NAME __________________________________ AP NOTES: UNIT 4 (9) SOLUBILITY EQUILIBRIA Ksp
XXX) Titration Curve of a polyprotic acid
XXXI) Solubility Equilibria and Ksp
From tooth enamel to precipitation in nuclear power plant cooling towers to the release of toxic Al3+
ion from the soil due to acid precipitation, to hand-warmers, to stalactite formation, to kidney stones, to
hard water in your well to building a coral reef … each finds an application in this part of our work on
equilibria.
This section of the unit is all about studying the degree of solubility of a less-than completely soluble
substance.
Thus far we have involved ourselves with the study of acids and bases. (And … as your text points out,
we have concerned ourselves with only homogeneous solutions … ). Yeah, the other shoe is about to
drop.
This portion of the unit deals with the dissolution or the precipitation of ionic
compounds … and perforce, these reactions are heterogeneous.
Hence never has it been more appropriate to write:
IF YOU ARE NOT PART OF THE SOLUTION … YOU ARE PART OF THE PRECIPITATE.
A) A compound is considered to be soluble in water, if it dissolves in water. Now that might mean
that is dissolves (and remains undissociated) OR that it dissolves and dissociates into ions.
1) Sucrose is an example of the former situation. It is soluble in water. It dissolves, but it
remains utterly molecular … only the intermolecular forces of attraction between the
molecules are disrupted by the polar water.
a) Sucrose is a non-electrolyte
2) Potassium Nitrate is an example of a very soluble of the latter situation. It dissolves and it
dissociates virtually completely into K+ and (NO3)-. You’ll notice that the water does not
seem to be able to attack the covalent bonds comprising the nitrate polyatomic ion.
a) KNO3 is a strong electrolyte:
KNO3(s) + H2O(ℓ) K1+(aq) + NO31-(aq)
3) Then there are the ionic compounds which establish equilibria in saturated solutions, for they
are not thoroughly soluble in water. AgI(s) with an equilibrium constant of 8.5 x 10-17. Yeah
… well …whatever … BUT… what does dissolve, dissociates completely into ions … hence
AgI is also a strong electrolyte ….Remember, there is a difference between “concentrated”
and “strong electrolyte”….and saturated does not necessarily imply a high concentration….
B) But, compounds like AgI with such small Ksp values (sp = solubility product) are labelled often as
insoluble in water … or nearly insoluble or worse yet and even more indeterminately as
slightly soluble. Ugh!
C) The key to understanding this issue of solubility is to go back to that which started it all…
Equilibrium … Hey, it’s no good trying to fight Mother Nature … Let’s just try to get Her
to a draw…and let’s try to understand each other.
409
D) The Solubility Product Constant Ksp
1) First things first … remember ONLY SATURATED solutions exhibit a solution equilibrium
AND, that a saturated solution is one, in which, the solution is in contact with undissolved
solution.
a) So what are the 2 conditions required to consider in a solution equilibrium?
i) *the solution must be saturated
and that means that *there must be some small amount of the solid present.
2) Since this is all about equilibrium … the same rules for the equilibrium expression apply…
We include only (aq) or (g) …. nehvah-ehvah include (s) or (l) in the expression.
a) Hence, since this sort of equilibrium is all about the dissolution of a solid, the
equilibrium constant *indicates how soluble the solid is in water.
b) The equilibrium constant is termed the Ksp … the solubility product constant.
Okay … the term, solubility you get … but, product? Take a look…
Consider a saturated solution of calcium fluoride: CaF2(s)  Ca2+(aq) + 2 F-(aq)
Ksp = [products]coefficient or Ksp = [Ca2+] [F-]2
[reactants]coefficient
Yeah, I saw
what you
did there
Notice that the extent of solubility is predicated upon the product of the aquated
ions of the solution … raised to their stoichiometric ratios. Okay?
This was all done way, way back in Unit 3 Equilibrium … Yeah … those were
good times … happy days…. Go visit those notes, if you have a need to do so.
To review … The solubility constant (Ksp) is NOT the molar solubility … but a
*temperature dependent constant indicating the extent of solubility of the
compound, before an equilibrium is established. The larger the constant, the
more soluble the compound… and of course, vice versa!
410
2) c) To press the point re: solubility vs. solubility constant take a look at what your text
emphasizes (p.749):

Solubility of a substance is the quantity that dissolves to form a saturated solution.

Solubility is often expressed in grams of solute / liter of solution.

Molar solubility is the number of moles of solute that dissolve in forming 1 Liter
of saturated solution.

Molar solubility has the unit of mol/L

Ksp is the equilibrium constant for the equilibrium between the ionic solid and its
saturated solution.

It is a unit-less value. Thus, the magnitude of the Ksp can help determine how
much of the solid dissolves to form a saturated solution
Be sure you grasp the differences … the nuances of meaning … I figure by this
time you are up to speed…but just in case you need some help, see me!
Now with all of that written … I gotta raise an important point, before I forget …
3) Put this idea of solubility together with our work with common ion effect and we can pretty
readily see that while the solubility constant has only one value at a given temperature,
the solubility of a salt can have different values in different kinds of solutions! HUH?
Well just think about it… the solubility of AgCl in distilled water, at a particular
temperature is in all likelihood different from its solubility in a NaCl solution
The presence of the common ion will suppress the dissolution of the lesser soluble
substance … Do you remember this idea? We saw it with strong acids and weak acids
and then more clearly with the common ion effect …. To completely pervert the Bob
Dylan lyric, The THEMES, they are a repeat’n!
Thus, the solubility constant is constant (well sort of) … but the actual degree to
which the constant can be “obeyed” at that specific temperature, is determined by



the pH of the solution for some compounds,
the presence of a common ion for others
… and concentration (but for our work will ONLY assume rather dilute solutions … and ignore
the effects of increasing concentration ssshh!!!) We will see this again in section XXX of these notes.
411
E) How to calculate molar solubility of a compound using the Ksp …. Piece o’ cake….
e.g.) Calculate the molar solubility of AgCl in water Ksp for AgCl = 1.77 x 10-10
Write the dissolution equilibrium equation:
AgCl(s)  Ag+(aq) +Cl-(aq)
Think about the stoichiometry… it’s: 1:1 or
Ag+ Clinitial 0.00 0.00
change +x
+x
equil
x
x
Write the equilibrium expression … plug and chug
Ksp = [Ag+][Cl-] ≈
1.77 x 10-10 = x2 ….
x = 1.33 x 10-5 mol/L …
Hence the molar solubility of AgCl = 1.33 x 10-5 moles per Liter of solution.
