Environmental Chemistry Spring 2005 Chapter 3

Transcription

1 Environmental Chemistry Spring 2005 Chapter 3 Ji Yang, Ph.D. Associate Professor School of Environmental Science and Engineering Shanghai Jiao Tong University

2 Aqueous Chemistry The complexity of the chemistry cannot be over emphasized. Gasses in water (O 2, CO 2, CH 4 etc.) All gasses in water are governed by Henry's Law - "the solubility of a gas in a liquid is proportional to the partial pressure of that gas in contact with the liquid"

3 Oxygen Oxygen (O 2 )is very important for higher aquatic life. It has its primary origin from the atmosphere 20.95% v/v max. limit = 8.32 mg/l at 25 C in equilibrium with air Solubility of O 2 in water Temperature of water mg/l 0 C 8.32 mg/l 25 C 7.03 mg/l 35 C

4 Oxygen 7-8 mg of organic matter will stoichiometrically consume all the O 2 in 1L of water. Oxygen is the final electron acceptor in biological energy systems for most higher animal and fish life. Higher life cannot exist without it! If O 2 concentration goes to Zero or become very small, fish and other marine organisms die.

5 Carbon Dioxide CO 2 is an increasing atmospheric gas Year CO 2 Level ppm ppm ppm

6 Carbon Dioxide A naturally occurring and biologically produced and sinks in natural waters From {CH 2 O} + O 2 + Bacteria ---- CO 2 + H 2 O Organic matter decomposition is also a significant source of CO 2 Natural Water Sinks for CO 2 In Oceans 1/3 of excess CO 2 produced by anthropogenic activity is absorbed by the Oceans CO 2 + Phytoplankton + Sun Light ---- (CH 2 O) Also Ca 2+ + CO 2 + H 2 O <---- CaCO 3 + 2H +

7 Carbon Dioxide Another important CO 2 sink is photosynthesis by plants, grasses, trees, weeds etc. in the lithosphere CO 2 + Plants + Sun Light ---- (CH 2 O) Ref. In the Introduction to Env. Sci. text. Living in the Environment, by Miller, 8th ed. Wadsworth Publishing Co. CA, Text pgs Carbon Cycle Fig. 4-28

8 Carbon Dioxide

9 Carbon Dioxide & Carbonate, ph & Alkalinity Carbon dioxide is a primary reagent in natural water buffer systems. In many surface water systems the primary source of CO 2 is the atmosphere at 0.035%. However, unlike O 2, in some systems the primary source of CO 2 may be the by product of biological reactions. Depending on the biosphere and the biological productivity.

10 Carbon Dioxide & Carbonate, ph & Alkalinity The solubility of CO 2 is primarily due to the formation of HCO 3- and CO 2-3. Greater than 25 mg/l of CO 2 in water will cause death to many aquatic life forms. CO 2 + H 2 O <---- HCO 3- + H + <---- H 2 CO 3 Different species (CO 2, HCO 3-, CO 2-3 ) predominate depending on the ph. ph 0-6 CO 2 ph 6-10 HCO 3 - Predominates in Most Natural Waters ph CO 3 2- The typical concentration of [HCO 3- ] is 1.0 x 10-3 mol/liter.

11 Carbon Dioxide & Carbonate, ph & Alkalinity The term Alkalinity is used to describe the particular water systems ability to accept H 3 O+ or H+ ions (buffer capacity). (sometimes called Total Alkalinity) Alkalinity is usually due to natural buffers such as HCO 3 - (the predominate CO 2 species in natural water) and is usually 1.0 x 10-3 mol/liter of water. (assume 1eq/mole) (also other we will investigate later such as PO 2-4, SO 2-4, and natural hydroxides {Al(OH) 3 })

12 Acids and Bases (review) The Bronsted - Lowery definition (1923) - "an acid is a substance capable of donating a proton and a base is a substance that can accept a proton." The Lewis definition (G. N. Lewis) - "an acid is an electron - pair acceptor and a Lewis base is an electron- pair donor"

