"the universal solvent"
Why?
Bond angle
105° between hydrogens and oxygen creates a permanent dipole
Dielectric Constant: of 78.5 (twice that of all other common solvents)
Measure of the ability of a solvent to cause charges of opposite charge to separate and behave
independently.
Will salt dissolve in oil?
Hydrogen bonding
Sets the pH scale for aqueous systems from 0-14.
What would the pH scale be in say, glacial acetic acid?
Solubilizes all phases
Liquids, solids, and gases dissolve in water.
Equilibrium Constant:
Along with temperature it sets the equilibrium constant.
Temperature is always stated
Aqueous system is always assumed
The complexity of the chemistry cannot be over emphasized.
Models are simplified systems to provide insight into the chemical reactions that occur.
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" (calculations in Ch. 5)
Oxygen (O2)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 O2 in water |
Temperature of water |
|
14.74 mg/L |
0 °C |
|
8.32 mg/L |
25 °C |
|
7.03 mg/L |
35 °C |
7-8 mg of organic matter will stoichiometrically consume all the O2 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 O2 concentration goes to Zero or become very small, fish and other marine organisms die.
CO2 is an increasing atmospheric gas
|
Year |
CO2 Level |
|
1958 |
314 ppm |
|
1978 |
332 ppm |
|
1992 |
356 ppm |
A naturally occurring and biologically produced and sinks in natural waters
From {CH2O} + O2 + Bacteria ---- CO2 + H2O
Organic matter decomposition is also a significant source of CO 2
Natural Water Sinks for CO2
In Oceans 1/3 of excess CO2 produced by anthropogenic activity is absorbed by the Oceans
CO2 + Phytoplankton + Sun Light ---- (CH2O)
Also
Ca2+ CO2 + H2O <---- CaCO3 + 2H+
Another important CO2 sink is photosynthesis by plants, grasses, trees, weeds etc. in the lithosphere
CO2 + Plants + Sun Light ---- (CH2O)
Ref. In the Introduction to Env. Sci. text.
Living in the Environment, by Miller, 8th ed. Wadsworth
Publishing Co. CA, 1994. Text pgs 95-96. - Carbon Cycle Fig. 4-28
Carbon dioxide is a primary reagent in natural water buffer systems.
In many surface water systems the primary source of CO2 is the atmosphere at 0.035%. However, unlike O2, in some systems the primary source of CO2 may be the by product of biological reactions. Depending on the biosphere and the biological productivity.
The solubility of CO2 is primarily due to the formation of HCO3- and CO32- .
Greater than 25 mg/L of CO2 in water will cause death to many aquatic life forms.
CO2 + H2O <---- HCO3- + H+ <---- H2CO3
Different species (CO2, HCO3- , CO32-) predominate depending on the pH.
|
pH 0 - 6 |
CO2 |
|
|
pH 6 - 10 |
HCO3- |
Predominates in Most Natural Waters |
|
pH 10 - 14 |
CO32- |
|
The typical concentration of [HCO3- ] is 1.0 x 10-3 mol/liter.
The term Alkalinity is used to describe the particular
water systems ability to accept H3O+
or H+ ions (buffer capacity).
(sometimes called Total Alkalinity)
Alkalinity is usually due to natural buffers such as HCO3 - (the predominate CO2 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 PO42-, SO42-, and Natural hydroxides {Al(OH)3})
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."
(Skoog or M & R)
The Lewis definition (G. N. Lewis) -
"an acid is an electron - pair acceptor and a Lewis base
is an electron- pair donor"
(McQuarrie and Rock {M & R})
Two competing reactions occur simultaneously:
Ka
HA + H2O <---- H3O+ + A -
Ka =
Kb
A- + H2O <---- OH- + HA
Kb =
=
The ion - product constant of water
2H2O <---- H3O+ + OH-
(Note products over reactants form the equilibrium constant)
K = = Kw
= [H3O+][OH-]
Kw is the "ion product constant for water" usually given at 25 °C
(standard) at 25 °C, Kw = 1.01 x 10-14 , actually K (55.6)2 = Kw
(nonstandard) at 0 °C, Kw = 0.114 x 10 -14, at 50 °C, Kw = 5.47 x 10-14
Why doesn't the liquid water appear in the equilibrium?
The concentration of liquid H2O(l) is relatively
constant compared to the ions and it is a liquid and liquids and
solids do not appear in equilibrium constants (H2O(l)
is a constant 55.6 mole)(and their activity is = to 1). A Proof is
provided on pg. 126 Skoog 6th ed.
Kw = [H3O+][OH-] = 1.0 x 10-14 (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+] = 10-5.1 = 7.9 x 10-6
Thus: (also useful relationship)
14 = pH + pOH
pOH = 14 - pH and pH = 14 - pOH
The range of pH values for common aqueous solutions.
Titration curve for the titration of 50.0 mL of 0.100 M HCl(aq) with 0.100 M NaOH(aq).
Note that the pH of the solution changes very slowly until the vicinity of the equivalence point is reached (50.0 mL of NaOH(aq)) and then changes very rapidly in this vicinity.
Carbon Dioxide and its hydrated equilibrium components have a unique relationship with natural waters
CO2(aq) + H2O <---- H2CO 3 <---- HCO3- + H+
CO2(aq) + H2O <---- H2CO 3 eq. 1
H2CO3 <---- HCO3- + H+ eq. 2
CO2(aq) + H2O <---- HCO3 - + H+ eq. 3 Sum
CO2(aq) + H2O <---- HCO3 - + H+
Keq =
But water in liquid, ([H2O(l)], liquids, and solids not included)
Equilibrium Constant - K (Keq)
K eq = = 4.45 x 10-7 pK
a1 = [-LogKa] = 6.35
HCO3- <---- CO32- + H+
K eq = = 4.69 x 10-11 pK
a2 = [-LogKa] = 10.33
Which equilibrium is most likely in natural water systems?? Why??
A distribution of species diagram for CO2, HCO3 -, CO32- is given:
The specie of carbonate found in most waters depends on the pH and since most natural waters are pH 6.8-8.2 HCO3- is the predominant form.
CO2 =
HCO3-
=
CO32-
=
Substitution of Ka1 and Ka2 from the
equilibrium into
yields a relationship as functions of
(1.) equilibrium constant and
(2.) hydronium ion concentration.
CO2 =
HCO3-
=
CO32-
=
[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
Total alkalinity (all contributing species) becomes:
[alk] = [HCO3-] + 2[CO32- ] + [OH-] = 1 x 10-3 mol/L
at pH 7 then [HCO3- ] ~= 1 x 10-3 mol/L
at pH 10 then [HCO3- ] + 2[CO3 2-] + [OH-] = ratio 4:4:1 = 1 x 10-3 mol/L
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.
CO2 + H2O + hv + algae ---- {CH2 O} + O2
HCO3- + H2O + hv + algae ---- {CH2O} + OH- + O2
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?
Given: One buffer present from gas Henry's Law of partial pressure
CO2(g) <--- CO2(aq)
CO2(aq) + H2O <--- H2CO 3 <--- HCO3- + H+
CO2(aq) + H2O <--- HCO3 - + H+
The product of two reactions
Ka1 = = [H+]2/(1.146 x 10-5)
= 4.45 x 10-7
[H+] = [HCO3-] = {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.
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.
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
CaCO3 rock and minerals
atmospheric CO2
etc.
Carbon dioxide-calcium carbonate equilibria.