Notice the unit for molar solubility = mol/L
1) Okay your turn … Calculate the molar solubility of CaF2 Ksp = 1.46 x 10-10
Write the dissolution equilibrium equation: * CaF2(s)  Ca2+(aq) + 2 F-(aq)
*remove
ICE it!!!
Ca2+ Finitial 0.00 0.00
change +x +2x
equil
x
2x
Write the equilibrium expression:
*Ksp = [Ca2+][F-]2 ≈
1.46 x 10-10 = (x)(2x)2
*1.46 x 10-10 = (x)(2x)2
* 1.46 x 10-10 = (4x3)
* remove
x = 3 Ksp
4
ARRRGH!!!!!!!!
or
x=
3
1.46 x 10-10
4
x = 3.3 x 10-4 mol/L
412
2) Switch it up … and apply what you understand ….
Solid silver chromate (Ag2CrO4) is added to pure water at 25°C and some of the solid
remains undissolved. The mixture is stirred for several days to ensure an equilibrium
exists between the undissolved Ag2CrO4(s) and the solution.
Analysis of the equilibrated solution shows that its silver ion concentration is
1.3 x 10-4 M (or moles/Liter).
Assuming that the silver chromate solution is saturated and that there are no other
important equilibria involving Ag+ or CrO42- ions in solution, calculate the Ksp of
the compound, from its concentration(s). (Brown and LeMay p. 749)
Write the dissolution equilibrium equation: * Ag2CrO4(s)  2 Ag+(aq) + CrO42- (aq)
Write the general equilibrium expression: * Ksp = [Ag+]2 [CrO42-]
Now, THINK: The Ksp depends upon the concentrations of silver ion and chromate ion.
We know from the problem that the [silver ion] = 1.3 x 10-4 g/mol
We do NOT know the [chromate]. How can we get a sense of the
chromate ion concentration, using the silver ion concentration?
*Look at the stoichiometry …..There are twice as many moles of silver
ion in solution, as there are chromate ions. Hence the [chromate]
must be ½ that of the silver ion concentration.
So *Ksp = (1.3 x 10-4)2 (6.5 x 10-5)
= 1.1 x 10-12
413
TRY THIS! 1.00 gram of copper(II) phosphate, Cu3(PO4)2 is added to 100.00 mL of distilled water at
25.0°C. The mixture is stirred occasionally over the period of a day. (Brown & LeMay p.750)
a) Write the equation representing the dissolution of copper(II) phosphate in water:
* Cu3(PO4)2(s)  3 Cu2+ (aq) + 2 PO43-(aq)
b) Write the equilibrium expression: *Ksp = [Cu2+]3 [PO43-]2
c) If, after a day of stirring, the equilibrium concentration of copper(II) were determined to
be 5.01 x 10-8 M, calculate the Ksp of the compound at 25°C
*Ksp = (5.01 x 10-8)3 (3.34 x 10-8)2 = 1.40 x 10-37
*The concentration of phosphate is calculated by recognizing that there is 1.5 times as much
copper(II) as phosphate (3:2) thus by dividing the copper(II) concentration by 1.5 we can
determine the stoichiometric equivalent for phosphate ion, in solution.
ans: 1.40 x 10-37
3) Ksp and Relative Solubility …
What would you answer, if asked: Of the choices, which of the following is the most soluble
ionic solid, in water, at 25°C?
1) Mg(OH)2
Ksp = 2.06 x 10
2) SrF2
Ksp = 2.5 x 10
-13
-9
3) PbBr2
Ksp = 4.7 x 10
4) Ba(OH)2
Ksp = 5.0 x 10
ans: *4
-6
-3
What would you answer, if asked: Of the choices, which of the following is the most soluble
ionic solid, in water, at 25°C?
Don’t answer this … ‘Cause frankly, ya got a problem. It looks simple enough
until you reflect upon how the Ksp is used to determine something like molar
solubility.
1) Mg(OH)2
2) CaF2
Ksp = 8.9 x 10-12
Ksp = 1.46 x 10-10
3) FeCO3 Ksp = 3.07 x 10-11
4) AgCl
Ksp = 1.77 x 10-10
Clearly, Ksp and molar solubility are related … but comparing two compounds is a
tricky business. Notice in the first question the dissolution stoichiometry for each
substance is: X(OH)2 → X2+ + 2 Y- which provides us with a general solution
of having to take the cube root of both the Ksp value and the concentrations to determine
the molar solubility …
414
So taking a look at something like Mg(OH)2 Ksp = 2.06 x 10
problem:
X2+ (OH)initial 0.00 0.00
change +x
+2x
equil
x
2x
-13
from the first
Ksp = (x)(2x)2 or Ksp = 4x3
The dissociation stoichiometry is the same for the other three choices: Compare
the dissociation stoichiometry of Mg(OH)2 to SrF2, PbBr2 and Ba(OH)2
But look at the choices from the second question: Which is most soluble?
1) Mg(OH)2
2) CaF2
Ksp = 8.9 x 10-12
Ksp = 1.46 x 10-10
3) FeCO3 Ksp = 3.07 x 10-11
4) AgCl
Ksp = 1.77 x 10-10
Notice the dissociation stoichiometries are different.
Two choices are: XY2 → X2+ + 2 YWhile the other two choices are:
yielding Ksp = 4x3
XY → X+ + Y-
yielding Ksp = x2
This difference will affect the molar solubility calculations … It is not so straight
forward to answer: Which is the most soluble?
Thus: * any direct comparison of Ksp values for different compounds can
be made only when the compounds have the same dissociation stoichiometry.
Given the above choices, we can compare magnesium hydroxide to calcium
fluoride (directly) OR we may compare directly iron(II) carbonate to silver
chloride … but any other comparisons should be avoided, without hitting the
ol’ mathematics route. Got it???
