CE 520 Lect7-prec-st..

CE 420/520 Environmental Engineering Chemistry
Dr. S.K. Ong
Precipitation and Dissolution Phenomena
 Extremely important in natural water, water and wastewater treatment processes
Natural systems – precipitation of calcium carbonate and sulfates in the formation of stalactities and
- precipitation of suspended matters at the mouth of estuaries
Engineered systems – use of aluminum sulfate and ferric chloride for the precipitation/coagulation of
suspended solids
- use of caustic soda for the precipitation of heavy metals, eg., zinc, copper and nickel
 Precipitation occurs in three steps
- _______________ – spontaneous formation of a fine particle, usually a cluster of few molecules or ion
pairs of component ions of the precipitate or precipitation of molecules or ion pairs on fine particles that are
unrelated chemically
- ______________________ – deposition of the precipitate constituent ions onto nuclei
- __________________________________ – initial solid formed may not be the most stable but over a
period of time the crystal structure may change to a stable phase
 Number of caveats to consider before embarking on the study of precipitation
i. Many precipitation-dissolution reactions are _______________________.
Example, large supersaturation of CaCO3 is often observed in natural waters before precipitation
of the solid occurs. Other factors such as temperature, biological activity and nucleating surfaces
can dramatically enhance precipitation-dissolution kinetics.
ii. Even when the kinetics of precipitation are fast, the solid formed is often not the most ___________
Example, amorphous ferrous hydroxide Fe(OH) 2 (amorp.) is typically precipitated first and
evolved to the more stable form of Fe(OH)2 crystalline.
iii. ______________________________ solids that initially precipitated are often “nonstoichiometric”.
Example, common form of manganese oxide “-MnO2” has the stoichoimetry of MnO1.3. Many
solids contain impurities and foreign ions incorporated in the matrix.
iv. The formation of pure solids phases is not the only or perhaps even the dominant processes by which
many solutes are removed from solution. Other processes – surface adsorption
Precipitation Equilibrium
 Assume NaCl is added to a beaker of water
- various observations
- NaCl will dissolve until a point is reached whereby crystal NaCl is observed. The NaCl has reached its
solubility limit, Cs.
Solubility is defined as the maximum amount of a substance in moles/L or mg/L that can dissolve in a
solution at a given set of conditions before the solid phase of the substance is evident.
- at times, more salt can be dissolved (beyond the solubility) – a supersaturated solution is obtained.
Chemical reaction can be expressed as
i.e., neglecting activity and using concentrations
by definition, the activity of NaCl(s) = 1, then K so  {Na  }{Cl  }
AzBy(s) < === > zAy+ + yBz-
In general, for a reaction
K so  {A y }z {B z }y
Note that if activities are used, then
Neglecting activity, then
K so 
K so
( A y ) z ( Az ) y
 [A y ] z [B z  ] y
K so  [A y ]z [B z ] y
Common Ion Effect
- when a solution contains an ion that is the same as one of the ions from the dissolution of a solid, the solubility of
the solid will be less than that when the solid dissolves in pure water
NaCl < === > Na+ + Cl-
If we add CaCl2, i.e., the Cl- concentration will increase and therefore drive the equilibrium to the left resulting in a
net decrease in dissolution of NaCl
Conditional Solubility Product
- Sometimes, the concentration of each individual ions cannot be measured. Instead the total concentration of the
species of a given element is measured. Under such situations, conditional solubility product is used.
Note that
= conditional solubility product
= total concentration of metal ion M in all of its complexed form
= total concentration of anion A in all its forms
= M [CT,M]
= A[CT.A]
= [M+][A-]
= M [CT,M] A[CT.A]
= MA [CT,M][CT.A] = MAPso
A community fluoridates its water by adding CaF 2 to water. What are the concentrations of Ca2+ and F- when
sufficient solids of CaF2 are added to water. If the water already contains 90 mg/L of Ca2+, what will be the
concentration of fluoride in the water? Given Kso for CaF2 = 3.0 x 10-11.
Effect of Complexation on Solubility
- most cations and anions in solution do not exist as free cations or anions but instead interact with other ions or
molecules to form complex ions or coordination compounds. Complex ions can be cationic, anionic or neutral.