Natural water in contact with a mineral calcium carbonate
CaCO3(s) (calcite, limestone, shells, coral...)
Ca(aq) + CO32-(aq) + <--- Ca2+ + H2CO3(aq)
Note: over simplified equation not balanced for H+ form water
Assuming: that [CO32-(aq) = [H 2CO3(aq)]
given Ka of H2CO3(aq) =
4.45 x 10-7
given Ksp of CaCO3(aq) = 4.8 x 10
-9
Water in contact with solid CaCO3 would have the calculated concentration of these species
[CO2] = 1.146 x 10-5 M
[HCO3-] = 9.98 x 10-4 M
[CO32-] = 8.96 x 10-6M
[Ca2+] = 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
Very useful for buffer calculations (a review)
(works in 90-95% of the time)
Ka =
Take the log of both sides of the equation
Log Ka = Log
and
Log Ka = Log[H3O+] + Log
Put log [H3O+] on the left of the
equation and Log Ka on the right
so that pH = -Log[H3O+] or -Log[H+
] and, -LogKa = pKa
-Log[H3O+] = -Log Ka - Log
pH = pKa + Log
(Final form of the Henderson-Hasselbach Equation)
The equation can be used for buffer calculations when:
Ka of the acid involved is small (Ka < 10-3).
* Dilution of a buffer does not change the pH at which it buffers within several orders of magnitude of dilution
Example (Skoog Figure 10-3 show dilution of a strong acid, weak acid, and a buffer.) Several examples of buffer titration curves.
Start with equilibrium equation and the equilibrium constant
HA + H2O <--- H3O+ + A -
(Products over reactants form the equilibrium constant)
Ka =
Take the log of both sides of the equation (when, Ka and Kb < 10-3 M)
Isolating the hydronium ion (it will become the pH)
= Ka
[H3O+] = Ka x [HA]/[A- ]
(take the -Log of both sides)
-Log[H3O+] = -Log Ka - Log[HA]/[A-]
p[H3O+] = pKa - Log[HA]/[A -]
(inversion of the Log changes the sign and it is usually written thus)
pH = Log Ka + Log
(Henderson-Hasselbach 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, 1991.
Analytical Chem., by Skoog et. al. 6th ed., Saunders, pg.
221, 1992.
Quantitative Analysis, by Day and Underwood, 6th ed.,
Prentice Hall, pg. 109, 1991.
To have a buffer both the weak acid and its conjugate base must be present to constitute that buffer pair.
Buffer Calculations
There are several general forms for buffer calculations.
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.
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.
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.
(Example Skoog - Figure 10-4 )
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.
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.
Natural water in contact with a mineral calcium carbonate
CaCO3(s) (calcite, limestone, shells, coral...)
Ca(aq) + CO32-(aq) + H20 <---- Ca2+ + H2CO3(aq) + 2OH -
This is mediated by the solubility of CaCO3 K sp = 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)
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?
If the first equilibrium lies far to the right and is dominant then the solution will be acidic.
If the second is dominant then the solution is basic.
Depending on the strength of these equilibrium, the solution will be acidic or basic.
The effect of dilution on the pH of buffered and unbuffered solutions. The dissociation constant for HA is 1.00 x 10-4. Initial solute concentrations are 1.00 M.
Titration curve for the titration of 50.0 mL of 0.100 M CH3COOH(aq) with 0.100 M NaOH(aq). The indicator used in this case is phenolphthalein, which changes from colorless to pink around pH=9. The relatively flat portion of the curve between 5 and 40 mL of added base illustrates graphically the CH3COOH(aq)/CH3COO-(aq) buffer action.
Titration curve for the titration of 0.100 M H3PO4(aq) with 0.100 M NaOH(aq). Note the three equivalence points. The third equivalence point is not sharply defined because HPO42-(aq) is a very weak acid, so HPO43-(aq) can compete with H2O(l) for protons. Note that the curve has three distinct sections: 0 to 50 mL, 50 mL to 100 mL, and 100 mL to 150 mL. In each section the behavior of pH versus volume of base is analogous to a titration curve for a monoprotic acid.
Titration curve for the titration of 50.0 mL of 0.100 M NH3(aq) with 0.100 M HCl(aq). The pH at the equivalence point is acidic because at that point we have a 0.050 M solution of NH4Cl(aq), and NH4+(aq) is an acidic ion while Cl-(aq) is a neutral ion.
Curve for the titration of acetic acid with sodium hydroxide. A: 0.1000 M acid with 0.1000 M base. B: 0.001000 M acid with 0.001000 M base.
Titration of 20 mL of 0.100 M H2A with 0.100 M NaOH. For H2A, K1 = 1.00 x 10 -3 and K2 = 1.00 x 10 -7. Method of pH calculations is shown for several points and regions on the titration curve.
The term Acidity in natural waters refers to the ability of the water to neutralize OH-.
Since the carbonate ion the bicarbonate ion and carbonic acid
are week acid and salts of week acids they constitute a buffer
and are largely responsible for the pH of natural waters.
(CO2, HCO3-, CO3
2-)
Other molecular species and ions contributing to alkalinity and
pH:
NH3, H3PO4, R-Carboxylic acids, H
4SiO4 (Silicic acid).
pH is an intensity function while
alkalinity is a capacity function
Each of these species can act as buffers and each produce pH stability and buffer capacity and contribute to the total alkalinity of the water solution.
When taken a group the term is frequently referred to as -
Total Alkalinity
Carbon dioxide-calcium carbonate equilibrium.
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)
HNO3 (acid) + H2O (base) <---- H3 O+ (c-acid) + NO3- (c-base)
H2O (acid) + NH3 (base) <---- NH4 + (c-acid) + OH- (c-base)
Strong acid or bases dissociate in water completely and leave no undissociated molecules.
Strong acids and bases react completely:
Acid + Base ---- salt + water
HNO3 + NaOH ---- NaNO3 + HOH
Weak acid or bases do not dissociate completely and leave varying amounts of both species in solution.
CH3COOH + H2O <---- H3O + + CH3COO-
In an equilibrium that can easily be shifted
Remember LeChatelier Principle
What will shift this to more CH3COOH form ??
Water is an Amphiprotic solvent acting as either a
proton donor or acceptor.
H3O+ , OH-
(A brief review)
Equilibrium Constant - K (Keq)
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 -
Kforward =
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]
Ca(aq) + CO32-(aq) + H2O <---- Ca2+ + H2CO3(aq) + OH -
at 25 °C, Ka of H2CO3
(aq) = 4.45 x 10-7
at 25 °C, Ksp of CaCO3(aq) = 4.8 x
10-9
Dissociation - Ion-Product constant
Example for Water:
2H2O <---- H3O+ + OH-
Kw = [H3O+][OH-]
Dissociation of a weak acid or base
Example Ka or Kb
CH3COOH + H2O <---- H3O + + CH3COO-
Ka =
CH3COO- + H2O <---- OH - + CH3COOH
Kb =
Also KaKb = Kw
(pH calculations simplified)
Assumptions of K1 dominance are usually acceptable when the K1 and K2 are separated by 103 or when K1/K2 103.
The pH of the weak acid can be calculated by the following
(ref. Skoog p 247).
This equation will take into account the influence of the hydrolysis of water.
Titration of polyprotic acids and bases can be treated as buffers between equivalence points.
An example of a diprotic acid is maleic acid titrated with NaOH.
The theoretical diprotic acid couple H2A and HA - are considered using the derived relationship for polyprotic acids.
