g 264g 108 g

It takes 192 g of oxygen to oxidize 180 g of glucose

mg/L = molarity X molecular weight X 103

= (1.67 X 10-3 M )(180 g/mole)(103 mg/g) = 301 mg/LSo the oxygen requirement would be300 mg/L X (192 g O2/180 g glucose) = 321 mg/L O2

Chemical Equilibrium• Different approaches may be used in predicting aqueous chemistry

speciation (i.e. concentration of products & reactants)• Chemical kinetics: The rate at which a reaction proceeds

toward equilibrium• Chemical equilibrium: tell you what the chemistry will be at

equilibrium, but not the kinetic rate at which the system reaches the equilibrium state. This is the most widely used method for obtaining chemical speciation.

• Most chemical reactions are, to some extent, reversible, proceeding in both directions at once. When products are being formed on the right at the same rate as they are being formed on the left, the reaction is said to have reached equilibrium .

aA + bB cC + dDThe small letters a, b, c, and d are coefficients corresponding to the number of molecules or ions

Chemical EquilibriumaA + bB cC + dD

At equilibrium we can write

where the [ ] represents the concentration of the substance at equilibrium, expressed in moles per liter (molarity). DO NOT USE mg/L.

K is the equilibrium constantK: Dissociation constant or ionization constantKs: Solubility product KH: Henry’s constant; equilibrium constant describing the concentration of gas in the water in equilibrium with the concentration of gas in the air.

ba

dc

[B][A][D][C]K =

logXpX10X pX

−== −

Precipitation ReactionsAll solids are to some degree soluble, although some are much more than others (e.g. NaCl is very soluble, whereas AgCl are very insoluble).

A generalized equation describing the equilibrium condition in which a solid is dissociating into its ionic components (dissolution) at the same rate that ionic compounds are recombining into the solid form (precipitation)

Solid aA + bB

As long as there is still solid present at equilibrium, its effect can be incorporated into the equilibrium constant

Ks values are often reported as pKs

[solid][B][A]K

ba=

ss logKpK −=

bas [B][A]K =

Selected solubility product constants at 25oC

Solid aA + bB• If we place solid in water, for every a moles of A that dissolves, b

moles of B will dissolve until equilibrium is reached.• When precipitating ions, it is possible to have a higher

concentration of ions in solution than dictated by the solubility product. This is called a supersaturated solution.

Solubility product constants at 25oC

Example 4How many mg/L of PO4

3- would be in solution at equilibrium with AlPO4(s)?

AlPO4(s) Al3+ + PO43-

pKs for aluminum phosphate is 20 (see table in previous slides)

For every mole of AlPO4(s) that dissolves, one mole of Al3+ and one mole of PO4

3- are released into solution. At equilibrium, the molar concentration of Al3+ and PO4

3- in solution will be equal.

Solving for X, we find PO43- = 10-10 moles per liter

mg/L = molarity X molecular weight X 103 = (10-10 moles/L)(95 g/mole)(103

mg/g) = 9.5 X 10-6

]][PO[Al10K 34

320s

−+− ==

X][PO][Al 34

3 == −+

220s X10K == −

Example 5If 50 mg of CO3

2- and 50 mg of Ca2+ are present in 1 L of water, what will be the final (equilibrium) concentration of Ca2+?

CaCO3(s) Ca2+ + CO32-

Molecular weight of Ca2+ is 40.08 and CO32- is 60.01, resulting in

initial molar concentration of 1.25 X 10-3 moles/L and 8.33 X 10-4

moles/L for Ca2+ and CO32-, respectively.