13 Acids and Bases (review) Many 3 dimensional structures of enzymes and other important biological molecules are ph dependent. Many (if not most) chemical reactions are ph dependent (remember your organic synthesis H + and OH - catalyzed reactions also Lewis defined acids and bases, etc.) Environmentally many leaching situations depend on the ph for movement of ions and molecules. Some we will study, complexation, precipitation, neutralization, adsorption, protonation, ionization, etc. Other examples of where ph is critical to other reactions or situations. Water soil interaction, solubility of particles in air (or more precisely water in air), shifting of equilibrium.

14 Acids and Bases (review) Two competing reactions occur simultaneously: K a HA + H 2 O H 3 O + + A - K a = + H A [ HA] K b A - + H 2 O OH - + HA K b = = A [ ] HA OH Kw K a

15 ph a brief review The ion - product constant of water 2H 2 O H 3 O + + OH - (Note products over reactants form the equilibrium constant) K w = [H 3 O + ][OH - ] K w is the "ion product constant for water" usually given at 25 C (standard) at 25 C, K w = 1.01 x (nonstandard) at 0 C, K w = x 10-14, at 50 C, K w = 5.47 x 10-14

16 Why doesn't the liquid water appear in the equilibrium? The concentration of liquid H 2 O(l) is relatively constant compared to the ions and it is a liquid, and liquids and solids do not appear in equilibrium constants. H 2 O(l) is a constant 55.6 mole)(and their activity is = to 1). Kw = [H 3 O + ][OH - ] = 1.0 x (at 25 C) Given: hydrogen-ion concentration of 7.9 x 10-6 ph = - log [H + ] = - log (7.9 x 10-6) = 5.1 Given: ph = 5.1 [H + ] = = 7.9 x 10-6 Thus: (also useful relationship) 14 = ph + poh poh = 14 - ph and ph = 14 - poh

17 Natural Waters & CO 2, Equilibrium Carbon Dioxide and its hydrated equilibrium components have a unique relationship with natural waters CO 2 (aq) + H 2 O H 2 CO 3 HCO 3- + H + CO 2 (aq) + H 2 O H 2 CO 3 eq. 1 H 2 CO 3 HCO 3- + H + eq. 2

18 Natural Waters & CO 2, Equilibrium CO 2 (aq) + H 2 O <---- HCO H + But water in liquid, ([H 2 O(l)], liquids, and solids not included)

19 Natural Waters & CO 2, Equilibrium Equilibrium Constant - K (K eq ) K eq = = 4.45 x 10-7 pk a1 = [-LogK a ] = 6.35 HCO 3- CO H + K eq = = 4.69 x pk a2 = [-LogK a ] = Which equilibrium is most likely in natural water systems?? Why??

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21 The specie of carbonate found in most waters depends on the ph and since most natural waters are ph HCO 3- is the predominant form. CO 2 = HCO 3- = CO 3 2- = Natural Waters & CO 2, Equilibrium

22 Substitution of K a1 and K a2 from the equilibrium into, yields a relationship as functions of (1.) equilibrium constant and (2.) hydronium ion concentration. CO 2 = HCO 3- = Natural Waters & CO 2, Equilibrium CO 3 2- = [Given: from air equilibrium, 0.035% of gas in air, Henry's Law of partial pressure] Natural water typically has an alkalinity of 1 x 10-3 mol/l

23 Natural Waters & CO 2, Equilibrium at ph 7 then [HCO 3- ] ~= 1 x 10-3 mol/l at ph 10 then [HCO 3- ] + 2[CO 3 2- ] + [OH - ] = ratio 4:4:1 = 1 x 10-3 mol/l

24 Natural Waters & CO 2, Equilibrium Also as the carbon source for the Bioshpere: While the carbonate system is the buffer for natural waters it is also the carbon source and a significant oxygen source when interacting with the algae in water. CO 2 + H 2 O + hv + algae ---- {CH 2 O} + O 2 HCO 3- + H 2 O + hv + algae ---- {CH 2 O} + OH - + O 2

25 The ph of Rain Water What is the natural ph of rain water? What determines the ph in rain water? Assuming no anthropogenic or natural pollution to the atmosphere, What is the ph of rain water?