Titration of 20 mL of 0.100 M H2A with 0.100 M NaOH. For H2A, K1 = 1.00 x 10 -3 and K2 = 1.00 x 10 -7. Method of pH calculations is shown for several points and regions on the titration curve.
The derived relationship can be used in this example until the second equivalence point is reached and a hydrolysis problem is created.
Diprotic acids have 2 buffering pairs and triprotic have 3.
(End of review and extension of buffer calculations)
Dissociation constants for acids.
Zwitterions - are ions containing both a positive and a negative charge simultaneously.
Amino acids are a particularly important class of amphiprotic compounds containing both weak acids (carboxylic acid) and weak bases (amines) (functional groups).
An example is glycine
three equilibria exist:
NH2CH2COOH <---- NH3+ CH2COO- (~ internal neutralization)
NH3+CH2COO- + H 2O <---- NH2CH2COO- + H 3O+
Ka = 2 x 10-10
NH3+CH2COO- + H2O <---- NH3+CH2 COOH + HO-
Kb = 2 x 10-12
This is an example of a Zwitterion.
Isoelectric point is the point where the ion has no attraction in an electromagnetic field. This occurs when the pH of the solution creates equal ionization on the molecule.
At the isoelectric point [NH2CH2COO- ] = [NH3+CH2COOH]
The relationship can be derived to predict this hydronium ion concentration and thus the pH:
example for glycine:
[H3O+] = ((KaKw)/K b)1/2
[H3O+] = (((2x10-10)1x10 -14)/2x10-12)1/2 = 1 x 10-6
pH = 6.0
= relative concentration
of the free acid and free anion as a function of pH
cT = sum of molar concentrations containing various species of the polyprotic acid.
0
1
2
1 = 0
+
1 +
2
Solving each of the equilibrium for the species of interest and plugging them in, results in alpha values that are independent of the total concentration.
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
BaSO4(s) <---- Ba+2 + SO4 -2
Ksp = [Ba+2][SO4-2]
Stoichiometry is the precise ratio of each of the elements (or complex ions) forming the molecule or solid precipitate.
1. Precipitation example assume - pH 7.0 in water
Anion: SO4-2
PbSO4(s) <---- Pb+2 + SO4 -2
Ksp = [Pb+2][SO4-2] = 1.6 x 10-8
similar reactions for Ba, Ca, Pb, Sr, etc., Cations
Anion: CO3-2
CaCO3(s) <---- Ca2+ + CO3 -2
Ksp = [Ca+ ][CO32- ] = 4.8 x 10-9
similar reactions for Ba, Cd. Ca, Pb, Mg, Ag
Anion: OH-
Al(OH)3(s) <---- Al+3 + 3OH-
Ksp = [Al+3][OH- ]3 = 2 x 10-32
similar reactions for Al, Cd, Fe (II) & (III), Pb, Mg, Zn
1. Precipitation example assume - pH 1.0 in water
Why?
Remember LeChatelier Principle
Anion: SO4-2
PbSO4(s) <---- Pb+2 + SO4 -2
Ksp = [Pb+2][SO4-2] = 1.6 x 10-8
SO4-2 + H+ <---- HSO 4-
Anion: CO3-2
CaCO3(s) <---- Ca2+ + CO3 -2
Ksp = [Ca+ ][CO32- ] = 4.8 x 10-9
CO32- + H+ <---- HCO 3-
For anion: OH-
Al(OH)3(s) <---- Al+3 + 3OH-
Ksp = [Al+3][OH- ]3 = 2 x 10-32
OH- + H+ <---- H2O
What happens to Al3+ ions and others like it?
Similar reactions for Al3+, Cd2+, Fe2+, Fe3+, Pb2+, Mg2+ , Zn2+
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.
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.
(not in text)
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)
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!
(not in text)
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.
(Skoog fig. 4-2 illustrates this phenomenon)
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 percipitate
(not discussed in text)
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.
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.
(Skoog Figure 4-5 illustrates 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.
Examples of Iron Hydroxide are used to illustrate the
phenomenon.
(References Skoog and Dzombak and Morel)
The solubility of Al(OH)3(s) as a function of pH. The amphoteric nature of Al(OH)3 is clearly shown by its solubility in both highly acidic and highly basic solutions. Note that Al(OH)3(s) is essentially insoluble over the pH range 4 to 10.
Solubilities of Fe(OH)3(s) and Zn(OH)2(s) as a function of pH. Note that a much lower pH is required to dissolve Fe(OH)3(s) than to dissolve Zn(OH)2(s). Therefore, at pH = 4.8, for example, Fe(OH)3(s) precipitates and Zn2+(aq) remains in solution. The Fe(OH)3(s) can be filtered off, thereby separating Fe3+(aq) from Zn2+(aq).
Solubility of ZnS and FeS in water saturated with H2S(atm) at various pH values. We can separate Zn2+(aq) from Fe2+(aq) by saturating a solution buffered at about pH = 1.5 with H2S(g). The ZnS precipitates and Fe2+(aq) remains in solution.
Solubility-product constants for various salts in water at 25 degrees C.
Solubility of silver benzoate in water as a function of pH. The addition of H3O+(aq) shifts the solubility equilibrium to the right (increase in solubility) as H3O+(aq) combines with C6H5COO-(aq) to produce C6H5COOH(aq) and thereby decrease [C6H5COO-].
EXAMPLE (Simple calculation and later Activity will be compared)
Solubility of a slightly soluble salt in water.
Ba(IO3)2(s) <---- Ba2+ + 2IO3-
Ksp = [Ba2+] x [IO3]2 = 1.57 x 10-9
How many grams of Ba(IO3)2(s) (fw = 487 g ) can be dissolved in 500 mL of water at 25 ýC ?
molar solubility of Ba(IO3)2 = [Ba2+ ]
[Ba2+] (2[Ba2+])2 = 4[Ba 2+]3 = 1.57 x 10-9
[Ba2+] = 7.32 x 10-4 M
amount of Ba(IO3)2 = 7.32x10-4 M x 0.5L = 3.66 x 10-4 moles of Ba(IO3) 2
3.66 x 10-4 moles of Ba(IO3)2 x 487{[g Ba(IO3)2]/[mole Ba(IO3)2 ]} = 0.178 g
Molar concentration expresses the number of a particular species present per liter of solution but the effective concentration may be different.
WHY?
This difference is call the activity - arising form electrostatic interactions in solutions.
It is usually only significant for concentrated solutions ~ 1M.
This is why the assumption of [concentration, M] is appropriate
below M < 10<sup-3
Cations and anions in solutions neutralize opposite charges and in concentrated solutions there is a greater probability that at any point in time more of them are interacting. In dilute solution the probability is decreased.
Activity - ax = f[X]
where
f - is called the activity coefficient
[X] - is the ionic concentration in solution
as [X] approaches lower concentrations, f approaches 1.
(in other words - as the solution gets more dilute f is less significant)
Therefore in dilute solutions activity - a - approaches ionic concentration [X] and ionic concentration is used to represent activity.
How significant can activity actually be? Compare at various concentrations.
Example Table 7-1 (Skoog) Activity Coefficient for Ions.