The water structure is given as shown:
- nonlinear molecule, H+ ions are 105o apart
- create dipole with positive ends on the hydrogen side and negative ends on the oxygen
- electrostatic attraction of the positive ends of one molecule for the negative end of
another. Bonds are called hydrogen bonds.
- tend to form an ice-like structure but degree of order extends only a short distance
- a flickering cluster from a monomer to up to (H 2O)40 molecules, with an average life of
10-10 to 10-11 seconds.
When a cation is added to the water, the “ice-like” structure is disrupted and the water molecules are oriented
towards the cation. This phenomenon is referred to as the hydration of the ion.
There are three types of water surrounding the cation
- first water layer – waters of hydration, strongly bonded to the cation, behaving as a single ion
- second layer – somewhat rigid and tightly bonded
- fluid layer - free water
The number of water molecules in the solvation sphere is called the primary hydration number.
See Figure 3.13 – shows that alkali metals are clustered near the bottom
- higher charged cations, trivalent metals have high heat of hydration, exist as highly hydrated
species such as Fe(H2O)63+ and Al(H2O)63+
- transition metals are clustered towards the center.
Commonly, a two step process takes place when a metal salt is added to water. This generally true for cation with
high hydration energy and with high ionic potential of the ion – whereby the OH bond is ruptured . This is called
hydrolysis or photolysis.
Two step process
Through a series of hydrolysis steps, the cation itself may become negatively charged. One can write an equivalent
equation of
As though OH- is added to the hydrated aquoaluminum (III) ion.
 A complex ion is formed when a cation or anion is covalently bonded to one or more coordinating groups. The
cation or anion is referred to as the central ion, (e.g., Zn2+) and the coordinating groups are called ligands (e.g.,
OH-, NH3).
 complexes can be cationic, anionic or nonionic depending on the charges of the central ion and the ligands.
 the total number of attachments or the total possible numbers of coordinated species to a central ion is referred to
as coordination number.
 Ligands are attached to the central ions by covalent bonds. Both electrons for the bond are from the ligand
 When a ligand is attached to the central ion at one point, it is referred to as a monodentate ligand.
 When the ligand is attached at more than one site (multisite), then it is called multidentate ligand
Example, carbonate, ethylenediaminetetraacetic acid (EDTA).
 Complexes containing one central ion are called mononuclear complexes
 more than one central ion – polynuclear complexes
Example, polynuclearhydroaluminium (III) complexes
mononuclear hydroxoaquoaluminum (III) complexes
polynuclear hydroxoaquoaluminum (III) complexes
 Number of ligands for a given complex ion is difficult to determine accurately
 A rule of thumb – the maximum number of ligands that can be associated with a metal ion is often twice the ionic
charge on the metal ion
 Maximum coordination number larger than 6 are very rare
 Quadrivalent metals such as Pt4+, Sn4+ generally have a maximum coordination number of six.
Complexation Reactions
 Reaction between a metal ion (Me) and a ligand can be written as follows (note that the charges have been omitted
to simplify the reaction):
 Note that for complex formation, the reaction is written as a formation constant rather than a dissociation constant
as in acid-base reactions or solubility equilibrium. The equilibrium constants are called formation constants or
stability constants.
Step-wise reaction is given by:
Or written in a cumulative form:
Note that ’s and k’s are related :
An example of complex formation is that between hydrated zinc ions and ammonia
Commonly written as shown by ignoring water of hydration
The effects of complex ions formation on solubility of a metal cation may be illustrated as shown:
Ammonia Complexes
To assist in the understanding of the effects of complex ions and pH on the solubility of a metal salt, a pC-pH
diagram is plotted. The purpose is to see how each complex will change with pH.