26 The ph of Rain Water Given: One buffer present from gas Henry's Law of partial pressure CO 2 (g) <--- CO 2 (aq) CO 2 (aq) + H 2 O <--- H 2 CO 3 <--- HCO 3- + H + CO 2 (aq) + H 2 O <--- HCO H + The product of two reactions K a1 = = [H + ] 2 /(1.146 x 10-5 ) = 4.45 x 10-7 [H + ] = [HCO 3- ] = {1.146 x 10-5 x 4.45 x 10-7 } 1/2 = 2.25 x 10-6 ph = -Log[H + ] = -Log[2.25 x 10-6 ] = 5.56 This would be the non-perturbed ph of rain water.

27 Alkalinity Alkalinity - is the capacity of water to accept H+, (H3O+, Hydronium ion) Alkalinity (or total alkalinity) is the buffer capacity of water for acid. This is also an indication of which streams are susceptible to acid rain and which have a buffer capacity sufficient to resist a ph change that could damage their ecosystems.

28 Alkalinity Alkalinity then becomes an important property of natural water stabilizing the ph for aquatic plant growth and aquatic life. [alk] = [HCO3-] + 2[CO3 2-] + [OH-] - [H3O+] Thus the lower the ph (< 7) the low the alkalinity typically and the lower the buffer capacity of the water and less tolerant to acids. Reservoirs of carbonate Decay CaCO 3 rock and minerals atmospheric CO 2 etc.

29 Alkalinity Natural water in contact with a mineral calcium carbonate CaCO 3 (s) (calcite, limestone, shells, coral...) Ca(aq) + CO 3 2- (aq) + <--- Ca 2+ + H 2 CO 3 (aq) Note: over simplified equation not balanced for H+ form water Assuming: that [CO 3 2- (aq) = [H 2 CO 3 (aq)] given K a of H 2 CO 3 (aq) = 4.45 x 10-7 given K sp of CaCO 3 (aq) = 4.8 x 10-9

30 Alkalinity Water in contact with solid CaCO 3 would have the calculated concentration of these species [CO 2 ] = x 10-5 M [HCO 3- ] = 9.98 x 10-4 M [CO 2-3 ] = 8.96 x 10-6 M [Ca 2+ ] = 4.99 x 10-4 M [H + ] = 5.17 x 10-9 M ph = 8.29 (the ph of some ocean water) These concentrations shift as the equilibrium relating them shift

31 Strong and weak acid - base ionizations, buffer behavior, solubility and precipitation Acid and bases Conjugate base is the entity that remains following the donation of a proton and a conjugate acid is what results after accepting a proton. (c- = conjugate) HNO 3 (acid) + H 2 O (base) <---- H 3 O + (c-acid) + NO 3- (cbase) H 2 O (acid) + NH 3 (base) <---- NH 4 + (c-acid) + OH - (cbase) Strong acid or bases dissociate in water completely and leave no undissociated molecules.

32 Strong and weak acid - base ionizations, buffer behavior, solubility and precipitation Strong acids and bases react completely: Acid + Base ---- salt + water HNO 3 + NaOH ---- NaNO 3 + HOH Weak acid or bases do not dissociate completely and leave varying amounts of both species in solution. CH 3 COOH + H 2 O <---- H 3 O + + CH 3 COO - In an equilibrium that can easily be shifted What will shift this to more CH 3 COOH form?? Water is an Amphiprotic solvent acting as either a proton donor or acceptor. H 3 O +, OH -

