Defesa

Chapter 14 Chapter 14 Chemistry of Natural Waters Chemistry of Natural Waters

WaterWater

What are the significant What are the significant physical/chemical properties physical/chemical properties

of water?of water?

Significant PropertiesSignificant Properties

• Intra-molecular bonding?Intra-molecular bonding?• Heat capacity?Heat capacity?• Polarity?Polarity?

• Solubility of gases – COSolubility of gases – CO22 and O and O22??

• Density as function (temperature?)Density as function (temperature?)• Presence of indigenous material?Presence of indigenous material?• Transparency?Transparency?

Stratification of a water body Stratification of a water body produced by solar radiation (Fig. produced by solar radiation (Fig.

14.2)14.2)

Depth(m)

5 10 15 20

5

4

3

1

Temperature (oC)

Water Surface

Thermocline

Epilimnion

Hypolimnion

TransparencyTransparency

• Sunlight can penetrate into Sunlight can penetrate into water at considerable water at considerable depths. depths.

• About 90% of the radiation About 90% of the radiation above 750 nm is absorbed above 750 nm is absorbed by the time sunlight has by the time sunlight has penetrated to 1 meter in penetrated to 1 meter in water. water.

• Transparency of water is an Transparency of water is an important property to important property to consider for chemicals in consider for chemicals in water such as oil-derived water such as oil-derived hydrocarbons and pesticides hydrocarbons and pesticides susceptible to degradation susceptible to degradation by sunlight (i.e. photolysis).by sunlight (i.e. photolysis).

Red

Blue

SalinitySalinity

• Salinity - 0.5 – 3.0 % or 3.5% salt, as a result Salinity - 0.5 – 3.0 % or 3.5% salt, as a result of its solvating capacity. of its solvating capacity.

• Water dissolves ionic compounds through ion-Water dissolves ionic compounds through ion-dipole interactions. dipole interactions.

• The Venice system which categorizes waters, The Venice system which categorizes waters, according to salinity zones. according to salinity zones.

– Seawater (>3.0%),Seawater (>3.0%),– estuaries (0.05 – 3.0%), andestuaries (0.05 – 3.0%), and– freshwater (<.05%)freshwater (<.05%)

Solubility of gases in waterSolubility of gases in water

• Solubility of gases is subject to Henry’s Law. Solubility of gases is subject to Henry’s Law. – [X] = K[X] = KHHPPXX

• Oxygen and carbon dioxide are the two most important Oxygen and carbon dioxide are the two most important gases dissolved in water - bar none!gases dissolved in water - bar none!– KKHH(CO(CO22) = 3.4x10) = 3.4x10-2-2mol Lmol L-1-1atmatm-1-1

– KKHH(O(O22) = 1.3 x10) = 1.3 x10-3-3 mol L mol L-1-1atmatm-1 -1

• COCO22 provides H provides H22COCO33, which serves as a buffering, pH effect , which serves as a buffering, pH effect on the aquatic environment, and Oon the aquatic environment, and O22, which the aquatic , which the aquatic organisms must have to survive. organisms must have to survive.

• Solubility of these gases is a function of temperature, partial Solubility of these gases is a function of temperature, partial pressure in the atmosphere, and salinity , e.g. In 0pressure in the atmosphere, and salinity , e.g. In 0ooC fresh C fresh water at saturation has 2 mg/L more dissolved oxygen than water at saturation has 2 mg/L more dissolved oxygen than does seawater (3.5% salt). Why is this? does seawater (3.5% salt). Why is this?

Dissolved Oxygen in Natural watersDissolved Oxygen in Natural waters

• Solubility of oxygen is a function of temperature. Solubility of oxygen is a function of temperature. [Henry’s Law][Henry’s Law]

• KKHH = 1.3 x10 = 1.3 x10-3-3 mol L mol L-1-1atmatm-1 -1 Therefore, thermal Therefore, thermal pollution is a problempollution is a problem

• ProblemProblem. Confirm by calculation the value of 8.7 . Confirm by calculation the value of 8.7 mg/L for the solubility of oxygen in water.mg/L for the solubility of oxygen in water.

