Proton Transfer Reactions Bronsted-lowry Theory Chemistry Lab Report In this module you learned about contaminant distribution between air-water, organic-a

Proton Transfer Reactions Bronsted-lowry Theory Chemistry Lab Report In this module you learned about contaminant distribution between air-water, organic-aquatic and organic acid/base-aquatic phases. For the module discussion, pose a question that you have about one of the topics.

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Module Reading material from Chapter 10 of Environmental Organic Chemistry. Rene P. Schwarzenbach, Philip M. Gschwend, and Dieter M. Imboden. 3rd Edition. Wiley Interscience (2016). ISBN: 978-1-118-76723-8.

The lecture notes are attached within CONTAMINANT FATE AND TRANSPORT
ORGANIC SOLVENT-WATER PARTITIONING
5.0
ADK
1
ORGANIC SOLVENT-WATER PARTTIONING
? In this section we are concerned with partitioning of organic
compounds between two immiscible liquids.
? This type of equilibrium is particularly useful when considering
distribution of nonpolar organics between water and natural solids (i.e.
soils, sediments, biosolids etc.).
? Substantial information exists in pharmacology studies of nonpolar
drug uptake by organisms and cellular material.
? n-Octanol has been used traditionally as a surrogate for organic matter.
? Kow values of organic compounds may differ by as much as 10 order of
magnitudes (see Fig. 1)
? Partitioning between an organic solvent and water is described by the
organic solvent-water distribution coefficient defined as:
Kosw ?
5.1
Cos mol / lorg. solvent
C w mol / lwater
[1]
THERMODYNAMIC BASIS
? EXPERIMENT: Consider the following experiment (see Fig. 2):
?
Add an immiscible solvent (i.e. n-octanol) in water.
?
Mix the two liquids well.
?
Allow some time for the system to reach equilibrium and for the
phases to separate.
?
Add a known amount of a compound in the aqueous phase and
allow the system to reach equilibrium.
CONTAMINANT FATE AND TRANSPORT
ORGANIC SOLVENT-WATER PARTITIONING
ADK
2
CONTAMINANT FATE AND TRANSPORT
ORGANIC SOLVENT-WATER PARTITIONING
Figure 2: Partitioning of a Chemical Compound between
Immiscible Organic Solvent-Water Phases
ADK
3
CONTAMINANT FATE AND TRANSPORT
ORGANIC SOLVENT-WATER PARTITIONING
ADK
4
OBSERVATIONS:
?
It is obvious that some water will dissolve in octanol (? 1
molecule H2O/4 molecules octanol) and some octanol will
dissolve in water (? 8 molecules octanol/100,000 molecules H2O).
?
The molar volume of the solvent may be altered if appreciable
amounts of water dissolve in the organic solvent. For example, the
molar volume of pure n-octanol Vos = 0.16 l/mol whereas the
molar volume of octanol saturated with water is 0.12 l/mol.
?
The molar volume effect is less pronounced for highly nonpolar
solvents.
?
The compound dissolved in the aqueous phase will gradually
dissolve in the organic solvent phase until equilibrium is reached.
? At equilibrium the fugacities of the organic compound in the aqueous
and organic solvent phase will be equal:
f os ? f w
[2]
? os x os ? ? w x w
[3]
? In terms of molar concentrations:
? osCosVos ? ? wCwVw
K osw ?
[4]
C os ? wV w
?
C w ? osVos
ln K osw ? ln ? w ? ln ? os ? ln
[5]
Vw
Vos
e
?Gosw
?
? cons tan t
RT
[6]
CONTAMINANT FATE AND TRANSPORT
ORGANIC SOLVENT-WATER PARTITIONING
ADK
5
? COMMENTS (on Eq. [6]):
?
Obviously the partition constant is related to the partial molar
excess free energy of the compound in the two phases (RTln?)
expressing the difference between the nonideality of the solution
of the compound in the two phases
?
For nonpolar compounds ?w is quite large (102-1011) due to the
cost of “cavity” formation and thus ?w dominates the magnitude of
Kosw for nonpolar compounds.
?
Since the organic solvent (os) is likely to be more chemically
related to the organic compound (oc) dissolved, the os-oc
molecule interactions are not dominant thus ?os = ?pure compound = 1.
? ASSUMPTIONS:
?
The activity coefficient of the partitioning compound is
independent of its concentration in the aqueous phase.
?
Organic solvent molecules present in the aqueous solution have no
effect on the activity coefficient of the compound. In other words
os-oc molecule interactions in the aqueous phase are neglected.