415
TRY THIS! From Nivaldo Tro’s Chemistry A Molecular Approach 3rd ed. 2014 p. 785
Questions: 1) Compare at 25°C, CaCO3 (Ksp = 4.96 x 10-9) and MgCO3 (Ksp = 6.82 x 10-6). Which is more
soluble in water? *magnesium carbonate … it has a greater (higher, larger) Ksp value,
suggesting a greater molar solubility at 25°C
2) Calculate the molar solubility of calcium carbonate and of magnesium carbonate at 25°C
For Calcium Carbonate
* CaCO3(s)  Ca2+(aq) + CO32-(aq)
For Magnesium Carbonate
* MgCO3(s)  Mg2+(aq) + CO32-(aq)
*Ksp = [Ca2+][ CO32-] ≈4.96 x 10-9 = (x2)
*Ksp = [Mg2+][ CO32-] ≈ 6.82 x 10-6 = (x2)
x = 7.04 x 10-5 mol/L
x = 2.61 x 10-3 mol/L
3) For those of us with well water in the Northeast, “hard water” can be an issue. According to the
reading what is the source of the calcium ion and /or magnesium ion in the water?
* Slight amounts of the solid carbonates are dissolved into water as it percolates and moves
through the soil. This water may be collected in wells.
keep going…..
416
4) Here in the northeastern USA, huge deposits of CaCO3(s) exist. The shelves of calcium carbonate
were probably developed by evaporating sea waters with carbonates reacting with the calcium and
left behind by retreating waters. In fact, much of our current topography can be seen due to the
Wisconsin Glacier (of some 12,000 years ago) having a “tough time” getting over these massive
shelves of limestone and depositing huge glacial erratics of dolomite limestone &/or granite (like the
Balanced Rock of North Salem?) as the glacier was stopped, melted, re-froze at these points of
limestone embedment.
Acid precipitation may be (but is not limited to) rain/snow/sleet with a pH less than 5.5. The rain
alone in New York State has an average pH of approximately 4.1 to 4.4 (a pH similar to that of
tomato juice and soda)
Hypothesize as to the effects of acid precipitation on the concentration of the hard water ion Ca2+
that may collect in people’s wells.
*Due to the large concentrations of calcium carbonate in the soil, and the rather acidic
condition of the precipitation, the well water has a very good chance of being quite hard …
and may be on the increase, as the pH levels drop. The acid rain could react with the calcium
carbonate along the general lines of HX(aq) + CaCO3 → CaX + H2O + CO2. This could
cause the calcium ions to become even more soluble in water (depending upon the nature of
CaX) and collect even more readily in the well-water.
XXX) Factors which Affect Solubility … We have been building to the idea that solubility of a solid is
dependent upon temperature and various other solutes dissolved in the solution. Three factors which
may affect the solubility in water, of an ionic compound are:
1. the presence of common ions
2. solution pH
3. the presence of complexing agents
417
A) Common-Ion Effect: The presence of additional calcium ion or carbonate polyatomic ion in a
saturated solution can reduce the solubility of calcium carbonate. This is an application of
Le Chatelier’s Principle
In general, the solubility of an ionic compound is lower in a solution containing a common
ion, relative to the compound’s solubility in pure water
For instance, study the graph of the molar solubility of calcium fluoride as a function of
sodium fluoride concentration. Fluoride is, clearly, the common ion….(Brown and LeMay p. 751)
1) Calculation of Molar Solubility
in the Presence of a Common Ion.
e.g.) Calclulate the molar solubility of
CaF2 (Ksp = 1.46 x 10-10) in a solution
containing 0.100 NaF
There is an implicit assumption here, that an
equilibrium between the calcium ion and fluoride
ion exists or develops … Hence the solution is
or becomes essentially saturated (for there is an
equilibrium) and this equilibrium will be disturbed
(perturbed) by the presence of the more soluble
sodium fluoride.
Write the dissolution equilibrium and the Ksp
CaF2(s)  Ca2+(aq) + 2 F-(aq) thus: Ksp=[Ca2+][F-]2
ICE it … But remember … While we are concerned ONLY with the ions of calcium and
fluoride there is 0.100 mol of F- already present due to the NaF….
Ca2+
Fwe can make a similar statement as we did in
initial 0.00
0.100
weak acids …. “x” is small compared to 0.100 M
change +x
+2x
and
it can be ignored (meaning 2x is ignored)
equil
x
0.100 + 2x
 Ksp=[Ca2+][F-]2
or 1.46 x 10-10 = [x][0.100]2 or
x = 1.46 x 10-10 = 1.46 x 10-8 M
0.0100
due to the squared value of 0.100
The molar solubility of CaF2 (a few pages back), in pure water was calculated to be
3.3 x 10-4 M … This means that the presence of NaF … with a common ion of F-1
reduced the solubility of CaF2 by over 20,000 times its solubility in pure water!!!!
418
TRY THIS! (No Math Required)… In which solution is BaSO4 most soluble?
1) in a solution that is 0.10 M BaNO3
2) in a solution that is 0.10 M Na2SO4
3) in a solution that is 0.10 M NaNO3
ans: *3 …. solubility will be affected by the presence of a common-ion … In choice 3 there is no ion, common to barium
sulfate. Additionally, sodium nitrate is a neutral salt … given that the sodium ion is the conjugate acid of a strong base and
the nitrate is the conjugate base of a strong acid. Hence, neither ion will have an affinity for H+ or OH- and thus there
should be no interference with the solubility of barium sulfate due to reasons of changing pH.
B) The Effect of pH on Solubility … It’s all about that anion … Keep this in mind….
Ultimately, the driving force for dissolution (and for all chemical processes) is determined by the Gibbs
free energy change. But that is for our next unit … Right now, let’s sum the dissolution of a salt as a two
step process which includes:
1. the breakup of the ionic lattice of the solid into cations and anions (electrolytes)
2. followed by the generation of molecule-ion attractions, meaning that the polar ends of water
molecules are attracted to and surround released ions.
The first step is rather endothermic … and one would think unlikely to happen.
But the second step is hugely exothermic and releases a large amount of energy.
Thus the net energy change depends on the sum of two large energy terms (often approaching
1000 kJ/mol) having opposite signs. Each of these terms will to some extent be influenced by the size,
charge, and polarizability of the particular ions involved, and on the lattice structure of the solid. This
large number of variables makes it impossible to predict the solubility of a given salt.
The solubility of a sparingly soluble salt of a weak acid or base will depend on the pH of the solution.