For every mole of Ca2+ that is removed from solution, one mole of CO3

2- is removed from solution. If the amount removed is given by Z, then

]][CO[Ca1010K 23

28.305pKs s −+−− ===

4

663

2

236

4398.305

108.282

)104(1.04104.34102.08

2a4acbbZ

0Z)Z10(2.08101.04

Z]10Z][8.3310[1.25104.9510

−

−−−

−−

−−−−

×=

×−×±×=

−±−=

=+×−×

−×−×=×=

Example 5So that the final Ca2+ concentration is

[Ca2+] = 1.25 X 10-3 – 8.28 X 10-4 = 4.22 X 10-4 M

or,

mg/L = molarity X molecular weight X 103 = (4.22 X 10-4

moles/L)(40 g/mole)(103 mg/g) = 16.9 mg/L

Acid-Base ReactionIn 1923, Bronsted and Lowry suggested a new conceptof acid and base behavior. They proposed that an acid is any substance that can donate a proton to another substance.

HNO3(aq) + H2O(l) H3O+(aq) + NO3-(aq)

HNO3(aq) H+(aq) + NO3-(aq)

And a base is a substance that can accept a proton from any other substance.

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

If a compound is stronger acid than water, then water will act as a base. If a compound is a strong base than water, then water will act as an acid.The acid/base chemistry centers on water and it is important to know how strong an acid water is

H2O(l) H+(aq) + OH-(aq)

O][H]][OH[HK

2

−+=

Acid-Base ReactionSince water dissociates only slightly, the molar concentration after ionization is not changed enough to be a of significance, so [H2O] is essentially constant and can be included in the equilibrium constant

where Kw is the dissociation constant of water (pKw = 14). Kw is temperature dependent.

A solution is said to be acidic if [H+] is greater than [OH-], neutral if equal, and basic if [H+] is less than [OH-].Neutral solution , then [H+] = [OH-] = 10-7 M.

A convenient expression of the hydrogen ion (proton) is pH.

C25at101K]][OH[H o14w

−−+ ×==

pH10][Hwhere]log[HpH −++ =−=

Acid-Base ReactionTherefore, a neutral solution at 25oC has a pH of 7, an acidic solution has a pH < 7, and a basic solution has a pH > 7.

Strong acids

Note the use of single arrow to signify that, for practical purposes, we may assume that the reaction proceeds completely to the right.

Acid-Base ReactionSelected weak acid dissociation constants at 25oC

Weak acids are acids that do not completely dissociate in water. An equilibrium exists between the dissociated ions and undissociated compound. The reaction of a week acid is

HA(aq) H+(aq) + A-(aq)

aaa logKpK[HA]

]][A[HK −==−+

Example 6If 100 mg of H2SO4 (MW = 98) is added to 1L of water, what is the final pH?

H2SO4(aq) 2H+ + SO42-

Therefore 2 X (1.02 X 10-3)M H+ is produced. The pH is:pH = -log(2.04 X 10-3) = 2.69.

mole/L101.02mg/g101

98g/mole1

H2O1L100mg 3

3−×=

Example 7If 15 mg/L HOCl (hypochlorous acid) is added to a potable water for disinfection and the final pH is 7.0, what percent of the HOCl is not dissociated ? Assume the temperature of water is 25oC.

HOCl(aq) H+(aq) + OCl-(aq)pKa = 7.54

Solving for the HOCL concentration[HOCl] = 3.47[OCl-]

87.54a 102.8810K −− ×==

[HOCL]]][OCL[HKa

−+=

[HOCL]]][OCL[10102.88

78

−−− =×

Example 7Since the fraction of HOCl that has not dissociated plus the OCl-that was formed by the dissociation must, by the law of conservation of mass, equal 100% of the original HOCL added[HOCl] + [OCl-] = 100%

then 3.47[OCl-] + [OCl-] = 100%

[OCl-] = 23.37% [HOCl] = 77.6%

The Carbonate SystemBuffer solution: A solution which resists large changes in pH when an acid or base are added.Atmospheric carbon dioxide, CO2(g), produces a natural buffer through the following reactions:CO2(g) CO2(aq) + H2O H2CO3 H+ + HCO3