? Thus the term ?w can be substituted by ?wsat:
Kosw
? wsatVw
1 1 1
?
? sat ? ?
? sVs
Cw ? s Vs
[7]
? CONCLUSIONS: From the above relationship and values of Table 1
we can conclude that:
?
Nonpolar compounds of low water solubility strongly favor the
organic over the aqueous phase.
?
Nonpolar organic compounds feel equally comfortable in the
highly nonpolar n-hexane and in the more polar n-octanol, as
CONTAMINANT FATE AND TRANSPORT
ORGANIC SOLVENT-WATER PARTITIONING
ADK
6
shown by the similar values of the partition constants (benzene,
toluene etc.).
Also note that nonpolar solvents have molar
volumes close to that of n-octanol (0.12 l/mol).
?
Polar substances (with oxygen or nitrogen functional groups) feel
more comfortable in octanol (-OH moieties create a friendly
environment for polar groups) than in nonpolar hexane. Note the
similar values of activity coefficients for benzene and water in
octanol (2.7, 3.6) and their difference in hexane (2.0, 2600)
CONTAMINANT FATE AND TRANSPORT
ORGANIC SOLVENT-WATER PARTITIONING
ADK
7
Table 1: n-Octanol-Water and n-Hexane-Water Partition Coefficients1
1
2
Partitioning
Compound
Kow
Khw
Cwsat
mol/l
?o
?h
n-Hexane
13,000
52,000
1.5?10-4
4.4
1.0
Benzene
130
170
2.3?10-2
2.7
2.0
Toluene
490
560
5.6?10-3
3.0
2.4
Chlorobenzene
830
810
4.5?10-3
2.2
2.1
Naphthalene
2,300
2,400
8.7?10-4
4.2
3.7
Benzaldehyde
30
13
3.1?10-2
8.9
19.0
Nitrobenzene
68
29
1.7?10-2
7.3
16.0
1-Hexanol
34
2.8
1.3?10-1
1.9
21.0
Aniline
7.9
0.8
3.9?10-1
2.7
25.0
Phenol
28
0.1
8.9?10-1
0.32
61.0
Water
0.04
0.00005
5.5?10-1
3.6
2600
Schwarzenbach R. P. et al., “Envronmental Organic Chemistry”, Wiley Interscience, 1993.
Value probably incorrect since ?wsat? ?w? because of intermolecular interactions of the phenol species at saturation concentrations
CONTAMINANT FATE AND TRANSPORT
ORGANIC SOLVENT-WATER PARTITIONING
5.2
ADK
8
EXPERIMENTAL DETERMINATION OF Kow
? “Shaker Flask” Method: The method yields direct measurements of
Kow. The procedure is similar to that described in section 4.1 above.
The method however is only applicable to compounds of low-tomedium hydrophobicity (Kosw < 105). Data for more hydrophobic compounds are very limited. 5.3 Kow ESTIMATION TECHNIQUES ? Aqueous Solubility: The octanol-water partition coefficient has been related to the aqueous solubility of organic compounds by empirical relationships such as the following equation: log K ow ? ? log C wsat ? log ? o ? log Vo [8] ? The above equation is based on the following assumptions: ? ?o is set equal to 1, 10, 100, 1000 (diagonal lines in Figure 3). ? the activity coefficient of the compound is independent of concentration in both phases ? the influence of octanol presence in water on ?w is negligible thus: ?w? ? 1 C wsatVw the concentration of the compound is low enough so that the molar volume of water-saturated octanol is not affected (Vo ? 0.12 l/mol) ? The deviations between the ?-lines and the actual experimental points in Fig. 2 represent estimates of log? for each compound. ? From the experimental data plotted on Fig. 2 we conclude that for most liquid compounds having water solubility greater than about 10-6 mol/l CONTAMINANT FATE AND TRANSPORT ORGANIC SOLVENT-WATER PARTITIONING ADK 9 the estimated values of ? are between 1 and 10, indicating nearly ideal solution behavior of these compounds in water-saturated octanol. ? If all else fails estimate Kow using web-based programs ? Virtual Computational Chemistry Labs ALOGPS Program ? Syracuse Research Corporation Interactive LogKow (KowWin) Demo CONTAMINANT FATE AND TRANSPORT ORGANIC SOLVENT-WATER PARTITIONING ADK 10 CONTAMINANT FATE AND TRANSPORT ORGANIC SOLVENT-WATER PARTITIONING ADK 11 EXAMPLE 1 10-6 mole of lindane is added to 100 mL separatory funnel containing 10 mL of octanol and 90 mL of water. Determine the concentration (mg/l) of lindane at equilibrium and 25?C. Kow of lindane is 6.025?103 IUPAC name (1r,2R,3S,4r,5R,6S)-1,2,3,4,5,6hexachlorocyclohexane Formula Mol. mass Chemical data C6H6Cl6 290.83 g/mol Production and agricultural use is banned the Legal 169 countries that parties to the Stockholm status Convention, but pharmaceutical use is allowed until 2015. SOLUTION 1. Mass balance for lindane nT ? no ? nw ? 