To understand the reason for this, consider a hypothetical salt MA which dissolves to form a cation
M+ and an anion A–which is also the conjugate base of a weak acid HA. The fact that the acid is weak
means that hydrogen ions (always present in aqueous solutions) and M+ cations will both be competing
for the A–:
The weaker the acid HA, the stronger the conjugate base and the more readily will it bond with
H+ reconstituting the weak acid HA. With an excess of H+ (as in an acidic solution) even more
anions (A-) will be removed from the solution and bonded up into HA. Ultimately this is a Le
Chatelier application. As the available anion (A-) is removed from solution, more of the solid will
dissolve.
Thus, In general, the solubility of an ionic compound with a strongly basic or even weakly basic
anion (conjugate base) increases, when dissolved in acidic solution as opposed to water.
419
Tro (p 788) has a terrific example as to how pH may affect the solubility of a weak base or salt.
Consider … of all things … the active ingredient in Milk of Magnesia!
Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq)
Hopefully you will recall that the “lighter” or less-massive
alkaline-earth bases are weak bases …like Mg(OH)2
You see, all of the (OH-) in a solution of high pH inhibits the dissolution of Mg(OH)2, a weak
base…. The solubility of this compound is highly dependent on the pH of the solution into
which it dissolves.
When the pH is high then common-ion and Le Chatelier’s Principle click into action.
As I wrote, … it is Le Chatelier and common-ion all over again….
This means that given the equilibrium:
Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq)
Adding to the high levels of (OH-) present, will cause a common-ion reduction in
solubility and shift the equilibrium to the left, lowering solubility.
However, when the pH is low, that must mean there is an acid about! As the solid magnesium
hydroxide dissolves, the acid will react with the free (OH-) driving the reaction to the right, and
thereby increasing the solubility of magnesium hydroxide.
Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq)
excess H+ of the acid will react with the (OH-) as it is
produced, via the dissolution of Mg(OH)2. As the OH- is
consumed the dissolution will continue, according to
Le Chatelier’s Principle
Common basic (alkaline) anions include (OH-), S2-, F- & CO32-, CN-1. Hence, hydroxides,
sulfides some fluoride salts, carbonates & cyanides are more soluble in acidic water, than in
pure water. Hopefully this makes perfect sense … since any conjugate base of a weak acid
or (OH-) will be affected by the presence of an acidic pH or added acid.
420
This leads us to the formation of stalactites (c for ceiling) and stalagmites (g for ground), as rainwater
(which is naturally acidic due to the formation of carbonic acid, from CO2 ), moves through limestone-rich
soils and rocks… altering solubilities of carbonates …
The Chemistry of Cave Formation
http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s21- solubility-and-complexation-eq.html#averill_1.0-ch17_s04_s01_f01
(a) This cave in Campanet, Mallorca, Spain, and its associated formations are examples of pH-dependent solubility equilibria.
(b) A cave (or sinkhole) forms when groundwater containing atmospheric CO2, forming an acidic solution, dissolves limestone
(CaCO3) in a process that may take tens of thousands of years. When water emerges from the ceiling of a cave that is open to
the atmosphere, some of the excess CO2 it contains is released as it equilibrates with the air. This raises its pH and thus reduces
the solubility of the carbonates, which precipitate as stalactites. Some of the water remains supersaturated and does not
precipitate until it drips to the cave floor, where it builds up the stalagmite formations.
Summary
TYPE OF IONIC COMPOUND
BECAUSE ….
base of a weak acid)
DISSOLVES
BETTER IN...
an acidic solution
(of any acid)
A weak base (e.g. transition metal -OH)
an acidic solution
The solid weak base dissolves slightly, releasing
(OH-) …but the anions begin to be neutralized by
the acid, and as the neutralization occurs, more of
the solid base dissolves
A neutral salt (A salt with the conjugate
a neutral water
The anion of the salt is the conjugate base of a
strong acid, and will not hydrolyze water, or react
with an acid …and pure water is the better solvent.
A basic salt (A salt with the conjugate
base of a strong acid)
the anion is the conjugate base of a weak acid. This
cb has the ability to react with the excess H+ of the
acid, reconstituting the weak acid, to some extent.
This consumes the conjugate base and shifts the
equilibrium to the right, dissolving more of the
solid
421
TRY THIS!
(No Math Required) …Imagine
putting the following SOLID salts into water into an acidic
solution. Explain whether it is more soluble in water or in acidic solution… Be sure to use
the anion of the compound in your response
1) BaF2(s) … this is more soluble in a(n) *acidic solution. F- is the conjugate base of a weak acid.
The F- will hydrolyze water, and produced HF, consuming the available (free) F- and as the [F-],
decreases, the barium fluoride will dissolve more….due to a shift in the equilibrium to the right.
BaF2  Ba2+ + 2F-
…as
F- decreases …. BaF2  Ba2+ + 2F-
more BaF2 will dissolve based on Le Chatelier’s Principle
2) AgI(s)… this is more soluble in a(n) *neutral solution …. iodide is the conjugate base of a strong acid
and thus has no affinity to hydrolyze water.
3) Mg(OH)2(s) ….This is more soluble in a(n) * acidic solution … because OH- is a good base. There
would be a neutralization reaction between the acid of the solution and the hydroxide ion. This
H+ and OH- neutralization would essentially consume the OH- and drive the dissolution of the
solid to the right …
4) FeCO3 …. This is more soluble in a(n) * acidic solution. The carbonate ion is the conjugate base
a weak acid (HCO3-). This conjugate base will react with the acid in the solution
driving the dissociation of iron(II) carbonate to the right, as the carbonate is consumed.
5) PbBr2 … This is more soluble in a(n) * neutral solution. Br- is the conjugate base of a strong acid
(HBr) and Br- will not hydrolyze water.
6) Cu(OH)2 …This is more soluble in a(n)* acidic solution, as the (OH-) ions will react with the acid
(be neutralized by) and as the hydroxide ions are consumed, more of the solid will
dissolve.
7) HgS …This is more soluble in a(n) *acidic solution, as S-2 is the conjugate base of a weak acid.
The cb will react with the excess H+ of the acid solution, reconstituting the weak acid.
This consumes the cb and thus more of the solid salt will dissolve
422
TRY THIS! (No Math Required)… Which of these compounds would be more soluble in a solution with
a pH of 3…. PbBr2 or PbF2?