The activity coefficient f can be related to the charge on an ion and the ionic strength of the solution through the Debye Huckel relationship (1923)
Where:
fi is the ionic strength, [µ] that the
Debye Huckel relationship depends on can be calculated in the
following manner:
The ionic strength [µ] that the Debye Huckel relationship depends on can be calculated in the following manner:
Where:
µ - is the ionic strength
Ci is the molar concentrations of ion i
Z i is the charge on the ion i
[[Sigma]] is the sum of (of each ion in solution)
[[alpha]] is the effective diameter of the hydrated ion i in
Angstrom units
Also written: µ = 1/2([A]ZA2 + [B]ZB2 + [C]ZC2 +...)
aA + bB <---> cC + dD
Equilibrium Constant - K
K is the thermodynamic equilibrium constant
K =
{Because - Activity = ax = f[X]; f is the activity coefficient}
K = x
= K'
K' =
K' is the concentration equilibrium constant
fi is the activity coefficient of i,
So:
Why do we use concentration and ignore activity 99% of the time?
Because As: u ----> 0, fx ----> 1, a x ----> [X], K' ----> K
* These relationships are important, know these relationships, these are conceptual and are what you should not forget so that the concept belongs to your understanding.
Activity coefficients for ions at 25 degrees C.
Activity coefficients for dissolved ionic species.
EXAMPLE:
Ionic Strength calculation
Calculate the Ionic Strength of (a) a 0.1M KNO3 and (b) a 0.1M solution of Na2SO4.
using:
u = 1/2([A]ZA2 + [B]ZB 2 + [C]ZC2 +...)
also
Where:
u - is the ionic strength
Ci is the molar concentrations of ion i
Z i is the charge on the ion i
A, B, C - represent ions
(a). the concentration of K+ and NO3 - are 0.1 M
u = 1/2(0.1 x 12 + 0.1 x 12) = 0.1
(b). the concentration of Na+ and SO4 -2 are 0.1 M
u = 1/2(0.2 x 12 + 0.1 x 22) = 0.3
There are 2 Na+ for each SO4-2 so that the Na+
conc. is 2 x 0.1 M or 0.2 M and
Zi = 2 for SO4-2 so Zi2 = 22
EXAMPLE:
Activity Coefficient calculation
(reference Skoog chapter 6 & 7)
Calculate the activity coefficient of Hg2+ in a solution that has an ionic strength of 0.085. Use 5 angstroms for the effective diameter of the ion. Compare the calculated value with fHg2+ in the table of activity coefficients for ions at 25 °C. (Skoog table 7-1).
from the Debye-Huckel equation
-Log fHg2+ = [(0.51)(2)2 (0.085)1/2]/[(1 + (0.33)(5)(0.085)1/2]= 0.4016
fHg2+ = 10-0.4016 = 0.397 = 0.40
From the Table of activity coefficients for ions at 25
°C. (7-1, Skoog)
We interpolate from the table the activities for f
Hg
2+ thus:
[(0.10 - 0.085)/(0.10 - 0.05)) x (0.46 - 0.38)] + 0.38 = 0.40
which agrees with our calculation for the Debye-Huckel equation
{Note: interpolation is a way to obtain a number between
two other
numbers on a table once popular due to log table calculations}
Example of Solubility calculation using activity
coefficients
(without common ion effect)
Reducing the concentration by an order of magnitude and the ionic strength with it.
Problem:
Using activities instead of concentration to
calculate the solubility of Ba(IO3)2 in a
0.033 M solution containing MgCl2.
The thermodynamic solubility product (Ksp),
for Ba(IO3)2 (barium iodate) is 1.57 x 10-9
(K'sp is the concentration constant corrected for activity)
Ba(IO3)2(s) <---> Ba2+ + 2 IO3-
Ksp = 1.57 x 10-9 = aBa2+ x aIO3-
Replacing activities in this equation by activity coefficients (f) and concentrations [X] yields -(from Activity -a x = f[X] );
Ksp = [Ba2+] fBa2+ x [IO3-]2 f2IO3 -
Ksp = [Ba2+] [IO3-] 2 (fBa2+ X f2IO3 -)
Rearranging
[Ba2+] [IO3-]2 = K sp/(fBa2+ X f2 IO3-) = K'sp
(K'sp the concentration-based solubility product, corrected for activity)
Ksp divided by the activity coefficients is equal to
the concentration based solubility product constant, K'sp
that is not constant but is dependent on the
specific ionic strength of the solution.
Both (fBa2+ X f2IO
3-) are dependent on the ionic strength of this particular
solution and the concentration of charged particles and their
specific character.
Example of Solubility calculation using activity
coefficients
(Cont. without common ion effect)
Using activities instead of concentration to
calculate the solubility of Ba(IO3)2 in a
0.033 M solution containing MgCl2
Ionic Strength due to MgCl2:
(ion effect assumes that ionic concentration is from soluble salt
and not from sparingly soluble precipitate)
using:
u = 1/2([A]ZA2 + [B]ZB
2)
where: A and B are both the molar concentrations and Z the Charges of the ions
u = 1/2([Mg2+] x 22 + [Cl- ] x 12) Therefore:
u = 1/2(0.033 x 22 + 0.066 x 12 ) = 0.099 = ~0.1
We must first adjust the thermodynamic Ksp for the concentration and type of ionic strength and ion concentration of our solution and turn it into the K'sp concentration specific for this unique ionic strength set of condition the molecule in question barium iodate is being calculated for.
Using the Table of Activity Coefficients (at 25 °C) for
Ionic Strengths:
From the table in Skoog pg. 152 an ionic strength of 0.1
fBa2+ = 0.38
&
f2 IO3- = 0.78
continuing: Correcting for activity effect in the constant
[Ba2+][IO3-]2 = K
sp/( fBa2+ x f2IO3
- ) = 1.57x10-9/(0.38)(0.78)2
= 6.8 x 10-9 = K'sp
K'sp for Ba(IO3)2 Barium Iodate corrected for ionic activity in solution for this specific set of ions MgCl2 Magnesium Chloride contributing to the ionic strength at their present concentrations of 0.033M in solution.
Considering ionic strength and activity:
Using activities instead of concentration to calculate the solubility of Ba(IO3)2 in a 0.033 M solution containing MgCl2.Using the corrected solubility for activity and ionic strength of the MgCl2 solution.
K'sp for Ba(IO3)2 Barium Iodate corrected for ionic activity in solution for this specific set of ions MgCl2 Magnesium Chloride contributing to the ionic strength at their present concentration of 0.033 M in solution.
Correcting for activity effect in the constant
[Ba2+][IO3]2 = Ksp/ (fBa2
+ x f2IO3-) = 1.57 x 10-9/(0.38)(0.78)
2 = 6.8 x 10
-9 = K'sp
K'sp = [Ba2+][IO3-] 2 = 6.8 x 10-9
[IO3-] = 2[Ba2+] and if [Ba 2+] = X
[Ba2+] (2[Ba2+])2 = 4[Ba 2+]3
[Ba2+][IO3-]2 = [X] 2[X]2 = 4[X]3 = 6.8 x 10-9
X3 = 6.8 X 10 -9/4 = 1.7 X 10-9
Where: X=1.2 X 10-3 M
Considering activity then: X = 1.2 x 10-3M
Soluble concentration of [Ba2+] = 1.2 x 10-3
M, Considering activity
Using concentration instead of activity to calculate the solubility of Ba(IO3)2 in a 0.033 M solution containing MgCl2. MgCl2 is not considered in the calculation.