Example: show how the concentrations of Fe(III) hydrocomplexes change with pH. Determine the pH where the
concentration of Fe(III) is the minimum. Given:
Fe(OH)3 (s)
< ==> Fe3+ + 3OH3+
Fe + H2O
< == > Fe(OH)2+ + H+
Fe + 2H2O < == > Fe(OH)2+ + 2H+
2Fe3+ + 2H2O < == > Fe2(OH)24+ + 2H+
Fe3+ + 3H2O < === > Fe(OH)3o + 3H+
Fe3+ + 4H2O
< == > Fe(OH)4- + 4H+
log k
Note Limitations:
(i) Assume pure solids
- most probably will not be pure in real situations,
- may have impurities such as Ca2+, Mg2+ which may change the activity of the solid phase
(ii) Surface area/mole of solid will affect the solubility, i.e., kinetics and therefore the K so
(iii) Precipitation forms "active" solids which is thermodynamically unstable. Usually the K so for "active" forms are
higher than the "inactive" forms. Active forms can be converted into inactive forms or return to solution.
(iv) To get to "stable" solute, need to exceed the solubility product Kso by a factor of 10 - 100
Precipitation of major cations
See listing of solubility products and stability constants for major cations in water
- oxides and hydroxides
- carbonates and hydroxide carbonates
- sulfides
- phosphates
Using pC-pH diagrams, similar diagrams can be plotted to show the domain of the solid phase with respect to pH.
Aluminum hydrocomplexes
 Alum or aluminum sulfate is commonly used in the coagulation of colloidal materials.
 Aluminum ion, Al3+, behaves very much like Fe3+ in that it is easily hydrolyzed. However, Al3+ has a greater
tendency to polynuclear species (see earlier notes on polynuclear hydroxoaquo aluminum complexes)
 See Table 5.1 or page 268 in Water Chemistry book
Al(OH)3 (amph) < === >
Al3+ + H2O
< === >
Al3+ + 2H2O
< === >
Al3+ + 3H2O
< === >
Al + 4H2O
< === >
2Al3+ + 2H2O < === >
3Al3+ + 4H2O < === >
13Al3+ + 28H2O < === >
Al3+ + 3OHAl(OH)2+ + H+
Al(OH)2+ + 2H+
Al(OH)30 + 3H+
Al(OH)4- + 4H+
Al2(OH)24+ + 2H+
Al3(OH)45+ + 4H+
Al13O4(OH)247+ + 32H+
log Kso = - 33
log K1 = -5
log 2 = -9.3
log 3 = -15.0
log 4 = -23.0
log 22 = -7.7
log 43 = -13.9
log 32,13 = -98.7
 Plotted as shown in Figure 6.8, page 269
- lowest solubility = 6 – 7
- below pH 6.5, solubility controlled by highly charged polymeric species
 The above diagram is for freshly precipitated Al(OH) 3 (amph)
 A more stable form of Al(OH)3, gibbsite has a similar formula but different solubility product, log Kso = -36.3,
which slightly less soluble than the freshly amorphous Al(OH) 3
 Similar pC-pH diagram can be plotted. Differences:
- gibbsite predominates over a larger pH region than freshly precipitated Al(OH) 3
- for pH < 5, Al3+ rather than polymeric Al species tends to dominate
 The above diagram can be used for the removal of colloidal materials in water treatment (see Figure 5- 1, page
- colloidal particles in water are negative
- some researchers have proposed that polynuclear species can act as short-chain polymer and are adsorbed
onto specific sites on the colloidal particles and destabilizing them. This will allow aggregation to happen.
Initial dose produces insufficient hydroxoaluminum (III) species to produce effective stabilization.
- further addition of alum allows complete stabilization to occur.
- further increase in alum dose will result in restabilization of the particles because of near complete
coverage of particles with hydroxoaluminum (III) species.
- further addition forms Al(OH)3 precipitate – since the solubility product of Al(OH)3 is exceeded. Here
the bulky precipitate enmeshes particles in the precipitate and settles rapidly to form the so-called “sweep
floc” region of coagulation. At most water treatment plants, coagulation/flocculation takes place in the
“sweep floc” range.
 A commonly used diagram that relates the above phenomenon with the pC-pH diagram is attached.
- Included on the diagram (lower) is the zeta potential (charge of the particle) vs pH
- points D and E are the isoelectric points of the colloidal particles. At D, charge neutralization occurs, any
addition of Al3+ will cause restabilization. This is followed by sweep floc at approx. 30 mg/L or 10 -4 M.