10?6 mol 2. The fractional distribution of lindane in the aquatic and organic phases can be defined as: fw ? nw n moles in water ? ? w?6 moles in octanol ? moles in water no ? nw 10 fw ? CwVw CoVo ? CwVw 3. From the definition of Kow: K ow ? Co Cw Co ? K owCw 4. Combination of the two equations above gives: CONTAMINANT FATE AND TRANSPORT ORGANIC SOLVENT-WATER PARTITIONING fw ? CwVw K owCwVo ? CwVw fw ? Vw K owVo ? Vw fw ? 90ml ? 0.0015 6.025 ? 103 ?10ml ? ? 90ml ADK 12 5.Determine the number of lindane moles in water nw ? f wnT ? 0.0015 ? 10?6 mol nw ? 1.5 ? 10?9 mol 6. Determine the molar concentration of lindane in water: Cw ? nw 1.5 ? 10?9 mol ? ? 1.667 ? 10?8 mol / l ?3 Vw ?90ml ? 10 l / ml ? ? ?? ? ? Cw ? 1.667 ? 10?8 mol / l 290.83 ? 103 mg / mol ? 4.848 ? 10? 3 mg / l ? 0.5 ppm EXAMPLE 2 Use web-based programs to estimate Kow for the following compounds 1. caffeine 2. theobromine 3. androstenedione 4. testosterone SOLUTION 1. Google caffeine 2. Go to the Wikipedia entry and pickup CAS number for caffeine: 58-08-2 3. Go to the Virtual Computational Chemistry Labs ALOGPS Program by following this link 4. On the first line of the dialog box key in the CAS number for caffeine 58-08-2 and click the button submit on the right hand corner. Allow the program to run calculations and voila: 5. It returns a logKow = 0.16 CONTAMINANT FATE AND TRANSPORT ORGANIC SOLVENT-WATER PARTITIONING ADK 13 6. Similarly go to the Syracuse Research Corporation Interactive LogKow (KowWin) Demo enter the CAS number and get 7. This returns a logKow = 0.16 8. Repeat for the other compounds. CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING 4.0 ? ADK 1 AIR-WATER PARTITIONING This section discusses equilibrium partitioning of organic chemicals between the gas phase and an aqueous solution. ? For neutral compounds at dilute solute concentrations in pure water the air-water distribution ratio is referred to as the Henry’s Law constant. ? Henry’s Law constant (KH) represents a measure of the relative abundance of a chemical in the gas phase (partial pressure Pi) to that in the aqueous phase (molar concentration, Cw), thus: KH ? ? Pi atm ? l / mol Cw [1] If the abundance of a compound in the gaseous phase is expressed as ? mol/lair (Ca) then a dimensionless Henry’s Law constant, K H is defined: ? C K H ? a mol / l air / mol / l w Cw ? [2] ? K H and KH are related by applying the ideal gas law to convert partial pressure ( Pi ? ? n i V ? RT ): ? K KH ? H RT ? [3] The air-water partition coefficient quantifies the relative escaping tendency of a compound to exist as vapor molecules rather than in the water solution. ? Compounds with high vapor pressures (low fugacity in the gas phase) and high activity coefficient in water (high fugacity in aqueous solution), partition is favorable for the gaseous phase. (see Fig. 1) CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING ADK 2 CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING 4.1 ? ADK 3 THERMODYNAMIC BACKGROUND The fugacity of a compound in an ideal gas and in aqueous solution has been defined as: f g ? Pi fw ? ? w xw P 0 ( l ) [4] where Pi = partial pressure of compound Po(l) = vapor pressure of the pure liquid compound ? At equilibrium: Pi ? ? w x w P 0 ( l ) ? [5] After expressing the concentration of the compound in the liquid phase as C w ? x w Vw , the partition constant is defined as: 0 Pi ? w x w P ? l ? KH ? ? ? ? wVw P 0 ? l ? [6] Cw x w Vw ? The above equation shows that KH is directly proportional to both the activity coefficient of the compound in water (?w) and the vapor pressure of the pure organic compound [P0(l)]. 