Neglecting ionic strength and activity:
Ksp = [Ba2+][IO3-] 2 = 1.57 x 10-9
[IO3-] = 2[Ba2+] and if [Ba 2+] = X
[Ba2+] (2[Ba2+])2 = 4[Ba 2+]3
[Ba2+][IO3-]2 = [X] 2[X]2 = 4[X]3 = 1.57 x 10-9
4[X]3 = 1.57 X 10-9 = Ksp
X3 = 1.57 X 10 -9/4 = 3.9 X 10-10
Neglecting activity X = [Ba2+] = Ba(IO3)2 = 7.32 x 10-4M
Soluble concentration of [Ba2+] = 7.32 x 10-4 M, Neglecting activity
This is the same as on page 31 where activity was neglected also but as you can see it is not correct if there are other ions present in solution with discrete charges that will influence the dissociation of this molecule.
Compare the two calculations
Neglecting activity conc./Considering activity conc. x 100 = % difference
(7.32 x 10-4)/(1.2 x 10-3) x 100 = 61% difference in the two calculations
The Activity calculation is 39% different and 39% more accurate when Activity is also evaluated.
Here in this example Activity makes the calculation 39% more accurate when evaluated considering the complex interactions.39% more barium iodate, Ba(IO3)2 is actually in solution than would be calculated using traditional calculations when these other ions are present in a very conservative example with low additional salt concentrations.
More of the (+) and (-) ions are associated with the other charges in solution and are thus not free to react to precipitate as would normally be predicted not considering other salts in solution. This is what makes environmental calculations very complex and requires the use of a more sophisticated approach.
Soluble ions like sodium chloride are electrostatically attracted to other ions and"...ions may be so close together that they behave as a neutral molecule..." and do not react as would be simply predicted by traditional equations.
In a solution of Na+ + Cl- + Ba2+ + 2(IO3-)Ba2+ (Cl-) behave as neutral and Na+ (IO3-) behave as neutral
Therefore the solubility
Ba(IO3)2(s) <---> Ba2+ + 2IO3-
does not hold as a simple relationship in salt solutions.
( Ref. Hargis, Analytical Chem. Principles and Techniques; or Skoog ch.7, 6th ed.)
Titration of 20 mL of 0.100 M H2A with 0.100 M NaOH. For H2A, K1 = 1.00 x 10 -3 and K2 = 1.00 x 10 -7. Method of pH calculations is shown for several points and regions on the titration curve.
Equilibrium
Ka =
If you assume all H+ came from HA
Ka = =
|
CH3COOH + H2O |
<----> |
H3O+ |
+ |
CH3COO- |
|
|
Initial Concentration |
CHA |
0 |
0 |
||
|
Reaction |
CHA- x |
+x |
+x |
Method 1. Successive Approximation
Method 2. Quadratic Formula
Use 5% rule to make assumption about denominator -x.
EXAMPLE
Conjugate base hydrolysis problem of the general form
Calculate the hydronium ion concentration and the pH in a
0.0100 M sodium hypochlorite (OCl-) solution at 25
°C.
Given Ka = 3.0 x 10-8.
OCl- + H2O <---> OH- + HOCl
Kb =
= Kw/Ka
Kb = = 3.3 x 10-7
|
OCl- + H2O |
<---> |
OH- |
+ |
HOCl |
|
|
Initial Concentration |
0.01-x |
x |
x |
Kb = = 3.3 x 10-7
[OH-] = 5.7 x 10-5
(5.7 x 10-4/0.010 M ) x 100 = 0.6%
OK test of assumption < 10%
[OCl-] = 0.001 - 5.7 x 10-4 ~= 0.010 M
pH
recall
Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25 °C)
[H3O+] = 1.0 x 10-14/5.7 x 10 -4 = 1.7 x10-10
pH = - Log [H3O+] = 10.75
or
pOH = - Log [OH-] = - Log 5.7 x 10-4 = 3.244
pH = 14 - pOH = 10.75
Titration curve for 25.00 mL of 0.100 M maleic acid, H2M, with 0.1000 M NaOH.
Curves for the titration of polyprotic acids. A 0.1000 M NaOH solution is used to titrate 25.00 mL of 0.1000 M H3PO4 (A), 0.1000 M oxalic acid (B), and 0.1000 M H2SO4
Composition of H2M solutions as a function of pH.
(frequently form acid in nature)
Metal ions in water are usually hydrated by water to stabilize the charge. They can also be coordinated with other anions or proteins or organic molecules to stabilize them. We will study these effects individually.
Metal ions can act as acids
Hydrated metal ions can act as acids as they are hydrated
Example: Fe(III) in water
(Same example as in text)
Fe(H2O)63+ <---> Fe(H2O) 5OH2+ + H+
Ka1 = 8.9 x 10-4
This metal ion is a strong acid compared to organic acids
Metal ion acids increase in acidity with increasing positive charge and decreases with increasing ionic radius.
Many metal ions form polymeric species with OH- as a bridging group. This is also another form of a complex formed with the hydroxide ions - 2(Fe(III) and 2(OH-) for a net charge of 4+ .
This is a process known as dehydration-dimerization
Other 2+, 3+, 4+, 5+ ,6+ ions form polymeric species with OH-
Al(III), Be(ii), Bi(III), Ce(IV), Co(III), Ga(III), Mo(V), Pb(II), Sc(II), Sn(IV) and U(VI)
Metal ions can also shift equilibrium and produce acidic aqueous environments such as in the case of Fe. The following is the reaction that occurs in the situation known as acid mine drainage.
Fe(H2O)63+ <---> Fe(OH) 3(s) + 3H+ + 3H2O
Why is this significant?
Hint:
Fe(OH)3(s) + 3H+ + 3H2O <---> Fe(H2O)63+
Ksp = 4 x10-38
Calcium Deposits, and Carbonates related theme
Minerals containing Ca and Carbonate in contact with water
CaCO3 (calcite, limestone, shells, coral...), CaSO 4, CaMg(CO3)2,
CaCO3(s) + CO2(aq) + H2O <---> Ca2+ + H2CO3
Water with high CO2 levels dissolve higher quantities of Ca minerals from geological materials in contact with the water. (formation of caves)
When the equilibrium is reversed large quantities of CaCO3
(s) are deposited thus giving rise to the term hardness.
(heating of water eliminating CO2 deposits Calcium
carbonate in pipes and boilers)
Many different species of the same element can exist in solution.
Different species have different chemical reactivity.
What is a species?
Why are they important in aqueous chemistry?
Generally -- species are different forms of an element that are chemically distinguishable or that can react in a different chemical manner. These include different oxidation states, metal complexes, organometallic compounds, hydrates, etc.
Table 2.2 form text (Manahan) gives some general species significant in water systems. Think also about Fe(II), Fe(III), Cr(III), Cr(VI), these are examples of oxidation state species. Different ligand or complexed species exist as well as different organic and organometallic species.
We will discuss this topic more extensively as a separate topic also and expand on this theme.
See table of species as examples:
82% of the 108 known elements are metals.
30 elements are of key importance and are usually evaluated.
100,000 different organic molecules are produced by industrialized countries. Each may have several environmentally derived conversions into metal binding compounds.
Thus several million species formed as complex ligands with other elements, ions, and organic ligands may result form the combination of these two sources.
Competing reactions from precipitation, multiple complexation and ligand formation and molecular bond formation are all being applied simultaneously.
Solvent and transport medium -
All of the fundamental environmental chemistry is water related as water forms the universal solvent for the planet and is the transport medium as well. It also enables most species formation.