*PbF2 is more soluble in an acidic solution than PbBr2 (Answer). PbF2 is a “basic salt”. It will ionize
yielding: PbF2  Pb(aq) + 2 F-(aq) and fluoride is the conjugate base of a weak acid (HF). (Setting)
The fluoride ion will react with the excess H+ to reform some HF, which consumes any of the fluoride
ion from lead(II) fluoride’s dissolution. As the fluoride ion concentration drops, more of the solid salt
will dissolve…. driving the dissolution to the right. (Proof)
Br- is the conjugate base of a strong
acid (HBr) and will neither hydrolyze water, nor does it have any ions that will react with an acid.
TRY THIS! Calculate the molar solubility of CaF2 at 25°C that is added to 0.010 M Ca(NO3)2 Brown and LeMay
The Ksp of calcium fluoride = 1.46 x 10-10
(p. 752 edited for Ksp value)
Hint 1: * Write the dissolution
equilibrium and the Ksp
expression for CaF2
* CaF2(s)  Ca2+(aq) + 2 F-(aq) thus: Ksp=[Ca2+][F-]2
* Ca2+
initial
0.010
change
+x
equil
0.010 + x
F0.00
+2x
2x
Hint 2:* ICE it … but remember
there is a [Ca2+] due to the
0.010 M Ca(NO3)2 …
*Ksp=[Ca2+][F-]2 or 1.46 x 10-10 = [0.010][2x]2 or 4x2 = 1.46 x 10-10
0.010
2
* x = 1.46 x 10
0.040
-10
or
2
x = 3.65 x 10
-9
or
-5
Hint 3:* The “x” is the
negligible amount of calcium
ion added to the solution due
to CaF2 …
ans: 6.05 x 10-5 M
x = 6.04 x 10 M
423
TRY THIS! Lead oxalate (PbC2O4), lead iodide (PbI2), and lead sulfate (PbSO4) are all rather insoluble,
with Ksp values of 4.8 × 10−10, 9.8 × 10−9, and 2.53 × 10−8, respectively. What effect does adding a strong acid,
such as perchloric acid, have on their relative solubilities? http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s21-04-solubility-andph.html
* The addition of a strong acid will affect a system with a conjugate base (or rather the conjugate base of a
weak acid. There are two in the above list that will be affected. The solubility of lead(II) oxalate will be
affected the most, since it is fully de-protonated (oxalic acid = H2C2O4.) Lowering the pH of a solution with
lead(II) oxalate will produce HC2O4- which will force the dissolution of the salt to the right:
PbC2O4  Pb2+(aq) + C2O42- (aq)
The other salt affected will be the lead(II) sulfate, for a very
(consumed forming the weak acid HC2O4-)
similar reason, but to a lesser extent as the anion HSO4- is not as weak an acid (it’s stronger than) as HC2O4-.
The solubility of lead(II) iodide should be relatively unaffected, as the iodide anion is the conjugate
of a strong acid, and has no affinity to re-constitute the molecular acid and to consume the I- anion
… forcing the dissolution to the right of the equation.
XXXI) Precipitation and the Separation of Ions
Qualitative analysis using
precipitation
What is precipitation?
& a review of Q
Selective Precipitation
A) Precipitation can occur when one of the possible cross products of the combination of a cation from
one solution & the anion from the other, is insoluble (or nearly insoluble, or slightly insoluble UGH!)
1) And, it is the range of possibilities … this concept that soluble and insoluble are simply
the capstones at either end to a rather expansive continuum….for solids have many levels
of solubility.
a) Hence understanding precipitation reactions may actually be helped by reviewing
the kinetics issue of Q…the reaction quotient.
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i) Q is different from Ksp because Q is the value of the product concentrations
under any conditions … whereas Ksp is the value of the product
concentrations at and ONLY at equilibrium.
b) Thus, revisiting our trusty CaF2 equilibrium: CaF2(s)  Ca2+(aq) + 2 F-(aq)
Q = [Ca2+][F-]2
2) Now, consider a solution of calcium fluoride in which Q < Ksp …. The dissolution should
continue to the right… All that is saying is that if a solution contains solid calcium fluoride
solid will continue to dissolve (it is unsaturated), until the solution saturates where Q= Ksp
When Q= Ksp we can assume the dissolution is at equilibrium. The
solution must contain a small amount of solid to maintain the equilibrium
CaF2(s)  Ca2+(aq) + 2 F-(aq) ….but that solid may be too
difficult to detect with the human eye …. or not….
Then there is the point when solid begins to precipitate out of solution,
where Q >Ksp …. precipitation will occur until a stable saturated solution
is reached.
However, under very special circumstances, Q can remain > Ksp for an
indefinite period of time … This is that rather unstable situation called
a supersaturated solution. The species will precipitate out when the
solution is disturbed sufficiently.
So, there is a quick and dirty look at unsaturated, saturated and supersaturated solutions,
with the twist of Q.
But, Q can be used to predict whether a precipitation will occur or not. Consider the mixing of two
solutions: KI(aq) and AgNO3(aq) …. Both are quite soluble in water.
KI(aq) + AgNO3(aq) →
AgI + KNO3 ….. We know the inorganic nitrates are soluble
so we have: K+(aq)+ I(aq) + Ag+(aq)+ NO3-(aq) → AgI + K+(aq) + NO3-(aq)
But, what about AgI? Will it precipitate out? According to the solubility rules, most of the halides of
silver, lead(II) and dimercury(I) are relatively insoluble in water. (Silver fluoride is a glaring exception
to this)
Well we might predict this by using Q… If Q were larger than Ksp (that is, if the concentrations of
Ag+ and I- were high enough), then a precipitate would form.
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Consider mixing of 0.0100 M AgNO3(aq) with 0.015 M KI(aq). Will AgI precipitate? (Ksp = 8.51 x 10-17)
Q = [Ag+][I-] = [0.0100][0.015] = 1.5 x 1004 … since Q is significantly larger than Ksp, there
most certainly will be a precipitate produced…..
Neat, Huh?
Conclusion: We can use Q to predict if the concentrations of species are large enough to produce
a precipitate in conjunction with our understanding of the solubility rules.
B) Selective Precipitation: Consider a solution mixture with multiple cations … Selective
precipitation is a process involving the addition of a reagent that forms a precipitate with one of the
dissolved cations of the solution but not with the other cations.