• [O[O22] = K] = KHHPPO2O2 = 1.3 x10 = 1.3 x10-3-3 mol L mol L-1-1atmatm-1 -1 X 0.21 atm = X 0.21 atm = 2.7 x102.7 x10-4 -4 mole/litermole/liter

• 2.7 x102.7 x10-4 -4 mole/liter X 32.0 grams/mole = 8.7 x10mole/liter X 32.0 grams/mole = 8.7 x10--

33grams/litergrams/liter• = 8.7 mg/liter= 8.7 mg/liter

Solubility of carbon dioxideSolubility of carbon dioxide

• COCO22 (g) in the atmosphere is in equilibrium with the (g) in the atmosphere is in equilibrium with the water system, in accordance with Henry’s law.water system, in accordance with Henry’s law.

COCO22 (g) ---> CO (g) ---> CO22 (l) (l)

[CO[CO22 (l)] = K (l)] = KHHPP CO2 CO2 (g)(g) = 3.4x10 = 3.4x10-2-2mol Lmol L-1-1atmatm-1-1PPCO2CO2

PPCO2CO2 = 350 ppm = 3.50 x10 = 350 ppm = 3.50 x10-4-4atm.atm.

Therefore, [HTherefore, [H22COCO33] =1.2x10] =1.2x10-5-5

pH of Water without CaCOpH of Water without CaCO33

• [H[H22COCO33] =1.2x10] =1.2x10-5-5

• COCO22 (l) + H (l) + H22O -> HO -> H22COCO33 -> H -> H++ + HCO + HCO33-- K Ka1a1 = 4.5 x10 = 4.5 x10-7-7

• HCOHCO33-- ---> CO ---> CO33

2-2- + H + H++ K Ka2a2 = 4.7 x10 = 4.7 x10-11-11

• [Don’t consider the second step of the diprotic acid, factor 10[Don’t consider the second step of the diprotic acid, factor 1044]]• pH of water in equilibrium with COpH of water in equilibrium with CO22(g)(g)• HH22COCO3 3 --> H--> H++ + HCO + HCO33

-- K Ka1a1 = 4.5 x10 = 4.5 x10-7-7

• 1.2x101.2x10-5-5 -x x -x x x x• 4.5 x104.5 x10-7 -7 = [x][x]/[1.2x10= [x][x]/[1.2x10-5-5 -x] -x]• xx22 = 5.4 x 10 = 5.4 x 10-12-12

• x = 2.3 x 10x = 2.3 x 10-6-6

• pH = 5.63pH = 5.63

Natural waters (acid-base chemistry Natural waters (acid-base chemistry of the carbonate system)of the carbonate system)

• Acid-base chemistry of Acid-base chemistry of natural water systems is natural water systems is dominated by the dominated by the carbonate ion, carbonate ion, COCO33

2-2-

• The carbonate ion is the The carbonate ion is the anion of a weak acid, anion of a weak acid, HCOHCO33

--, and this makes it , and this makes it a moderately strong a moderately strong base.base.

Understand this problem from two Understand this problem from two directions:directions:

COCO22 (g) in the atmosphere is in equilibrium with the water (g) in the atmosphere is in equilibrium with the water system, in accordance with Henry’s law.system, in accordance with Henry’s law.

COCO22 (g) ---> CO (g) ---> CO22 (l) (l)

[CO[CO22 (g)] = K (g)] = KHHPP CO2 CO2 (g)(g) = 3.4x10 = 3.4x10-2-2mol Lmol L-1-1atmatm-1-1PPCO2CO2

PPCO2CO2 = 350 ppm = 3.50 x10 = 350 ppm = 3.50 x10-4-4atm. [Hatm. [H22COCO33] =1.2x10] =1.2x10-5-5

COCO22 (l) + H (l) + H22O -> HO -> H22COCO33 -> H -> H++ + HCO + HCO33-- K Ka1a1 = 4.5 x10 = 4.5 x10-7-7

pH = pH = 5.65.6

HCOHCO33-- ---> CO ---> CO33

2-2- + H + H++

KKa2a2 = 4.7 x10 = 4.7 x10-11-11

[Don’t consider the second [Don’t consider the second step of the diprotic acid, factor 10step of the diprotic acid, factor 1044]]

CaCOCaCO33(s) is in equilibrium with Ca(s) is in equilibrium with Ca2+2+ and COand CO33

2-2-

• CaCOCaCO33 --> Ca --> Ca2+2+ + CO + CO332-2- K Kspsp = 4.6 x10 = 4.6 x10-9-9

• s = solubility of Cas = solubility of Ca2+ 2+ ss22 = 4.5 x10 = 4.5 x10-9-9 • s = [COs = [CO33