4.2 ? EFFECT OF CONCENTRATION ON KH As the concentration of the compound in the liquid phase increases towards saturation is KH affected? ? Suppose that a pure compound in its gaseous phase comes into contact with pure water. If the system is allowed to reach equilibrium, then the fugacity of the compound in the gas phase is equal to its vapor pressure thus: P 0 ? ? w x w P 0 ? l? ? [7] Also at equilibrium the liquid phase will be saturated with the compound, thus: CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING x ? sat w ? 1 ? sat w P0 ? 0 P ? l? [9] If the natural state of the compound is gas then: P 0 ? 1 atm ? [8] Note that if the natural state of the compound is liquid then: P 0 ? P 0 ? l? ? ADK 4 [10] Thus it can be stated that the equilibrium partial pressure in air above a saturated aqueous solution of a compound is equal to the vapor pressure of the pure compound at the same temperature. ? Assuming that the molar volume of the aqueous solution is not altered considerable by the organic compound, under aqueous phase saturation conditions the distribution ratio is: K Hsat ? ? P0 ? ? sat Vw P 0 ? l ? w sat Cw [11] So the question of how different KH is from Khsat comes down to how ?w differs from ?wsat. ? Research indicates that for several compounds studied (benzene, toluene, chlorinated solvents, organosulfur compounds) up to 3% molar solutions solute-solute interactions are negligible thus KH is not affected by concentration. 4.3 ? EFFECT OF TEMPERATURE Assuming that over small temperature ranges: ? the molar volume of water remains constant ? vapor pressure dependence on temperature is given by the equation: CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING ln P 0 ? ? ? ?H vap RT for a liquid compound that ? ADK 5 ? cons tan t sat w ? [12] 1 the temperature dependence x wsat is: ln ? ? sat w ?? ? [13] Thus the temperature dependence of KHsat is given by the equation: ln K H ? ln K 4.4 ?H se ?? ? cons tan t RT sat H ? ?H vap ? ?H es ? ? ?? ? ? cons tan t RT ? ? ?H Henry RT [14] ? cons tan t EFFECT OF SALTS Dissolved salts affect the activity coefficient in aqueous solutions but not the fugacity of the compound in the gaseous phase. Thus air-water partitioning varies directly with changes in ?w. ? In general values of the air-water partition coefficient are greater in seawater than in pure water (see Table 1) CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING ADK 6 Table 11 Dimensionless Henry’s Constants for Distilled Water and Seawater Compound KH ’ KH’ KH’(sw)/ (dist. water) (seawater) KH’(dw) CCl3F 3.6 5.0 1.4 CCl4 0.98 1.5 1.5 CH3CCl3 0.53 0.94 1.8 Hexachlorobenzene 0.054 0.07 1.3 2,4’-Dichlorobiphenyl 0.00713 0.079 1.4 2,4,4’-Trichlorobiphenyl 0.00595 0.00885 1.5 2,5,3’,4’-Tetrachlorobiphenyl 0.00357 0.00461 1.3 Dimethyl sulfide 0.075 0.089 1.2 Thiophene 0.095 0.11 1.2 1 Cited by Schwarzenbach R. P. et al. “Environmental Organic Chemistry”, Wiley, (1996). CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING ADK 7 EXAMPLE 1 (Air-Water Partitioning) The atmospheric concentration of the herbicide atrazine is 75.9 ng/m3, what is the partial pressure of atrazine in air at 27.5 oC? Atrazine, 2-chloro-4-(ethylamino)-6-(isopropylamino)-s-triazine Properties C8H14ClN5 215.68 g mol?1 colorless solid 1.187 gcm?3 Molecular formula Molar mass Appearance Density Melting point 175 °C Boiling point 200 °C Solubility in water 7 mg/100 mL SOLUTION 1. Convert atrazine concentration to mol/l: 75.9ng / m3 C? ? 10?9 g / ng ? 10?3 m3 / l ? 3.52 ? 10?13 mol / l 215.68 g / mol ? ? ? ? 2. Apply the ideal gas law: Pi ? CRT ? 