- restabilization zone is a function of the surface area of the colloid
- high concentration of anion such as phosphate, silicate (SiO(OH) 33-) or sulfate may cause suppression of
charge reversal and restabilization
- significant amount of natural organics in the form of humic substances could control the alum doses
required for coagulation and would alter the zones of coagulation shown on the diagram.
Systems other than oxides and hydroxides
 For systems such as carbonates, carbonate complexes are not formed but rather the precipitate of the compound.
In natural waters, the concentrations of some metal ions in water are controlled by the solubility of the metal
 In dealing with the heterogeneous equilibria system of Me 2+ - H2O – CO2, it is important to distinguish two cases:
- systems that are closed to the atmosphere
- systems that include a gas phase in addition to the solid phase and solution phase
System 1: Closed to the atmosphere
 What is the soluble metal-ion concentration as a function of CT and pH. This question will provide us with
information on whether the water is oversaturated or undersaturated with respect to a solid metal carbonate.
 Using calcium carbonate as an example (interest is in precipitation in pipes)
Using equation
K so 
CaCO3 (s) < == > Ca2+ + CO32-
]{CO 32 ]
[CaCO 3 (s)]
[Ca 2 ] 
K so
[CO 32 ]
K so
 T CT
The above equation provides the equilibrium saturation value of Ca 2+ as a function of CT and pH
If CT CO3 = 10-3 M, then
K so
10 8.42 10 5.42
[Ca 2 ] 
 T C T 10 3  2
log [Ca2+] = log [10-5.42] - log 2
 For high pH, 2 has slope of zero, therefore log [Ca2+] will change with a slope of zero.
 For pH between pK1<pH<pK2, log [CO32-] has a slope of +1. Correspondingly, log [Ca 2+] will have slope of –1
 For pH < pK1, log [CO32-] has a slope of +2, log [Ca2+] will have a slope of –2.
See Figure 5.5 (attached), saturation values for various metal ions with respect to their carbonates
System II: System in equilibrium with CO2(g): CaCO3(s) –CO2-H2O system
 This is the open system we have discussed earlier. It is similar to preparing a solution by adding CaCO 3 (s)
calcite to pure H2O and exposing the solution to a gas phase containing CO2
 For a partial pressure of 10-3.5 atm and using the equilibrium equations for H2CO3*, HCO3- and CO32-, a diagram
can be plotted as shown for the CO2 system.
 With CaCO3 < == > Ca2+ + CO32-, the solubility is given by Kso = [Ca2+][CO32-].
 The change in calcium concentration with pH is as shown. Note that for different metal-carbonate compounds,
similar lines can be drawn.
Combined Systems
 Thus far we have examined systems with a single precipitate. In many systems there may be more than one solid
phase present. To capture information various dominant species, two types of diagram may be drawn. They are:
- pC-pH diagram
- provides concentration of species as a function of pH
- predominance area diagram
- provides information on the domain where each species will be significant or dominating
- diagram does not provide concentration of each species for a given pH.
 Taking the Fe(OH)2 (s) and FeCO3(s) system as an example and assuming a closed system with CTCO3 = 10-3 M, a
pC-pH diagram can be drawn. In this particular system OH - will be one of species involved in changing the
solubility of CaCO3.
For Fe(OH)2
Fe(OH)2 (s) < == > Fe + 2OH
log Kso = - 14.5
Fe(OH)2 (s) < == > Fe(OH)+ + OHlog Ks1 = - 9.4
Fe(OH)2 (s) + OH-< == > Fe(OH)3log Ks2 = - 5.1
There are other species but the above three will be sufficient to illustrate the plotting of a combined diagram
For Fe CO3 (s)
FeCO3(s) < == > Fe2+ + CO32FeCO3(s) + OH- < == > FeOH+ + CO32FeCO3(s) + 3OH- < == > Fe(OH)3- + CO32-
log Kso = 10.7
log K = -5.6
log K = -1.3
Approach - draw separate diagrams
 Draw diagram for Fe(OH)3 (s) system (Figure 6.9)
- Fe2+ line, Fe(OH)+ line and Fe(OH)3 Draw diagram for FeCO3(s) system (Figure 6.10)
- CO32- line and Fe2+ line
- have two other lines for this diagram, Fe(OH) - and Fe(OH)3For Fe(OH)-, use
FeCO3(s) + OH- < == > FeOH- + CO32-
To draw this, take the log [CO32-] from the diagram and compute log[Fe(OH)+]
For pH > 10.3, - log [CO32-] is a constant, therefore
For pK1 < pH < pK2, - log [CO32-] is decreasing with a slope of 1, equation is a function of pH with a slope of 1,
therefore, log[Fe(OH)+] will not change - remains constant
Similar lines can be drawn for Fe(OH)3- where
For pH > pK2, slope is equal to 3
For pK1 < pH < pK2, - log [CO32-] is decreasing with a slope of 1, equation is a function of pH with a slope of 2,
therefore, log[Fe(OH)+] will change with a slope of 1.