Chelation is based on the formation of complex ligands
Metal Ions react with electron-pair donors to form:
1. Complex Ions.
(example Cu(NH3)42+
(example Cd2+ + CN- <--- CdCN+
or
2. Coordination Compounds
(example Cu(NH2CH2COO)2
Ligands are ions or molecules that form covalent bonds with cations by donating a pair of electrons which are then shared by the ligand and the cation.
The number of bonds formed per molecule is referred to a "Dentate" meaning tooth-like.
Unidentate reagents form a single bond per molecule.
Examples of Ligands
|
Neutral |
Anionic |
|
H2O * |
F-, Cl- *, Br-, I- |
|
NH3 * |
SCN- |
|
RNH2 (aliphatic amines) |
CN- |
|
HCO3- * |
|
|
CO32- * |
|
|
SO42- * |
|
|
OH- |
|
|
RN(COOH)2 |
|
|
RCO2- (Carboxylate) |
|
|
S2- |
* Major forms in natural waters
Radioactive elements, and essential and toxic elements in the environment.
Occurance of organometallic compounds in the environment.
Results of an equilibrium computation for freshwater system containing four organic ligands
Some chelates are built up in a stepwise manner with successive additions of ligand bonds and successive stability constants that generally decrease in stability with the number of ligands.
Cu2+, Cu(NH3)2+, Cu(NH3 )22+, Cu(NH3)3 2+, Cu(NH3)42+, and Cu(NH3)52+
Example - (form text pg. 51)
Zn2+ + NH3 <---- ZnNH3 2+
K1 =
= 3.9 x 102
ZnNH32+ + NH3 <---- Zn(NH3)22+
K2 =
= 2.1 x 102
Overall formation constant -
Zn2+ + 2NH3 <---- Zn(NH3) 2 2+
B2 =
= K1K2 = 8.2 x 104
For:
Zn(NH3)32+
B3 = K1K2K3
AND
Zn(NH3)42+
B4 = K1K2K3K4
(For stepwise formation constants see Skoog Appendix 5 page A-11)
Rule of thumb - Take note. -
Complexes with monodentate ligands are usually less stable than
those with multidentate ligands.
The formation of complexes occur when electron pair donors contribute electrons to a metal ion through the ligand bond filling empty orbitals of the metal ion.
(pronounced kee'late)
A chelate is a cyclic complex formed by a cation bonded by two or more donor groups contained in a single ligand.
One of the most important group of complexometric agents are the Aminopolycarboxylic and Polyprotic Acids
These are large organic compounds that posses amines and carboxylic acid groups.
They tend to form 1:1 complexes with metal ions and usually form 4 or 6 ligand bonds (tetra- or hexa- dentates).
Example of the most studied of these is EDTA or Ethylenediaminetetraacetic acid.
It is a Hexadentate ligand with Six potential bonding sites two amino groups and four carboxyl groups.
EDTA is water soluble and has four ionizable groups
It is a zwitterion of the form H4Y
It is usually used in the salt form: Na2H2 Y€H2O
The Ionization of EDTA is pH dependent
The dissociation constants are:
K1 = 1.02 x 10-2,
K2 = 2.14 x 10-3,
K3 = 6.92 x 10-7,
K4 = 5.50 x 10-11
EDTA combines in a 1:1 ratio with all metals regardless of the charge of the particular cation.
What about Na and H then (with 2 each) ?
Note: chelators can also be ion exchangers
It forms a stable complex with almost all metal cations (group or= II)
The steps in its progressive ionization are illustrated in
Figure 13-2, 3, 4. Fundamentals of Analytical Chemistry,
Skoog, West, Holler, Saunders College Publishing, NY, NY, pg.
288-289, 1992.
Structure of a metal/EDTA chelate
EDTA forma 1:1 complex with metal ions regardless of the charge on the cation.
Mn+ + Y4- <---- MY(n-4)+
The Formation Constants KMY are given in Table 11-1 (13-1) for many metal ions.
For
KMY =
Table of Formation Constants for EDTA Complexes
|
Cation |
KMY |
Log KMY |
|
Ag+ |
2.1 x 107 |
7.32 |
|
Mg2+ |
4.9 x 108 |
8.69 |
|
Ca2+ |
5.0 x 1010 |
10.70 |
|
Sr2+ |
4.3 x 108 |
8.63 |
|
Ba2+ |
5.8 x 107 |
7.76 |
|
Mn2+ |
6.2 x 1013 |
13.79 |
|
Fe2+ |
2.1 x 1014 |
14.33 |
|
Co2+ |
2.0 x 1016 |
16.31 |
|
Ni2+ |
4.2 x 1018 |
18.62 |
|
Cu2+ |
6.3 x 1018 |
18.80 |
|
Zn2+ |
3.2 x 1016 |
16.50 |
|
Cd2+ |
2.9 x 1016 |
16.46 |
|
Hg2+ |
6.3 x 1021 |
21.80 |
|
Pb2+ |
1.1 x 1018 |
18.04 |
|
Al3+ |
1.3 x 1016 |
16.13 |
|
Fe3+ |
1.3 x 1025 |
25.1 |
|
V3+ |
7.9 x 1025 |
25.9 |
|
Th4+ |
1.6 x 1025 |
23.2 |
Data from G. Schwarzenbach, Complexometric Titrations, p. 8.
London: Chapman and Hall 1957. Contents are valid at 20deg.C and
ionic strength of 0.1. at appropriate pH.
Fundamentals of Analytical Chemistry, Skoog, West, Holler,
Saunders College Publishing, NY, NY, pg. 288-289, 1992.
Figure 13-8 (Skoog) Depicts the pH range that EDTA functions over and the ions it complexes from pH 2 to 11.
Calculated titration of a model freshwater with cysteine (a) Speciation of the metals. (b) Cysteine speciation.
The hexadentate EDTA complex surrounds the ion and shields it from solvent interaction
Figure 13-4 (Skoog)
These laboratory complexes are pH dependent and must be carried out in Buffered solutions.
Natural complexes are also pH dependent and rely on naturally occurring buffers in natural water such as the carbonate or phosphate systems. (in biological systems such as Fe in hemoglobin)
Because the complexation is pH dependent a new term the Conditional or Effective Formation Constant (dependent on pH) apply at a single pH.
They modify
Mn+ + Y4-<---- MY(n-4)+
KMY =
Structure of H4Y and its dissociation products.
They take the following form
Mn+ + Y4- <---- MY(n-4)+
KMY =
(not corrected of pH)
Where:
4 =
cT = [Y4-] + [HY3-] + [H2 Y2-] + [H3Y-] + [H4Y]
Therefore solving for the product of the two constants yields:
=
4KMY = K'
MY
where:
K'MY is the conditional formation constant at a
specific pH where
4 applies.
4 is
calculated by applying the following relationship involving the pH
through the hydronium concentration.
4 =
4 has values
of 3.7 x 10-14 at pH 2.0 to 9.8 x 10-1
as given in Table 13-2 (Skoog).
Composition of EDTA solutions as a function of pH.
Minimum pH needed for satisfactory titration of various cations with EDTA
1. Example of titration curve using Ca2+
General rule of thumb for the calculation of concentration pM n+ for a complexometric titration.
| Region | Major Constituent | Major supplier of Ca2+ |
|
1. Before the Eq. Pt. |
Ca2+ + CaY2- |
Ca2+ |
|
2. At Eq. Pt. |
CaY2- |
CaY2- |
|
3. After the Eq. Pt. |
CaY2- + Y4- |
CaY2- |
Figure 13-5 (Skoog) Is the titration curve from Example 1 (Ca).