3.52 ? 10?13 mol / l ?0.0821atm ? l / mol ? K ??300.5 K ? ? 8.68 ? 10?12 atm ? ? EXAMPLE 2 The Henry's law constant for atrazine at 25 oC is 6.2 X 10-6 atm-L/mol, estimate the air/water partition coefficient, KAW, for atrazine at 25 oC. SOLUTION KAW is the dimensionless Henry’s constant thus: KH 6.2 ? 10?6 atm ? l / mol ? K AW ? K H ? ? ? 2.53 ? 10? 7 RT ?0.0821atm ? l / mol ? K ??298 K ? CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING ADK 8 EXAMPLE 3 Estimate the Henry's law constant for the herbicide metolalchlor (FW = 283.8 g/mol; mp < 25 oC; Sw = 530 mg/L; Pol = 1.7 X 10-8 atm). IUPAC name (RS)-2-Chloro-N-(2-ethyl-6-methyl-phenyl)-N-(1methoxypropan-2-yl)acetamide Empirical names Dual, Pimagram, Bicep, CGA-24705, Pennant. Molecular formula Molar mass Appearance Properties C15H22ClNO2 283.79 g mol?1 Off-white to colorless liquid Boiling point 100 °C at 0.001 mmHg Solubility in water 530 ppm at 20 °C SOLUTION 1. Convert aquatic solubility on a molar basis: ?530mg / l ? 10?3 g / mg ? 1.87 ? 10?3 mol / l Cwsat ? 283.8 g / mol 2. Estimate KH from solubility and vapor pressure: Po 1.7 ? 10?8 atm K H ? lsat ? ? 9.1 ? 10? 6 atm ? l / mol ?3 Cw 1.87 ? 10 mol / l ? ? CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING ADK 9 EXAMPLE 4 200 mL of water containing 10-10 mol of compound X dissolved is added into a sealed container with 9.5 L of headspace. The KH of X is 0.369 atm-L/mol. What is the final aquatic concentration of X in the container and in the headspace at 25 oC after equilibrium is established? NOTE: Ignore water evaporation. SOLUTION 1. Determine the initial molar concentration of compound X in the aqueous phase: 10?10 mol i Cw ? ? 5 ? 10?10 mol / l 0.2l 2. Write a mass balance for X after equilibrium is established: Cwi Vw ? Cwf Vw ? CaVa [1] where Cwf ? molar concentration of X in the aqueous phase at equilibrium, mol/l Ca = molar concentration of X in the air phase at equilibrium, mol/l Vw = volume of water, l Va = volume of headspace, l 3. Calculate the dimensionless Henry’s constant: 0.369atm ? l / mol K C [2] K H? ? H ? ? 0.015 ? af RT ?0.0821atm ? l / mol ? K ??298K ? Cw 4. Solve Eq. [2] for Ca, substitute in Eq. [2] and solve Cwf ? ? Cwi Vw 5 ? 10?10 mol / l ?0.2l ? ? ? 2.92 ? 10?10 mol / l K H? Va ? Vw 0.015 ? 9.5l ? 0.2l 5. Determine Ca Ca ? 0.015Cwf ? ?0.015? 2.92 ? 10?10 mol / l ? 4.38 ? 10?12 mol / l Cwf ? ? ? CONTAMINANT FATE & TRANSPORT AIR-WATER PARTITIONING 4.5 ADK 10 ESTIMATION METHODS OF KH ? Most data in the literature are based on estimation methods ? The most common estimation method involves calculation of KH from capor pressure and solubility data. ? Table 2 shows experimental and estimated Henry’s constants. ? An alternative method has been proposed by Hine and Mookerjee2 based on structural contributions. ? The underlying idea of the method is that every subunit of organic compounds has a constant effect on air-water partitioning regardless of the compound in which it occurs. ? Table 3 shows the various subunit contributions. ? Calculation of Henry’s constant is based on linear algebraic equations of the following form: log K H ? a 1 subunits of type a 1 ? a 2 subunits of type a 2 ? ... [16] EXAMPLE 5 Estimate KH for bromodichloromethane and phenol using the Hine and Mookerjee method. SOLUTION 1. For bromodichloromethane (CHBrCl2) we have the bollowing subunit contributions: C-H = +0.11 C-Br = -0.87 C-Cl = -0.30. Thus: log K H? ? 1? ?0.11? ? 1? ?0.87? ? 2? ?0.30? ? ?1.36 ?K? ? ?K? ? 2 H estimated H observed ? 0.044 ? 0.085 Hine, J. and Mookerjee P. K., “The intrinsic hydrophilic character of organic compounds. Correlations in terms of structural contributions”, J. Org. Chem., 40, 292-298, 1975. CONTAMINANT FATE &... Purchase answer to see full attachment

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