33 Henderson-Hasselbalch Equation Very useful for buffer calculations (a review) (works in 90-95% of the time) K a = Take the log of both sides of the equation Log K a = Log and Log K a = Log[H 3 O + ] + Log

34 Henderson-Hasselbalch Equation Put log [H 3 O + ] on the left of the equation and Log Ka on the right so that ph = -Log[H 3 O + ] or -Log[H + ] and, -LogK a = pk a -Log[H 3 O + ] = -Log K a - Log ph = pk a + Log (Final form of the Henderson-Hasselbach Equation)

35 Henderson-Hasselbalch Equation The equation can be used for buffer calculations when: K a of the acid involved is small (Ka < 10-3 ). Note: This automatically makes the assumption that the contribution of the natural dissociation of water into hydronium and hydroxide ions concentrations is insignificant. Also the concentration of the buffer components should be >10-6 M Why? * Dilution of a buffer does not change the ph at which it buffers within several orders of magnitude of dilution

36 Alternate Derivation of the Henderson-Hasselbach Equation Start with equilibrium equation and the equilibrium constant HA + H 2 O <--- H 3 O + + A - (Products over reactants form the equilibrium constant) K a = Henderson-Hasselbalch Equation Take the log of both sides of the equation (when, K a and K b < 10-3 ) Isolating the hydronium ion (it will become the ph)

37 Henderson-Hasselbalch Equation = K a [H 3 O + ] = K a x [HA]/[A - ] (take the -Log of both sides) -Log[H 3 O + ] = -Log K a - Log[HA]/[A - ] p[h 3 O + ] = pk a - Log[HA]/[A - ] (inversion of the Log changes the sign and it is usually written thus) ph = Log K a + Log (Henderson-Hasselbach Equation)

38 Henderson-Hasselbalch Equation The equatiion can be used for buffer calculations when Ka of the acid involved is small (Ka < 10-3 ). Refs. General Chem., by McQuarrie and Rock, Freeman Press, NY, pg. 677, Analytical Chem., by Skoog et. al. 6th ed., Saunders, pg. 221, Quantitative Analysis, by Day and Underwood, 6th ed., Prentice Hall, pg. 109, 1991.

39 Buffers To have a buffer both the weak acid and its conjugate base must be present to constitute that buffer pair.

40 Buffers 1. The weak acid is present in a solution and an equilibrium is established that produces the ionized weak acid as well as the weak acid. This produces the buffer pair and buffers the system at the midpoint of the pka of the weak acid. ph = Log Ka + Log At the midpoint of the buffer range the weak acid concentration and conjugate base ion concentrations are equal and the log of 1 which is = to 0. [H+] = [A-] from [H+]/[A- ] =1 Thus the second term goes to 0 and ph = pka + 0 Of course this can be the only buffer present and how likely is this in nature? But it is an approximation for the dominate buffer.

41 Buffers 2. The weak acid and conjugate base are both present in solution and some volume of acid or base solution is added to the water changing the hydrogen ion concentration. 3. Only the weak acid or the conjugate base is present. Then a strong acid or a strong base is added to produce the other to provide the buffer pair.

42 Buffers

43 Buffer Capacity Buffers resist changes in ph when significant quantities of both weak acid and conjugate base are present. The amount of acid or base that can be added without changing significantly the ph depends on the concentration of the Buffer Pair. The greater the concentration the more acid or base that can be tolerated. Definition - Buffer capacity is the number of moles of strong acid or base that result in a 1 ph change of a 1.0 L buffer solution.

44 Buffer Capacity The point at which the buffer can take the maximum amount of added acid or base is where there is an equal amount of both acid and conjugate base in solution. This corresponds to the pka of the acid. Buffers are generally calculated to buffer around their pk a where the maximum tolerance to adjustment is. They are then adjusted with either their pair concentration or with acid or base to attain the exact buffer ph for specific purposes.

45 Buffer Capacity Rule of thumb- Choose a buffer with a pka within +/-1 ph unit of the desired buffer ph. If you want it to buffer there. The shape of the titration curve is controlled by the completeness of the dissociation and the reaction with the strong acid or base.