Ca has a larger Formation Constant than Mg and this shows up in the titration curve just as the titration of a weaker acid would in an acid base titration.
This illustrates the selectivity of a complexing agent like EDTA for specific ions in solution.
The variability of the complex with different metal ions at different pH values results in a different formation constant at different pH values. A minimum pH needed for an acceptable complex of various metal ions is in the range of pH 6-8.
Besides competing for different ions other competing reactions must also be considered in environmental complexation reactions.
Multiple competing reactions are occurring.
Hydroxides of iron and aluminum start forming at a pH of 3.5.
Multiple complexing agents such as monodentate Cl-, F-, OH-, NH3, or H2O
Chelating agents are found in natural waters, as part of industrial waste waters and sewage. They are ubiquitous in nature.
What are some of them?
Humic acids, Fulvic acids, Phosphates, anions, biological waste
products (NH3), Industrial molecules, etc, and water
itself.
Example - Case study of EDTA at Oak Ridge National Laboratory -
Oak Ridge National Laboratory in Oak Ridge, TN used EDTA to
chelate metals to make them more soluble in the clean-up and
disposal process and pumped them into disposal pits and disposal
wells at concentrations of 10-7 M. Some 12-15 years
after burial EDTA is making metals mobile in the environment that
would not be mobile. This case study shows the slow degradation
rate of EDTA and other similar chelators and the role they play in
transport of metals in the environment. (The use of specific
isotopes permitted the positive identification of these metal ions
in the environment and identified them as anthropogenic and coming
form the study)
{pg. 66 text, 6th ed. Env. Chem., by Manahan}
Others involving detergents (water softeners), Natural materials - later.
A Common Water Hardness Testing
Calcium Carbonate equivalency is calculated by adding drops of EDTA solution to a measured Buffered solution of the water to be tested. The standard usually used is 1 drop per 0.065 g (1 grain) of CaCO3.
Epilog (as an analytical tool)
Several decades ago the titration of common metal ions was performed using EDTA on a routine basis. Many of the complexometric procedures are 30 to 100 years old. In the past decades most of the tests have been transferred to spectroscopic methods.
Another form of complex formation -
Many metal ions form polymeric species with OH- as a
bridging group. This is also another form of a complex formed with
the hydroxide ions - 2(Fe(III) and 2(OH-) for a net
charge of 4+.
Case Study -
The sodium salt of NTA is used as a detergent or detergent
additive and is a substitute for phosphate detergents. It is
present in natural waters as a contaminant in mg and greater
quantities. It has a relatively long environmental lifetime in the
environment being broken down by a dominantly bacterial mechanism
almost exclusively.
The text goes through the ionization of Nitrilotriacetic acid (NTA).
NTA has 3 hydrogens to dissociate with Kas and pK as as follows.
Ka1 = 2.18 x 10-2 and pKa1 = 1.66
Ka2 = 1.12 x 10-3 and pKa2 = 2.95
Ka3 = 5.25 x 10-11 and pKa3 = 10.28
Ka2 dominates the species range for all normal natural waters due to its pH range. Figure 3.4 (Manahan) demonstrates this with the display of HT2- as the predominant species from pH 3 to pH 10.
Example from the book of a series of complex interactions of -
Ksp, Ka3, Kf and Kw.
Hydroxides such as Pb(OH)2 are solids and form with many metals at natural water pH's (usually at relatively low pHs).
Many metal ions are suspended in water or precipitated from water as hydroxides. An examination of the relationship of a chelate and a hydroxide.
Which are dominant at pH 8.0 in natural water?
Fractions of NTA species at selected pH values.
(Env. Chem. 6th ed. text, pg. 72-73)
Pb(OH)2 is a species in suspension in a fresh natural water at pH 8.00 NTA is present and at pH 8 is predominantly HT2- ion.
Pb(OH)2(s) + HT2- <---- PbT- + OH- + H2O
Equations describing the solubilization are:
adding them algebraically
Pb(OH)2(s) <---- Pb2+ + 2OH-
Ksp = [Pb2+][OH-]2 = 1.61 x 10-20
HT2- <---- H+ + T3-
Ka3 = = 5.25 x 10-11
(not adjusted for alpha [H+])
Pb2+ + T3- <---- PbT-
Kf =
= 2.45 x 1011
H+ + OH- <---- H2O
Pb(OH)2(s) + HT2- <---- PbT- + OH- + H2O
A question - which is the predominant species? - complexed lead as PbT- or uncomplexed chelating agent NTA as HT2- .
Thus
= K/[OH-] =
(2.07 x 10-5)/(1.00 x 10-6)
Thus is 20/1 in
favor of the complex over the free form of the chelating agent.
Also note that the hydroxide form of the solid precipitate was solubilized.
What does this tell you about the interaction of chelation in natural waters?
The solubility of this complex decreases with increasing pH
Apply Le Chatelier's principles to the overall reaction -
Pb(OH)2(s) + HT2- <---- PbT- + OH- + H2O
What would increasing [OH-] do to the [PbT- ]?
What would increasing [HT2-] do to the [PbT- ]?
What would decreasing [OH-] do to the [PbT- ]?
This is another very important ubiquitous component in natural waters.
A second Example with the chelator NTA.
Now lets do the same examination using lead carbonate (PbCO 3 ).
PbCO3 is in equilibrium with the CO2, HCO 3, CO32- system and the NAT species. At pH 7.0 the HT2- species is again the dominant species.
CO2(aq) + H2O <---- HCO3 - + H+
Keq = = 4.45 x 10-7 pKa1=[-LogKa
]= 6.35
HCO3- <---- CO32- + H+
Keq = = 4.69 x 10-11 pKa2=[-LogKa
]= 10.33
PbCO3(s) + HT2- <---- PbT- + HCO3-
Components of equilibrium (problem from text):
PbCO3(s) <---- Pb2+ + CO3 2-
Ksp = [Pb2+}[CO32- ] = 1.48 x 10-13
Pb2+ + T- <---- PbT-
Kf =
= 2.45 x 10-11
HT2- <---- H+ + T-3
Ka3 = = 5.25 x 10-11
CO32- + H+ <---- HCO 3-
PbCO3(s) + HT2- <---- PbT- + HCO3-
Given:
In natural waters HCO3- frequently has a
concentration of 1.0 x 10-3 M/L
Thus:
Thus is thus a
41 to 1 lead complexed as (PbT-) to uncomplexed free
NTA from of (HT2-).
As can be seen from Le Chatelier's principle as applied to the overall equation
PbCO32-(s) + HT2- <---- PbT - + HCO3-
the higher the HCO3- concentration the less of the complexed lead (PbT- ) and the lower the HCO3- the greater the complexed lead (PbT- ).
Thus in this case the complex depends on the HCO3 - concentration but both are dependent on the pH. Go back to the alpha table and adjust the pH and look at the effect of pH on complexation of NTA and see that the lower the pH the more the solubility and the higher the concentration of the complex PbT-.
Since competing complexes are always present lets examine one of the most common ions, calcium, and its influence on the NTA complex of lead.