2-2-] = 6.8 x10] = 6.8 x10-5-5 moles/liter moles/liter• but, CObut, CO33

2-2- is the anion of a weak acid, thus is the anion of a weak acid, thus• COCO33

2- 2- + H + H22O ---> HCOO ---> HCO33-- + OH + OH--

• K = [HCOK = [HCO33--][OH][OH-]-]/[CO/[CO33

2-] 2-] K = KK = Kww/K/Ka2a2 = 1 x10 = 1 x10-14-14/4.7x10/4.7x10-11-11 = 2.1x10 = 2.1x10-4-4

• Therefore, Therefore, CaCOCaCO33 + H + H22O --> CaO --> Ca2+2+ + HCO + HCO33-- + OH + OH--

• K’ = K x Ksp = 2.1x10K’ = K x Ksp = 2.1x10-4-4 x 4.6x10 x 4.6x10-9-9 = 9.7x10 = 9.7x10-13-13

• S = [OHS = [OH--] = [ Ca] = [ Ca2+2+] = ] = 9.9 x109.9 x10-5-5 m/liter m/liter • compared to compared to 6.8 x106.8 x10-5-5 moles/liter moles/liter • when we considered solubility withoutwhen we considered solubility without• hydrolysis.Also, gives us a pH = 10hydrolysis.Also, gives us a pH = 10

CaCOCaCO33(s) is in equilibrium with Ca(s) is in equilibrium with Ca2+2+ and COand CO33

2-2-

• For the more adventurous, what we have here is coupled For the more adventurous, what we have here is coupled equilibria with two interdependent variables; solubility and equilibria with two interdependent variables; solubility and degree ionization; i.e. s and x, respectively.degree ionization; i.e. s and x, respectively.

• CaCOCaCO33 --> Ca --> Ca2+2+ + CO + CO332-2-

S S-xS S-x• COCO33

2- 2- + H + H22O ---> HCOO ---> HCO33-- + OH + OH--

S -x x xS -x x x

• 2.1x102.1x10-4 -4 = [x][x]/[S-x] and 4.6 x10 = [x][x]/[S-x] and 4.6 x10-9-9 = [S][S-x] = [S][S-x]• when you change one, you affect the other. Two equations, two when you change one, you affect the other. Two equations, two

unknowns, but both quadratic equations. unknowns, but both quadratic equations. • Set-up an iterative approach, starting with S = 6.78x10Set-up an iterative approach, starting with S = 6.78x10-5-5..

Comparing the iterative method to Comparing the iterative method to the approximation method.the approximation method.

• Approximation method:Approximation method:• S = 6.8x10S = 6.8x10-5-5 and [OH and [OH--] = 9.9x10] = 9.9x10-5-5

• pOH = 4.00 pH = 10.0pOH = 4.00 pH = 10.0• Iterative methodIterative method• S = 1.25x10S = 1.25x10-4-4 and [OH and [OH--] = 8.8x10] = 8.8x10-5-5

• pOH = 4.05 pH = 9.95pOH = 4.05 pH = 9.95

Now consider the total systemNow consider the total system

• CaCOCaCO33 +CO +CO22 (l) + H (l) + H22O --> CaO --> Ca2+2+ +2 HCO +2 HCO33--

• K = 1.5x10K = 1.5x10-6 -6 = [Ca= [Ca2+2+ ][ HCO ][ HCO33-- ] ]2 2 /P/PCO2CO2

• [S][2S][S][2S]22/0.00035 =/0.00035 = 1.5x101.5x10-6 -6

• S = S = 5.1 x105.1 x10-4-4 M = [Ca M = [Ca2+2+]] and the amount and the amount of COof CO22 dissolved. dissolved.

• This is This is 35 times35 times the amount of CO the amount of CO2 2

[1.2x10[1.2x10-5 -5 moles/liter] that would dissolve moles/liter] that would dissolve without the presence of CaCOwithout the presence of CaCO33 and and five five times times the calcium concentration the calcium concentration calculated without the involvement of calculated without the involvement of carbon dioxide [1.25x10carbon dioxide [1.25x10-4-4 moles/liter]. moles/liter]. This is a This is a synergisticsynergistic effect. effect.