 Overlap diagrams, the following can be observed:
- Two areas of different solids that do not overlap (the solid here will be dominant and will precipitate first)
- One area where the two solids overlap.
Need to find the dividing line where one solid will dominate over the other
Fe(OH)2 (s) < == > Fe(OH)+ + OHlog K -9.4
FeCO3(s) + OH- < == > FeOH+ + CO32log K = -5.6
(ii) - (i)
FeCO3(s) + 2OH- < == > Fe(OH)2 (s) + CO32log K = 3.8
Therefore we have
solve for [H+], gives pH = 10.5
or we can assume that since the pH is around 10.3 (where the areas overlap), then assume that CO32- is 10-3.
At this pH, the system is in equilibrium with both phases.
For pH < 10.5, _______ will dominate.
For pH > 10.5, _______ will dominate
Predominance Area Diagram
 Shows only the predominant species at any particular pH, not the concentration of each soluble species
 To draw the diagram need to fix the total concentration of your species, for example Fe(II). For book example,
use CT Fe(II) = 10-5 M
 List all species - 5 species:
 Pair them up, 10 possible combinations
 For each pairing, write the stoichiometric equation that relates them
 Draw line on the pC-pH diagram to indicate at which pH or area each species will dominate
 Example:
Fe(OH)2 - Fe(OH)3use
Fe(OH)2 (s) + OH-< == > Fe(OH)3log Ks2 = - 5.1
Since CTFe = 10-5 M, substitute [Fe(OH)3-] = 10-5 and setting the activity of the solid phase = 1, we have
 This is vertical line that divides the species Fe(OH)3- and Fe(OH)2(s). pH less than 14.1, Fe(OH)2(s) dominates,
pH greater than 14.1, Fe(OH)3- dominates.
 Usually the next step is to take the next pairing that involves one of the above two species. In this case, Fe(OH) 2,
since this species dominates for lower pH.
Fe(OH)+ Fe(OH)2 (s)
Fe(OH)2 (s) < == > Fe(OH)+ + OHlog Ks1 = - 9.4
If Fe(OH)+ = 10-5 M, then pH = 4.6 + 5 = 9.6
 Have a dividing line between the two species.
 This is performed for each pairing. Will do more when we get into redox reactions.
Specific Applications
- calcium carbonate solubility
- phosphate solubility
Calcium solubility
Recall the different type of systems
Closed System
Open System
Open System with Acid or Base Added
These systems can be solved analytically or graphically.
Charge balance
2[Ca2+] + [H+] = [OH-] + [HCO3-] + 2[CO32-]
Kso = [Ca2+][ CO32-]
Kw = [H+][OH-]
2[Ca2+] + [H+] = [OH-] + [HCO3-] + 2[CO32-] + CA
= 2[SO42-] for H2SO4 added
= [Cl-] for HCl added
2[Ca2+] + [H+] = [OH-] + [HCO3-] + 2[CO32-] + CB
= [Na+] if NaOH is added
Ka1 = [HCO3-][H+]/[H2CO3]
Ka2 = [CO32-][H+]/[HCO3-]
CTCO3 = [H2CO3] + [HCO3-] + [CO32-]
Closed System
Open System
Open System with 1 x 10-3 M
[H+] added
Total Alk. (eq/L)
Total Hardness
pH 
Ca2+ 
Ca2+ 
 Interested in knowing whether a solution in equilibrium with solids is oversaturated or undersaturated, i.e. will
provide information on whether CaCO3 will precipitate out.