-2H+ + CO3

2-

where H2CO3 = carbonic acidHCO3

-= bicarbonate ion

CO32- = carbonate ion

CO2(g) = carbon dioxide gas CO2(aq) = aqueous carbon dioxide

Aqueous CO2(aq) is formed when atmospheric CO2(g) dissolves in water; we can find its concentration in fresh water using Henry’s lawWhere KH = Henry’s law constant (mol/L . atm); PCO2 = partial pressure of CO2 in the atmosphere

2COHaq2 PK][CO =

The Carbonate SystemThe carbonate system is the most important buffer system in water and wastewater treatment. Also, it is the most important acid-base system in natural waters because it largely controls its pH.

CO2(g) CO2(aq) + H2O H2CO3 H+ + HCO3-

2H+ + CO32-

Any change in the system components to the right of CO2 causes the CO2 either to be released from solution or to dissolve. Le Chatelier’s Principle: A system at equilibrium, when subjected to a perturbation, responds in a way that tends to minimize its effect.

Examine the character of the buffer system in resisting a change in pH assuming 4 cases (system exposed to atmosphere, system is open)Case I: Acid is added to the system (it unbalances the system by

increasing H+ concentration)Case II: Base is added to the system (base consumes H+)Case III: CO2 is bubbled into the carbonate systemCase IV: carbonate buffer is stripped of CO2

Behavior of carbonate buffer system with the addition of acids and bases or the addition or removal of CO2

Case I and II are common in natural settings when a reaction proceeds over a relatively long period

Case III and IV are not common in natural settings. They are used in water treatment to adjust the pH

The pH will change more dramatically because the atmosphere is no longer a source or sink of CO2

The Carbonate SystemIn natural waters in equilibrium with atmospheric CO2, the amount of CO3

2- in solution is quite small in comparison to the HCO3- in solution.

The presence of Ca2+ in the form of limestone rock or other naturally occurring sources of calcium results in the formation of calcium carbonate CaCO3, which is very insoluble. As a consequence it precipitates from solution

CaCO3(s) Ca2+ + CO32-

The Carbonate SystemAlkalinity: is the sum of all titratable bases down to about pH of 4.5. It is found experimentally by determining how much acid it takes to lower the pH of water to 4.5. In most waters the only significant contributions to alkalinity are the carbonate species and any free H+ or OH-. The total H+ that can be taken up by a water containing primarily carbonate species is

In most natural water situations (pH 6 to 8), the OH- and H+ are negligible such that

Note that [CO32-] is multiplied by two because it can accept two

protons.

H2CO3 H+ + HCO3- pKa1 = 6.35

HCO3- H+ + CO3

2- pKa2 = 10.33

][H][OH]2[CO][HCOAlkalinity 233

+−−− −++=

]2[CO][HCOAlkalinity 233

−− +=

pC-pH diagram for carbonate system

H2CO3 HCO3-

CO32-

OH - H +

1. Below pH of 4.5, essentially all the carbonate species are present as H2CO3, and the alkalinity is negative (due to H+).

2. At pH of 8.3 most of the carbonate species are present as HCO3-, and

the alkalinity equals HCO3-.

3. Above pH of 12.3, essentially all of the carbonate species are present as CO3

2-, and the alkalinity equals 2[CO32-] + [OH-].

pKa1 pKa2< 4.5 > 12.3

Titration curve for a hydroxide-carbonate mixture

The figure shows the change of species described above as the pH is lowered by the addition of acid to a water containing alkalinity. At about pH 8.3 the carbonate is essentially all converted to bicarbonate

Source: Sawyer, McCarty, and Parkin, 1994

The Carbonate SystemBy convention, alkalinity is not expressed in molarity units, but rather in mg/L as CaCO3.

The alkalinity is then found by adding all the carbonate species and the hydroxide, and then subtracting the hydrogen ion. When using the units “mg/L as CaCO3” the terms are added directly. The multiple of two CO3

2- has already been accounted for in the conversion.

