Remember the solubility
rules on page 164 of your Moore et al (p. 118 of Brown, Lemay,
and Bursten, 9th Ed., p. 111 of 8th
Ed.) textbook. Alkali metal ions,
ammonium, nitrates and acetates are soluble, halide
compounds are soluble with exceptions. Phosphates,
carbonates, hydroxides and sulfides are insoluble with the exceptions of alkali
and ammonium ions.
A saturated solution with
solid in contact with water will come to equilibrium with ions dissolving and
precipitating with equal rates.
AgCl(s) = Ag+(aq) +
Cl-(aq)
The equilibrium expression
for this heterogeneous reaction would be (leaving out the [AgCl] since the
concentration of a solid is constant);
Ksp = [Ag+] [Cl-]
The constant Ksp
is called the solubility product. The reaction showing silver phosphate
dissolving and the solubility product expression would be
Ag3PO4 (s) = 3 Ag+ + PO4-3
Ksp = [Ag+]3
[PO4-3]
Please note that Ksp
calculations work best for solutes that are considered insoluble. As we get to higher concentrations of solute
ions, a lot of non-ideal behavior makes these simple calculations invalid. Ksp also works better with ions with low
charges (+ or -1) and when no other ions are dissolved. Other dissolved ions can greatly change
solubility.
Hopefully we learned many chapters ago that solubility is the
grams of solute per 100. g of solution. To convert this into a solubility product
we would first determine the molar solubility (moles solute per 1.00 L of
solution) and then using the stoichiometry of the equation showing dissolving,
determine the ions concentration and calculate the Ksp
e.g. g solute
® (use molar mass to convert to moles) ® moles solute
100. g soln. (use density to convert to V) 1. L soln.
®(use stoichiometry to
determine [ ] ions in soln.) then substitute in
Ksp = [Mn+]x [X-m]y
Given a
solubility of silver chloride of .00019 g/ 100. g solution at 25°C, what is the molar solubility and Ksp? Since this is a very dilute solution, we can assume
the density is close to that of pure water at 25°C
solubility = .00019 g X 1mol
/143.3209 g AgCl X 1 = 1.325696 X10-5mol/L
100. g soln. 1 mL/1g 1L/1000 mL
So the molar solubility is to
2 significant figures 1.3 X 10-5mol/L (don't round that number yet)
Since dissociation of silver
chloride will have equal concentrations of silver and chloride
AgCl(s) = Ag+
+ Cl-
|
Initial |
0 |
0 |
|
Change |
+x |
+x |
|
Equilibrium |
x |
x |
Ksp = x2 = (1.325696 X 10-5)2
= 1.8 X 10-10 (if you had squared 1.3 X 10-5 you would
get 1.7 X 10-10).
To go from Ksp to
molar solubility, since x2 = 1.8 X 10-10, the square root
of Ksp = molar solubility in this example (it will not always be so
easy).
The solubility of lead (II)
fluoride is 0.0510 g / 100.g solution.
Assuming density = 1.00 g / mL, What is the
molar solubility and Ksp? (molar mass =
245.2 g/mol)
You should be able to get
molar solubility = 2.08 X 10-3 M
Now for PbF2 (s) = Pb+2 + 2 F-
|
Initial |
0 |
0 |
|
Change |
+x |
+2x |
|
Equilibrium |
X |
2x |
Ksp = x (2x)2 = 4x3
Ksp = 3.6 X 10-8
To go backward, Ksp =4x3,
and x will = molar solubility. A common
mistake above is to forget to square the 2.
For practice, go
backward and forward on these 2 calculations.
Factors
affecting Solubility
1.
Temperature – We learned earlier that for most ionic solids, as temperature increases,
solubility increases.
2.
Acids and pH
5.
Amphoterism
Acids
and pH (lower pH will increase solubility of basic anions)
For
salts that contain a basic anion, such as hydroxide, the equilibrium
involving that anion will affect solubility.
For example, as pH decreases, ([H+] increases), solubility of
a metal hydroxide will increase because the hydrogen ion will consume the
hydroxide, shifting the solubility equilibrium reaction to the right to
dissolved ions. A salt with a basic
anion (e.g. carbonate, phosphate, cyanide, and sulfides) will show an increased
solubility as [H+] increases (pH decreases). Salts derived from strong acid and bases are
negligible conjugates bases and acids, and will not be affected as much by pH
changes.
e.g.