Selective precipitation is fairly straight forward… A sample containing a mixture of metal
cations is subjected to the addition of several precipitating agents. At each step, some of the
metal cations (those forming precipitates with the precipitating agent) can be separated out as
solids. The remaining solution mixture is subjected to other precipitating agents … and so forth.
Let’s look at a very general Qualitative Analysis scheme. There are dozens of variations to study
…. this is just one very general scheme. From Brown and LeMay p. 793
426
To see it another way … and to introduce us to “group 1 -5 cations” ….
Add HCl(aq)
Precipitate group 1 cations
Separate solution from precipitate (filter)
Add H2S (g)
Precipitate group 2 cations
Separate solution from precipitate (filter)
Add NaOH(aq)
or NH3 or
(NH4)2S
Precipitate group 3 cations
Separate solution from precipitate (filter)
Add Na2CO3(aq)
or
(NH4)2HPO4(aq)
Precipitate group 4 cations
Separate solution from precipitate (filter)
from: http://www.smc.edu/AcademicPrograms/PhysicalSciences/Documents/Chemistry_12_Experiments/Group%201%20Qualitative%20Analysis.pdf
You will notice that Ag+, Pb2+, and Hg22+ are called group1 cations … because they are the first group to be
assayed for and removed from a mixture of metal cations. You will notice that the alkali metal cations are the
remnant group … group 5 cations … because these ions are notorious for failing to make any sort of precipitate
and to remain in solution as hydrated ions.
As a little history … this was (and in some areas, still is) a hot, hot debate. In the early 1980s IUPAC officially
implemented the re-design of the Periodic Table. In those days, the alkali metals were Group IA, and the
alkaline-earth metals were Group IIA. The scandium family was known as IIIB… Chemists around the world argued
the change to group 1, group 2, group 3 …. fearing confusion with the qualitative analysis scheme and its “groups”.
Assignment: Read page 762 in your Brown and LeMay text and familiarize yourself with the
qualitative analysis methodology for cations….
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B) 1) Group 1: Insoluble Chlorides … The addition of HCl precipitates out the silver, lead(II) and
dimercury(I) ions. The absence of a precipitate upon adding HCl(aq) indicates that the
solution is free of these ions.
connection to Ksp and how pH affects solubility
Group 2: Acid-insoluble sulfides…. The filtrate is treated with H2S …a weak acid, but a source
of sulfide ion. Only the most insoluble sulfides will precipitate…
Group 3: Base-insoluble sulfides and hydroxides… Here’s the heavy hitter … to make the
rather acidic filtrate, alkaline, ammonium sulfide is added … and under these basic
conditions … Q exceeds Ksp and the more soluble sulfides are forced to precipitate.
Other cations (like Al3+, Fe3+, and Cr3+) will precipitate as hydroxides
Group 4: Insoluble phosphates: Add diammonium hydrogen phosphate (a.k.a. ammonium
phosphate dibasic or ammonium monohydrogen phosphate [yuck])
This addition will precipitate out most of the alkaline-earth cations
Group 5: The alkali metal ions and ammonium …. You must use flame tests to detect their
presence …
428
Here is a slightly more difficult qualitative analysis scheme, for anions, from Western Oregon University at
http://www.wou.edu/las/physci/poston/ch223/AnionAnalysis.htm
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XXXII) Complex Ions: Essentially a complex ion contains a central metal ion bound to one or more ligands.
And, a ligand is a neutral molecule or an ion that acts as a Lewis base with the central
metal ion.
A) The number of ligands attached to a metal ion = coordination number. The most common
coordination numbers are 2,4 and 6 (others are known)
1) Common ligands = NH3 (called ammine with a double m), H2O, Cl- CNa) e.g. Co(H2O)62+, Ni(NH3)62+, Cu(NH3)42+, CoCl42-,
coordination number = 6
coordination number =4
Ag(NH3)2+
coordination number = 2
b) Metal ions add ligands one at a time in steps characterized by an equilibrium constant
called formation constants (Kf) or stability constants.
B) Transition metal ions tend to be good Lewis Acids….They readily accept electrons due to vacant
orbitals or the ability to re-arrange orbitals so that they become vacant. Additionally, they tend to be smaller in
radius and have a high + charge
1) For instance, silver ion is hydrated by water (a reasonable Lewis base) to form Ag(H2O)22+(aq)
… which we simply have been writing as Ag+(aq). So, the ‘nekked silver ion does not really
exist all on its own in water…. It forms a complex ion.
In the above example, silver ion is the transition metal ion acting as the Lewis acid, and water
is the ligand.
2) Were a stronger Lewis base added to the system, the water would be displaced.
a) For example (as you will recall from our Tollen’s Test Lab ….(the silver mirror) ….
When we add concentrated ammonia (sti-n-ky!) to a solution containing silver ion
(or rather …Ag(H2O)22+(aq) we produced the silver diammine complex ion which
was then reduced to silver metal.
Ag(H2O)22+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) + 2 H2O(ℓ)
b) The equilibrium constant associated with the formation of a complex ion if Kf
or the formation constant…. It is determined by the Law of Mass Action … and
for silver diammine it looks like:
Kf = Ag(NH3)2+
[Ag+][NH3]
Kf for silver diammine = 1.7 x107 … a huge value, indicating a highly favored
formation, prior to the establishment of the
equilibrium.
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The following is edited but uses the work of Stephen and Susan Zumdahl from Chemistry 9 th ed. p. 775-7
C) Let’s look at that last formation a bit more closely and work with the Kf a bit more…
When a solutions containing Ag+ is mixed with a solution containing NH3 molecules, in excess,
are mixed the following reactions take place:
Ag+(aq) + NH3(aq)  Ag(NH3)+(aq)
(K1 = 2.1 x 103)
Ag(NH3)+(aq) + NH3(aq)  Ag(NH3)2+(aq)
(K2 = 8.2 x 103)
All 4 species are in equilibrium with each other: Ag+(aq) Ag(NH3)+(aq)  NH3(aq) Ag(NH3)2+(aq)
Calculation the equilibrium concentrations is a bit hair-raising … BUT!!!!! Under most
circumstances, the concentration of the ligand (NH3, in this example) is much larger than the
total concentration of the metal ion, and we can make certain approximations….