 Applications: precipitation of CaCO3 and CaSO4 in pipes at homes, in boilers which may clogged up pipes.
 To prevent precipitation, pH can be adjusted as shown above.
 Usually will treat the system as a closed system. But estimation is somewhat in error as it is never a closed
 Two approaches in predicting whether the solution will result in CaCO 3 precipitation.
- Use Gibbs free energy, set up equation CaCO3 (s) + H+ < = > Ca2+ + HCO3compute G
- Use of the Langelier Index (LI) or Saturation Index (SI)
hypothetical pH the water will
have if in equilibrium with CaCO3
pHs is derived from CaCO3 (s) + H+ < = > Ca2+ + HCO3- and is given by
If LI > 0 oversaturated CaCO3 will precipitate
If LI < 0 undersaturated CaCO3 will dissolve
If pH < 9, total alkalinity  [HCO3-]
If silica contributes or other weak acids are present, must find the true [HCO 3-]
Also, [Ca2+] is the free ion present. If calcium complexes are present such as CaCO30, CaHCO3+
and CaOH+, then need to compute [Ca2+] using
= [Ca2+] + [CaOH+] + [CaOH-] + [CaHCO3+] + [CaCO3o]
Example: Water pH = 8.0
[Ca2+] = 1 x 10-3
[HCO3-] = 2 x 10-3
temp = 10o C
 = 5 x 10 –3
pHs = pKa2 – pKso + p[Ca2+] + p[HCO3-] – log Ca2+ - log HCO3with  = 5 x 10 –3, and from Figure 3-4, Ca2+ = ______, HCO3- = _______
pHs = ________________________________________________________
LI = _________________ = _____________
Same water but in water heater at temp = 60o C (see Table 4.7 for pKso and pKa2 values)
pHs = ____________________________
LI = ___________________________________________________________
Practical application – usually maintain LI slightly positive, to allow a thin layer of precipitate to coat surface of
Precipitation Potential
If solution is supersaturated, need to know how much CaCO 3 (s) will precipitate. Use the following principles:
(i) Total acidity does not change as CaCO3 precipitates or dissolves
Total Acidity = 2[H2CO3*] + [HCO3-] + [H+] - [OH-]
If carbonates are removed no change in total acidity
(ii) Total alkalinity (eq/L) - Calcium (eq/L) = constant, as precipitation or dissolution occurs
Alk (eq/L) – [Ca2+ (eq/l)] = __________________________________
= ___________________________________
Example in Book, page 293
Calculate the amount of CaCO3 in eq/L and mg/L as CaCO3 which will precipitate for a solution with the following
initial conditions
pH = 8.7
[Ca2+] = 1 x 10-3
[HCO3-] = 2 x 10-3
temp = 10o C
 = 5 x 10 –3
See Table 4.7 and Table 4.2, at 10o C, Ka1 = 10-6.46, Ka2 = 10-10.49, Kso = 10-8.15, Kw = 10-14.5
Assume close system, find CTCO3 and compute [H2CO3*] = 1.1 x 10-5, [CO32-] = 3.2 x 10-5
Initial Values:
Acidity = 2[H2CO3*] + [HCO3-] + [H+] - [OH-]
= __________________eq/L
Total Alk. – [Ca2+] = = [HCO3-] + [OH-] - [H+]
= ___________________eq/L
After precipitation and equilibration, have the following:
Total Acidity = unchanged = 2.02 x 10-3 eq/L
Total Alk. – [Ca2+] = unchanged = 7.0 x 10-5
Ka1 = [HCO3-][H+]/[H2CO3]
Ka2 = [CO32-][H+]/[HCO3-]
CTCO3 = [H2CO3] + [HCO3-] + [CO32-]
Kw = [H+][OH-]
Kso = [Ca2+][ CO32-]
Solve by trial and error for [H+].
[H+] = __________ and [Ca2+] = _______________
Amount precipitated = [Ca2+] initial – [Ca2+]final
= ___________________= ____________moles/L
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