The pH of River and Lake Water The pH of River and Lake Water Saturated with COSaturated with CO22 and CaCO and CaCO33

• The bicarbonate ion can act as either a base or an acidThe bicarbonate ion can act as either a base or an acid• HCOHCO33

-- --> H --> H++ + CO + CO332-2- K Kaa = 4.7 x10 = 4.7 x10-11-11

• HCOHCO33-- + H + H22O -----> HO -----> H22COCO33 + OH + OH-- K Kbb = 2.2 x10 = 2.2 x10-8-8

• {Consider only the base reaction because of magnitude {Consider only the base reaction because of magnitude difference in K’s}difference in K’s}

• [HCO[HCO33--] = 1.02x10] = 1.02x10-3-3 moles/l from previous slide, and [H moles/l from previous slide, and [H22COCO33] ]

= 1.2x10= 1.2x10-5-5 from Henry’s law calculation from Henry’s law calculation..• Therefore: 2.2x10Therefore: 2.2x10-8-8 = [OH = [OH--][1.2x10][1.2x10-5-5 ]/[1.02x10 ]/[1.02x10-3-3 - [OH - [OH--]] - ]] -

{neglect addend}{neglect addend}• solving for [OHsolving for [OH--] = 1.9 x10] = 1.9 x10-6-6 and pH =8.27. Thus, river and and pH =8.27. Thus, river and

lake water at 25lake water at 25ooC whose pH is determined by saturation C whose pH is determined by saturation with COwith CO22 and CaCO and CaCO3 3 should be slightly alkaline with a pH of should be slightly alkaline with a pH of about 8.3.about 8.3.

Carbonate system in waterCarbonate system in water

• Remember that you have two equilibria taking Remember that you have two equilibria taking place, and the extent is dependent upon pH.place, and the extent is dependent upon pH.

Calculating the fraction of each Calculating the fraction of each species present in water.species present in water.

• In the equations listed below, the fraction of each In the equations listed below, the fraction of each species present in the carbonate system is shown.species present in the carbonate system is shown.

Expressing the fractions as functions Expressing the fractions as functions of [Hof [H++] and equilibria constants; K] and equilibria constants; Ka1a1

and Kand Ka2a2

• Converting the fractions to expressions of [HConverting the fractions to expressions of [H++], ], and Kand Ka1a1 and K and Ka2a2. Note the variance in [H. Note the variance in [H++] ] dependence .dependence .

• As system becomes acidic, what would you As system becomes acidic, what would you expect to happen? expect to happen?

Concentrations of species in Concentrations of species in carbonate system as function of pHcarbonate system as function of pH

• The points to note here are (1) the variance in the The points to note here are (1) the variance in the concs. , and (2) which specie is dominate at specific pH concs. , and (2) which specie is dominate at specific pH values.values.

Carbonate system in waterCarbonate system in water

Organic Compounds in Organic Compounds in WaterWater

Humic substancesHumic substances• Humic substances are the Humic substances are the

most abundant polymers in most abundant polymers in naturenature

• They are brown, acidic They are brown, acidic material, formed by material, formed by oxidation and condensation oxidation and condensation among polyphenols.among polyphenols.

• Structures will vary in Structures will vary in different areas of the different areas of the country.country.

• Polymers form from the Polymers form from the breakdown of plant material. breakdown of plant material.

Carbon CycleCarbon Cycle

• Figure 14.9. Figure 14.9. Increase in Increase in substances rich in substances rich in organic carbon organic carbon can increase can increase demand on demand on dissolved oxygen– dissolved oxygen– significant significant implications on implications on aquatic organismsaquatic organisms..

Inorganic aqueous wastesInorganic aqueous wastes

• Ammonia Ammonia - problem in - problem in both sewage and both sewage and waste water waste water treatment because of treatment because of its toxicity. its toxicity.

• NitrificationNitrification - - microbial microbial oxidationoxidation of of ammonia, ammonia, nitrate ion nitrate ion is the final product. is the final product.

• In soil, final oxidation In soil, final oxidation products are products are nitrous nitrous oxide and nitrogen. oxide and nitrogen.

N2, N2O(atmosphere)

Nitrogen fixation

OXIDATIONAND

REDUCTION

NO2-

(nitrite dissolved in water)

RESPIRATIONPHOTOSYNTHESIS aminoacids

Proteins(animals and plants

NH3 (NH4OH)(ammonia dissolved in water)

Figure 14.1

Denitrification

NO3-

(nitrate dissolved in water)

Nitrification

Estuary and Oceanic Estuary and Oceanic systemssystems

Omit pages 327 – 332Omit pages 327 – 332

Also, Note Henry’s Law is given asAlso, Note Henry’s Law is given as

H = P/S (solubility)H = P/S (solubility)

We have used it as S = H*P, note unitsWe have used it as S = H*P, note units