The following is all about using the larger [ligand] to help make certain assumptions
re: complex ion formation…
e.g.) Consider a solution made my mixing 100.0 mL of 2.0 M NH3 with 100 mL
of 1.0 x 10-3 M AgNO3. Assuming the diammine complex ion is the dominant ion after
reaction, compare the concentrations of NH3, Ag+ , Ag(NH3)+(aq) and Ag(NH3)2+(aq) after
the complex ions have been produced.
Consider the difference in concentrations … and the Kb of NH3 (1.8 x 10-5). The
ammonia is in excess … the amount of ammonia which reacts with water is utterly
negligible … so there is no need to account for that (NH3 + H2O  NH4+ + H3O+)
SO, IGNORING REACTION OF AMMONIA WITH WATER …
WHAT IS IN THE BEAKER? …And recall we are interested in complex ion formation
* Ag+ , NO3-, NH3, H2O
Before any complex ion formation, what are the initial concentrations of Ag+ and NH3?
a) Moles of Ag+ = 0.001 M = mol / 0.100 L or 0.0001 mol … but this was diluted
to a total volume of 200. mL, upon the addition of the ammonia
[Ag]initial = 0.0001 mole/0.200 L or 5.4 x 10-4 M
b) Mole of NH3 = 2.0 M = mol / 0.100 L or 0.20 mol …. but this was diluted to a
total volume of 200. mL, upon the addition of the silver nitrate
[NH3]initial = 0.20 mole/0.200 L or 1.0 M
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Remember the equilibria present:
Ag+(aq) + NH3(aq)  Ag(NH3)+(aq)
(K1 = 2.1 x 103)
Ag(NH3)+(aq) + NH3(aq)  Ag(NH3)2+(aq)
(K2 = 8.2 x 103)
Assumption: Given the equilibrium constants and the large excess of NH3 we can
assume that both reactions go (essentially) to completion … This (again) is essentially
equivalent to writing a net reaction of:
Ag+ + 2 NH3 → Ag(NH3)2+
HEY! This assumption essentially provides us with a stoichiometric ratio … in which twice
as much ammonia is required as silver ion ….
Wait a minute … This is a stoichiometric ratio …. and we are speaking of equilibria ….
Yeah, …. Run a stoich table
But … before we do that, there are a few items to be addressed:

For this type of stoich table, molarities can be be used ..because ….. (next point)


We assume the reaction goes to completion, using all the Ag+ to form Ag(NH3)2+(aq)
Virtually none of the silver monoammine Ag(NH3)+(aq) is present, because we are
assuming a complete reaction … and given the larger K value for the formation of
Ag(NH3)2+(aq) silver diammine, we can also assume that none of it dissociates into
silver monoammine Ag(NH3)+(aq)

Given the huge excess of NH3, the concentration at equilibrium is essentially, still
1.0 M, given that so little of it is consumed … Once we account for sig figs … there
is no effective change.

Recall the original problem was about our assumption that the diammine ion is the
dominant ion … and that our job has been to compare the concentrations of NH3,
Ag+ , Ag(NH3)+(aq) and Ag(NH3)2+(aq)

Given all of the above, we can use the K1 and K2 expressions to calculate the
concentrations of Ag(NH3)2+(aq)
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Ag+
before
after
+ 2 NH3
5.0 x 10-4 M
0 (limiting reagent)
Ag(NH3)2+
→
1.0 M
1.0 – 2(5.0 x 10-4) ≈1.0 M
0
5.4 x 10-4 M
Use K2 to determine the monoammine concentration at equilibrium
Ag(NH3)+(aq) + NH3(aq)  Ag(NH3)2+(aq)
K2 = [Ag(NH3)2+]
[Ag(NH3)+][NH3]
(K2 = 8.2 x 103)
or [Ag(NH3)+] = [Ag(NH3)2+]
K2[NH3]
[Ag(NH3)+] = 5.0 x 10-4
or = a very small 6.1 x 10-8 M
(8.2 x 103)(1.0)
(it looks as the assumption that the diammine complex ion as the dominant
species … may be half-proven….)
Use K1 to determine the silver ion concentration ….
Ag+(aq) + NH3(aq)  Ag(NH3)+(aq)
(K1 = 2.1 x 103)
[Ag+] = 6.1 x 10-8
= 2.9 x 10-11 M …. um, yeah these reaction really
3
(2.1 x 10 )(1.0)
did go to completion, if this were
all of the Ag+ left in solution….
Hence combining the stoich table results, and our calculations, the formation of a
complex ion really drove this reaction to the right….
[Ag(NH3)2+ ] = 5.0 x 10-4 M
[Ag(NH3)+] = 6.1 x 10-8 M
[Ag+]
= 2.9 x 10-11 M
On the following page you find a process for working out a complex ion equilibrium problem … I can’t
really see it on the AP – but I’m giving it to you none-the-less. I frankly see a much more important
concept regarding this issue … beyond the math…. (more on my personal thoughts, after the next page)
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B) Complex-Ion Equilibria: From Tro p. 798
Note: The following process could apply to any combination of virtually any good Lewis acid
(such as a solution containing a transition metal ion) and a good Lewis base (such as ammonia)
200.0 mL 0.0015 M Cu(NO3)2(aq) is added to a 250.0 mL solution of 0.20 M NH3(aq). The
reaction produces Cu(NH3)42+(aq) (Kf = 1.7 x 1013)
After the solution reaches equilibrium, what is the concentration of Cu2+(aq) which remains?
(Note: The Kf is huge … the remaining Cu2+ ion will be incredibly small, because virtually all
of the ion will be consumed, prior to the establishment of equilibrium)
 Write the balanced equation for the complex ion equilibrium …essentially a net-ionic equation.
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)42+(aq) and write Kf = [Cu(NH3)42+]
[Cu2+] [NH3]4
 Use M = mol/L to calculate the new molarities of Cu2+ and NH3, due to the mixing of the two
solutions.