Calculate the molar solubility of Mg(OH)2 in pure water and a
buffered pH=9.0 solution. Ksp = 1.8 X 10-11 (page 665)
Mg (OH)2 (s)
= Mg+2 +
2
|
Initial |
0 |
0 |
|
Change |
+x |
+2x |
|
Equilibrium |
x |
2x |
Ksp = [Mg+2]
[
x = 1.7 X 10-4
M= molar solubility and [
At a pH = 9.0, pOH = 5.0 and [
[Mg+2] = Ksp
/ [
x = 0.18 M
molar solubility
in pH = 9.0 = 0.18 M
Mg(OH)2
is used to neutralize stomach acid without burning your throat on the way
down. In un-buffered liquids, magnesium
hydroxide is very insoluble, and so not much dissolves and it doesn’t hurt your
throat. In the stomach where there is
lots of acid, it will dissolve and the hydroxide can neutralize some of the
stomach acid.
In
acid base reactions, we have said if one of the reactants is strong the
reaction will go to completion. If both
are weak, you will get an equilibrium. Consider tooth enamel which consists mainly
of a mineral called hydroxyapatite, Ca10(PO4)6(OH)2. Bacteria in your mouth convert sugar into a
weak acid such as lactic acid, and this will react completely with the strong
base.
Ca10(PO4)6(OH)2
(s) + 2 HLactic Acid (aq) -> 10 Ca2+ (aq) + 6 PO4-3
(aq) + 2 H2O (L) + 2 Lactates-
With
fluoridated water and fluoride toothpaste, the hydroxide in hydroxyapatite
is replaced by the weak base, F‑. Instead of reacting completely with lactic
acid, you will get an equilibrium, and thus less of
your enamel will dissolve, leading to less cavities.
Common ion effect (decreases solubility)
The
presence of a common ion will reduce the solubility by forcing the equilibrium
back to the left toward the precipitate (Hip, Hip, Hooray
for Le Chatelier). It will not change
the Ksp, but it will change the two solubilities. This can be used to reduce the concentration
of a toxic heavy metal in water.
Compare
for silver chloride. Say we had a .010
M NaCl solution.
AgCl
(s) = Ag+ +
Cl-
|
Initial |
0 |
0.010 |
|
Change |
x |
+ x |
|
Equilibrium |
x |
0.010 + x |
Ksp = 1.8 X
10-10 = x
(.010 + x) ~ 0.010x (since x will
probably be very small compared to 0.010 M)
x = 1.8 X 10-8
M (see, I told you it would be small)
The
molar solubility of AgCl in 0.010 M NaCl would be 1.8 x 10-8 M. Compare this to the molar solubility of AgCl
in pure water of 1.3 X 10-5 M.
Formation of complex
ions (increases solubility)
The
combination of a metal ion as a Lewis acid and a Lewis base forms a complex
ion, such as Ag(NH3)2+ ,
or Fe(CN)6-4. The
following equilibrium involves an equilibrium constant called the formation
constant, Kf.
Ag+
+ 2 NH3 = Ag(NH3)2+ Kf
= 1.6 X 107
The
large equilibrium constant tells us this is a favorable reaction. This will increase the solubility of a
compound such as silver chloride by reacting and reducing the dissolved silver
ion concentration.
Consider
the simultaneous equilibrium
AgCl(s)
= Ag+ + Cl- Ksp
= 1.8 X 10-10
Ag+
+ 2 NH3 = Ag(NH3)2+ Kf
= 1.6 X 107
The
sum of the two equilibrium will have an equilibrium expression equal to the
product of Ksp and Kf .
AgCl(s)
+ 2 NH3 = Ag(NH3)2+
+ Cl- Kxp x Kf = 2.9 X 10-3
This
is a much more favorable reaction than the dissolving of silver chloride in
pure water.
Some
metal hydroxides are Amphoteric, they can react as an acid or base. The metal hydroxide can act as a
Bronstead-Lowery base
Zn(OH)2
(s) + 2 HCl → ZnCl2
+ 2 H2O
Or
the metal can act as a Lewis acid, accepting a pair of electrons from a
hydroxide, forming a complex ion.
Zn(OH)2
(s) + 2