46 Buffer Capacity Natural water in contact with a mineral calcium carbonate CaCO 3 (s) (calcite, limestone, shells, coral...) Ca(aq) + CO 3 2- (aq) + H 2 O <---- Ca 2+ + H 2 CO 3 (aq) + OH - at 25 C, Ka of H 2 CO 3 (aq) = 4.45 x 10-7 at 25 C, Ksp of CaCO 3 (aq) = 4.8 x 10-9 This is mediated by the solubility of CaCO 3 Ksp = 4.8 x 10-9 This is why the ph of sea water is around 8.2 What does concentration provide if not ph?? (Buffer Capacity or Total Alkalinity)

47 Buffer Capacity Buffers and their Importance in Natural Waters and Environmental Systems What is a buffer? - - Buffers are solutions that contain both a weak acid and its conjugate base in solution at the same time. Why are buffers so important?

48 Buffer Capacity Example: Calculate the ph of a buffer solution containing 0.01M acetic acid and 0.01M sodium acetate. Then calculate the ph after enough HCl is added to give a concentration of 0.001M. (pka=4.74)

49 Chemical Equilibrium:(A brief review) Equilibrium Constant - K (K eq ) Notes: Solution problems usually use concentrations of 1 M or less. As the term implies the reactions go in both directions Direction of the reaction and the equilibrium constant K forward =

50 Chemical Equilibrium:(A brief review) Keq is a temperature dependent constant. At dilute concentrations ( < or = to 1 x 10-3 Molar) - ion activity (a) is approximately = to ion Concentration [x] Did the Previous example meet these criteria?? Ca(aq) + CO 3 2- (aq) + H 2 O <---- Ca 2+ + H 2 CO 3 (aq) + OH - at 25 C, Ka of H 2 CO 3 (aq) = 4.45 x 10-7 at 25 C, Ksp of CaCO 3 (aq) = 4.8 x 10-9

51 Review of important and common types of equilibrium constants. Dissociation - Ion-Product constant Example for Water: 2H 2 O <---- H 3 O + + OH - K w = [H 3 O + ][OH - ] Dissociation of a weak acid or base Example K a or K b CH 3 COOH + H 2 O <---- H 3 O + + CH 3 COO - K a = CH 3 COO - + H 2 O <---- OH - + CH 3 COOH K b = Also K a K b = K w

52 Solubility and Precipitation in Water Ions are converted to an insoluble precipitate, the precipitate is in equilibrium with the ions and a constant concentration is maintained at a given temperature and with constant ion concentration as long as there is a solid precipitate present. The precipitate molecule and ions are in a saturated solutions defined by the equilibrium Example Ksp - solubility product constant usually written from the solid side of the equilibrium

53 Solubility and Precipitation in Water BaSO 4 (s) <---- Ba 2+ + SO Ksp = [Ba 2+ ][SO 4 2- ] Stoichiometry is the precise ratio of each of the elements (or complex ions) forming the molecule or solid precipitate.

54 Anion: OH - Al(OH) 3 (s) Al OH - Ksp = [Al 3+ ][OH - ] 3 = 2 x similar reactions for Al, Cd, Fe (II) & (III), Pb, Mg, Zn Solubility and Precipitation in Water 1. Precipitation example assume - ph 7.0 in water Anion: SO 4 2- PbSO 4 (s) Pb 2+ + SO 4 2- Ksp = [Pb 2+ ][SO 4 2- ] = 1.6 x 10-8 similar reactions for Ba, Ca, Pb, Sr, etc., Cations Anion: CO 3 2- CaCO 3 (s) Ca 2+ + CO 3 2- Ksp = [Ca 2+ ][CO 3 2- ] = 4.8 x 10-9 similar reactions for Ba, Cd. Ca, Pb, Mg, Ag

55 Solubility and Precipitation in Water 1. Precipitation example assume - ph 1.0 in water What happens to the equations? Why? Anion: SO 4 2- PbSO 4 (s) Pb 2+ + SO 4 2- Ksp = [Pb 2+ ][SO 4 2- ] = 1.6 x 10-8 SO H + HSO 4 - SO 4 2- is removed from solution which way will the reaction be driven?