Calcium also forms the NTA complex from the dominant ionic form of NTA (HT2- ). (given a pH of 7.0)
Ca2+ + T3- <---- CaT-
Kf =
= 1.48 x 108
and
HT2- <---- H+ + T3-
Ka3 = = 5.25 x 10-11
Ca2+ + HT2- <---- CaT- + H +
K' = = 1.48 x 10
8 x 5.25 x 10-11 = 7.75 x 10-3
K' is the product of Kf for CaT- and the dissociation of NTA Ka3 as before in the Pb example. Again in water Ca2+ is typically in natural fresh water at a concentration 1.00 x 10-3 M (remember pH at 7.0).
Thus [CaT-]/[HT2-] is 76/1 thus a 76 to 1 calcium complexed as (CaT- ) to uncomplexed free NTA from of (HT2- ).
What about the competition the carbonate?
Combining equilibria:
K" =
Remember -
PbCO32-(s) + HT2- <---- PbT - + HCO3-
and the revers of -(Ca2+ + HT2- <---- CaT- + H+) or
CaT- + H+ <---- Ca2+ + HT 2-
K' = = 1.48 x 10
8 x 5.25 x 10-11 = 7.75 x 10-3
PbCO32-(s) + CaT- + H+ <---- Ca+2 + HCO3- + PbT-
K" = K/K' = (4.06 x 10-2)/(7.75 x 10-3) = 5.24
Thus the distribution between the Pb complex and the Ca complex is described by
So twice the Ca chelate is present as Pb chelate. Thus the amount of Ca will influence the amount of PbCO3 solubilized form the solid and we have seen that many other equilibrium effect the amount of an element mobilized. In summary influences from
(a.) Carbonate and
(b.) Carbon dioxide,
(c.) Other metal ions,
(d.) pH (amount of H+ and OH- ), ionization
of NTA,
(e) extent of hydroxide
Thus many thing effect the complexation and equilibrium in water.
One final example of competing complexes, solubility's, metal chelates, pH and other equilibrium -
Graphical representation of the developed example -
This is one reason why we will need mathematical models to assist in combining all these competing equilibrium simultaneously.
Orthophosphate ion PO43-
Phosphoric acid H3PO4
pKa1 = 2.17
pKa2 = 7.31
pKa3 = 12.36
Phosphate is one of the essential molecules of living things and of industry and agriculture.
Orthophosphate (tetrahedral) (a resonance structure with one
double bond between phosphorous and oxygen in resonance),
(McQurrie and Rock pg. 398)
Holds Calcium in solution by chelating it and solubilizing it from CaCO3 geological formations.
It is also a buffer with a pKa at 7.31 and adds to the total alkalinity or total buffer capacity of a natural water system.
Phosphate in many forms is a complexing agent and a buffer.
Different forms have very different Ka1 , Ka2 ... Kan for polyphosphates
Phosphate can exist in straight chain polymeric forms of commonly 4 to 18 Phosphate monomers such as the:
Pyrophosphate ion P2O74- or
Pyrophosphoric acid H2P2O7
Triphosphate ion P3O105- or
Triphosphoric acid H5P3O10
Phosphate also forms ringed polyphosphates such as-
Trimetaphosporic acid and Tetrameaphosporic acid
These acids have different acid character from the monomer of phosphoric acid.
In natural waters all polymeric phosphates act as if they have
two distinct hydrogen types that have specific pKa
values of
pKa1 at pH 4.5 and
pKa2 at pH 9.5.
at pH 4.5 all the linear-chain hydrogens are lost (3H+ , hydronium ions)
at pH 9.5 all the end-chain hydrogens are lost (2H+, hydronium ions)
All of these polymers and ring structures eventually hydrolyze in natural waters to orthophosphoric acid.
Phosphoric acid linear- chain polyphosphates and ring phosphates along with its various salts are complexing agents along with SO42-, NH3, Cl- , CN-, EDTA, NTA, F-, OH-, CH3COO- and other naturally occurring molecules such as humic acids.
All function simultaneously in natural water systems.
One of the most important types of complexing agents that are naturally occurring are the humic acid and fulvic acids which are a class of compounds of similar origin and similar structural design.
These acids are not single molecules but a Class of Molecules.
Fulvic acids is generally accepted to be the soluble form of a large complex molecule class of 500 to 2,000 molecular weight.
Humic acids are generally accepted to be less soluble and exist in solid or particulate form in water and soil.
They have been recognized since 1800 but have only been recently studied extensively (since ~ 1970), due to there role in the formation of trihalomethanes {THMs} during chlorinating of water in purification.
Both forms ion exchange and complex metal ions accumulating and releasing these metals in complex interactions with pH, ionic strength, temperature, specific ionic content and other interactions.
Routine composition overviews suggest compositions of -
45-55% C; 30-45% O; 3-6% H; 1-5% N; and 0-1% S
Examples of these ions show their very complex structure.
You can recognize the usual chelating functional groups in abundance.
A possible schematic pathway for the formation of terrigenous humic acid from lignins.
Three proposed structures for terrigenous humic acids.
A possible pathway for the formation of marine humic acids.
Complexing functionalities of some biogenic ligands. (a) hydroxamate siderophores; (b) catechol siderophores; (c) phytochelatins.
The binding characteristics of the fulvic and humic acids follow roughly this pattern:
Fe3+ A3+ Fe2+ , Ni2+ , Pb2+ , Ca2+ , Zn2+ Mg2+, Ca2+ Na+ , K+
Charge, ionic radius, pH and other factors effect the complexation of these chelating agents.
Lignite is high in humic acids and is an example of a largely nonsoluble humic acid type. It is known for ion exchange of metals from percolating water.
In the environment all of the effects discussed so far and several more to come are acting simultaneously and therefore the most complex interactions known are what we are trying to study.
The complexity of natural phenomenon - a perspective.
In living systems (a group of cells, for example) certain patterns are set up, maintained by energy input by the system, ordered and others prohibited by the system (while it is alive) and patterns are discovered and identified. In each living system we looked for and identified these patterns in a particular order and establish the norm for this healthy living system.
In nature patterns are governed by the interaction of all equilibrium simultaneously and the results are up to us to predict and understand. The health of the system has many more natural patterns and is more difficult to study as it has more variables. If you are looking for a challenge you have found it.
References
References used in preparation of these lectures.
These are additional references of interest and for future
inquiry.
Aquatic Chemistry (An Introduction Emphasizing Chemical Equilibria inNatural waters), 2nd ed., Warner Strumm and James Morgan, John Wiley & Sons, Inc., NY, NY, 1981.
Environmental Inorganic Chemistry, Eds Kurt Irgolic and Arthur Martell, VCH Publishers, Deerfield Beach, FL, pgs. 1, 2, 4, 1985.
Principles and Applications of Aquatic Chemistry, Francois Morel and Janet Hering, John Wiley & Sons, Inc., NY, NY, pgs. 3, 369, 377-379, 394, 1993.
Surface Complexation Modeling - Hydrous Ferric Oxide, David Dzombak and Francois Morel, John Wiley & Sons, Inc., NY, NY, pgs. 104-105, 192, 1990.
Environmental Chemistry, Nigel Bruce, Wuerz Publishing Ltd. Winnipeg, Canada, 1991.
General Chemistry, Donald McZuarrie and Peter Rock, 3rd ed. Chapter, 20 (Solubiligy & Precipitation Reactions), Intrerchapter I & J, W. H. Freeman and Company NY, 1987.
Fundamentals of Analytical Chemistry, Skoog, West,
Holler, Saunders College Publishing, NY, NY, pg. 288-289, 1992.
Environmental Chemistry Home Page
Revised 4/23/99.