0.0015M = x
0.200L
0.20 M = x
=
0.250 L
= 0.00030 mol … thus
0.50 mol
thus
M = 0.00030 = 6.7 x 10-4 M
0.450 L
M = 0.050 = 1.1 x 10-1 M
0.450 L
ICE it … use the coefficients of the balanced equation and know that the Cu+2 goes virtually to 0
initial
change
equil
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)42+(aq)
6.7 x 10-4
0.11
0.0
-4
-4
≈-6.7 x 10
≈4(-6.7 x10 ) +6.7 x 10-4
x
0.11
6.7 x 10-4
Normally we let x represent the change
in [ ] goint to equilibrium, but here we
let x represent the small amount of ion
remaining once equilibrium is reached
Plug and Chug into Kf …. 1.7 x 1013 = (6.7 x 10-4) rearranged: x = 6.7 x 10-4
(x) (0.11)4
1.7 x 1013(0.11)4
x= 6.7 x 10-4 = 2.7 x 10-13
2.49 x 109
Brown and LeMay page 757-58 has a second example … with an interesting twist to the approach.
Check it out….
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C) The Effect of Complex Ion Equilibria on Solubility
So let’s kick it back for a second and ask … When it comes to doing all of this math …So what?
How does the formation of complex ions affect our study of chemistry?
Well throughout all of this math, there is an underlying idea, that formation of a complex ion can drive
a rather insoluble compound into the soluble region … and this allows us to analyze a mixture for
ion concentrations
Essentially the overall concept is:
The solubility of an ionic compound containing a metal cation that forms complex ions
increase in the presence of Lewis bases that complex with the cation. The most common
Lewis bases that increase the solubility of metal cations are NH3, CN- and OHLet me set this up for us: Imagine a mixture of the group1 precipitates AgCl, PbCl2 and Hg2Cl2 … how
can we analyze the mixture for silver?
AgCl(s)  Ag+(aq) + Cl-(aq)
Well treating with a strong acid does nothing for us … It will not dissolve out the AgCl
because the Cl- has virtually no affinity for H+ and we can’t affect the position of the
equilibrium… We can’t affect the position, by getting Cl- to bond with the added H+ and
pull the dissolution equilibrium to the right ….
So, how can we pull this equilibrium to the right? …. Create a complex ion …
For instance, treat the silver chloride with an excess of ammonia
Here is a passage from Nivaldo Tro’s Chemistry A Molecular Approach 3rd ed. 2014 p 797
For example, silver chloride is only slightly soluble in pure water:
A change in solubility of 17 million times more soluble … is pretty impressive….
435
Questions:
1) Given: Cu2+(aq) + 4 NH3(aq)  Cu(NH3)42+(aq) what is the ligand?
1) H2O
3) Cu2+
2) NH3
3) Cu(NH3)42+
*ans: 2 … ammonia is the ligand it is acting as a Lewis base, in that it has a pair of electrons to dontate.
The Cu2+ is the transition metal acting as a Lewis acid, with its vacant orbitals, and Cu(NH3)42+ is the
complex ion, made by the reaction of the Lewis acid and Lewis base
2) Which compound when added to water, is most likely to increase the solubility of CuS?
1) NaBr
2) KNO3
3) NaCN
4) MgBr2
ans: *3 it is a Lewis base which
can form a soluble complex ion with
Cu2+
3) Given the complex ion name: hexaaquacopper(II) ion
a) What is the probable ligand?
i) cyanide
ii) ammonia iii) hydroxide
iv) water
ans: * iv water
b) What is the coordination number?
i) 6
ii) 2
iii) 3
iv) 4
ans: *i 6
4) Given the complex ion name: tetraamminecopper(II) ion
a) What is the probable ligand?
i) cyanide
ii) ammonia iii) hydroxide
iv) water
ans: * ii ammonia
b) What is the coordination number?
i) 6
ii) 2
iii) 3
iv) 4
ans: *iv 4
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D) Amphoteric Metal Hydroxides
1) It is not surprising that insoluble (or slightly soluble) metal hydroxides become more soluble
as pH of the solution decreases, because they can act as a base and react with the H+ of the
acidic solution.
2) Surprisingly, some metal hydroxides can also act as acids … They are amphoteric
3) This makes metal hydroxides like Al(OH)3 incredibly interesting …
Aluminum hydroxide is soluble at high pH and low pH, but insoluble in a pH-neutral
solution.
e.g.) At high pH it can act as an acid…by reacting with OHAl(OH)3(s) + OH-(aq) → Al(OH)4-(aq)
e.g.) At low pH it can act as a base …by reacting with the H+
Al(OH)3(s) + 3H3O+(aq) → Al3+(aq) + 6 H2O(ℓ)
a) For instance, if we look at aluminum ion as its fully-fledged Lewis acid …The
complex then acts as an acid because the complex ion acts as an acid by losing a
proton from one of ligand water molecules.
Al(H2O)63+(aq) + OH-(aq)  Al(H2O)5(OH)2+(aq) + H2O (ℓ)
hexaaquaaluminium
Adding base to the solution drives the reaction to the right and continues to remove
protons from the ligand water molecules:
Al(H2O)5(OH)2+(aq) + OH-(aq)  Al(H2O)4(OH)2+(aq) + H2O (ℓ)
Al(H2O)4(OH)2+(aq) + OH-(aq)  Al(H2O)3(OH)3 (s) + H2O (ℓ)
which is = to Al(OH)3
The solution is now pH neutral and the hydroxide is insoluble … add more (OH)- and
the solid will begin to dissolve as the solution becomes more basic.
Al(H2O)3(OH)3 (s) + OH-(aq)  Al(H2O)4(OH)4- (aq) + H2O (ℓ)
tetrahydroxidoaluminate
b) Metal cations that form amphoteric hydroxides include:
Al3+, Cr3+, Zn2+, Pb2+, & Sn2+ …. generally due to (but not conclusively)
high charge and empty orbitals … but that is pretty tentative.
437
4) Since only certain cations show amphoteric behavior, this can be exploited in separating
cations. If, for example, NaOH is added in small amounts to solutions of Fe3+ and Al3+,
both will initially form precipitates: rust colored iron(III) hydroxide
[Fe(H2O)6]3+(aq) + 3 OH-(aq) → Fe(H2O)3(OH)3(s) + 3 H2O(ℓ)
and white aluminum hydroxide, Al(OH)3.
Because Al(OH)3 is an amphoteric hydroxide, while Fe(OH)3 is not, further treatment
with NaOH will re-dissolve Al(OH)3 and leave Fe(OH)3 as an insoluble solid. If you
then centrifuge the mixture, and decant off the solution, you will have separated Al3+
from Fe3+.
citation: http://www.oneonta.edu/faculty/kotzjc/LAB/Complexation.pdf
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