56 Solubility and Precipitation in Water Anion: CO 3 2- CaCO 3 (s) Ca 2+ + CO 3-2 Ksp = [Ca 2+ ][CO 3 2- ] = 4.8 x 10-9 CO H + HCO 3 - CO 3 2- is removed from solution which way will the reaction be driven? For anion: OH - Al(OH) 3 (s) Al OH - Ksp = [Al 3+ ][OH - ] 3 = 2 x OH - + H + H 2 O What happens to Al 3+ ions and others like it? Similar reactions for Al 3+, Cd 2+, Fe 2+, Fe 3+, Pb 2+, Mg 2+, Zn 2+

57 Precipitation Mechanisms Three stages of precipitation 1. Nucleation 2. Crystal Growth 3. Aggregate Particle Growth Nucleation - the formation of a small aggregate of molecules in a supersaturated solution Crystal Growth - the addition of molecules or ions to nuclear aggregate particles. Aggregate Particle Growth - the joining of large numbers of crystals to form large aggregate particles that settle out of solution.

58 Precipitation Mechanisms Colloidal suspensions can form with particle sizes of 10-6 to 10-4 mm diameter. They are to small too filter using conventional filter material. Colloidal particles are not retained by conventional filter paper and do not settle from solution but remain in suspension due to brownian movement. Particles 10-4 to 10-1 mm tend to settle from solution as precipitates spontaneously. Non-homogeneous composite solutions are refereed to as crystalline suspensions until they settle as precipitates.

59 Precipitation Mechanisms Particle size has been related to the relative supersaturation of the solution where: RS - Relative Supersaturation = (Q - S)/S Q - is the concentration of the solute at any instant S - is the equilibrium solubility Q - S is a measure of the degree of supersaturation High Relative Supersaturation tend to produce colloidal suspensions (small crystals, colloidal suspensions - particle size < 10-4 mm) Low Relative Supersaturation produce more crystalline particles with greater diameter (particle size 0.1 mm)

60 Precipitation Mechanisms The rate of nucleation increases exponentially with RS while the rate of crystal growth increases linearly Note: All generalities assume that we are using WATER as the solvent!

61 Stability of colloid Stability of colloids can be explained by associated charge and is related to the ionic strength of the solution. Charged ions associated with each colloidal particle repel other particles tending to keep them small enough to be influenced by brownian movement. To optimizing precipitation conditions - Minimize Q (concentration of solution) using more dilute solutions S is Maximized when precipitation is from hot water solutions or with specific ph change Increasing the total ionic content of a solution often coagulates the colloid and permits it to precipitate

62 Other Relevant Precipitation Events Coprecipitation Mechanisms: a. Surface Adsorption, (Al & Fe (OH)x are very efficient) b. Inclusion c. Occlusion and Mechanical Entrapment Coprecipitation is the inclusion in the precipitate of the insoluble compound of normally soluble substances. Precipitates tend to carry down other normally soluble components from the solution.

63 Other Relevant Precipitation Events Adsorption is the process of association on the surface such as the ions about a colloidal particle. This is a surface effect and surface area influences this effect. When particles are included into the lattice sites of the crystal the mechanism is referred to as Inclusion. Physical entrapment of other material (not in the crystal lattice) is Occlusion. Adsorption is very important in natural water especially due to the amount of iron {Fe(OH) 3 and Fe(OH) 2 }and aluminum hydroxide {Al(OH) 3 }that are in the environment. Iron and Aluminum hydroxides are flocculent (fluffy non-crystalline) precipitates that scavenge by adsorption most other metal ions.