=

species

CaCO3 EW

EWspecies)as(mg/LCaCOasmg/L 3

Example 8A water contains 100.00 mg/L CO3

2- and 75.0 mg/L HCO3- at a pH

of 10. Calculate the alkalinity exactly at 25oC. Approximate the alkalinity by ignoring [OH-] and [H+].

The equivalent weight are:CO3

2-: MW = 60, n = 2, EW = 30HCO3

-: MW = 61, n = 1, EW = 61H+: MW = 1, n = 1, EW = 1

OH-: MW = 17, n = 1, EW = 17

pH = 10; therefore [H+] = 10-10 M.mg/L = (10-10 mole/L)(1 g/mole)(103 mg/g) = 10-7

[OH-] = Kw/[H+] = 10-14/10-10 = 10-4 moles/Lmg/L = (10-4 mole/L)(17 g/mole)(103 mg/g) = 1.7

Example 8mg/L as CaCO3 is found using the equation

The equivalent weight of CaCO3 = MW/n where n = 2EW CaCO3 = 100/2 = 50

CO32- = 100.0(50/30) = 167

HCO3- = 75.0(50/61) = 61

H+ = 10-7(50/1) = 5 X 10-6

OH- = 1.7(50/17) = 5.0The exact alkalinity (in mg/L) is found byAlkalinity = 61 + 167 + 5.0 – (5 X 10-6)

= 233 mg/L as CaCO3

Alkalinity = 61 + 167 = 228 mg/L as CaCO3

=

species

CaCO3 EW

EWspecies)as(mg/LCaCOasmg/L 3

Solubility of Gases in Water• When air comes in contact with water, some of it dissolves into the

water.• The relationship between the equilibrium concentration of gas

dissolved in solution and the partial pressure of the gas defined by Henry’s law:

where[gas] = concentration of dissolved gas (mol/L)

KH = Henry’s law constant (mol/L . atm)Pg = the partial pressure of the gas in air (atm)

Pg is simply the volumetric concentration times the atmospheric pressure. For example, oxygen makes up about 21 percent of the atmosphere, so at 1 atm: Pg would be 0.21 X 1 atm = 0.21 atm.

Henry’s law constant varies both with temperature (solubility decreases as temperature increases) and with concentration of other dissolved gases and solids (the solubility decreases as other dissolved material in the liquid increases).

gH PK[gas] =

Solubility of Gases in WaterBecause atmospheric pressure varies with altitude, Pg will change as well

(Thomson and Mueller, 1987)

WhereP = atmospheric pressure at altitude H (atm)H = altitude (m)P0 = atmospheric pressure at altitude H (atm)

H101.15PP 40

−×−=

Example 9By volume, the concentration of oxygen in air is about 21 percent. Find the equilibrium concentration of O2 in water (in mol/L and mg/L) at 25oC and 1 atm of pressure. Recalculate it for Denver at an altitude of 1,525 m.

Pg = 0.21 X 1 atm = 0.21 atm.At 25oC, KH for O2 = 0.0012630 mol/L . Atm

[O2] = KHPg = 0.0012630 X 0.21 = 2.65 X 10-4 mol/L

In Denver, at 1,525 m, atmospheric pressure can be estimated using the equation

Pg = 0.21 X 0.825 atm = 0.17325 atm.[O2] = KHPg = 0.0012630 X 0.17325 = 2.19 X 10-4 mol/L

atm0.8251,525101.15-1H101.15PP -440 =××=×−= −

Alkalinity vs. Alkaline•Buffer solution: A solution which resists large changes in pH when an acid or base are added.

•Alkalinity serves as a measure of buffering capacity.

•The greater the alkalinity, the greater the buffering capacity.

•Alkaline water: water that has a pH greater than 7.•High alkalinity water: water with high buffering capacity.

•An alkaline water may or may not have a high buffering capacity. Likewise, a water with high alkalinity may or may not have a high pH.