Transcript Soluble - HCC Learning Web

Chapter 16
Solubility and
Complex Ion Equilibria
Section 16.1
Solubility Equilibria and the Solubility Product
 Solubility equilibrium is base on the assumption
that solids dissolve in water to give the basic
particles from which they are formed.
 Molecular solids dissolve to give individual
aqueous molecules.
 Ionic solids dissociate to give their respective
positive and negative ions:
Section 16.1
Solubility Equilibria and the Solubility Product
 The ions in formed from the dissociation of ionic
solids can carry an electrical current. Salt
solutions, therefore, are good conductors of
electricity. Molecular solids, however, do not
dissociate in water to give ions, so no electrical
current can be carried.
Section 16.1
Solubility Equilibria and the Solubility Product
Solubility
 The ratio of the maximum amount of solute to
the volume of solvent in which this solute can
dissolve.
 Generally expressed in two ways:
 grams of solute per 100 g of water
 moles of solute per Liter of solution
Section 16.1
Solubility Equilibria and the Solubility Product
Definition of Solubility
 A salt is considered soluble if it dissolves in water
to give a solution with a concentration of at least
0.1 M at room temperature.
 A salt is considered insoluble if the concentration
of an aqueous solution is less than 0.0001 M at
room temperature.
 Salts with solubilities between 0.0001 M and 0.1
M are considered to be slightly soluble.
Section 16.1
Solubility Equilibria and the Solubility Product
Solubility Rules
Soluble: Dissolve - Do NOT form a solid precipitate.
1.alkali metal ions and ammonium ion: Li+, Na+, K+, NH4+
2.acetate ion: C2H3O213.nitrate ion: NO314.halide ions (X): Cl-, Br-, I- (Exceptions: AgX, HgX, and PbX2 are insoluble)
5.sulfate ion: SO42- (Exceptions: SrSO4, BaSO4, and PbSO4 are insoluble;
AgSO4, CaSO4, and Hg2SO4 are slightly soluble)
Section 16.1
Solubility Equilibria and the Solubility Product
Solubility Rules
Insoluble: Do NOT Dissolve - Do form a solid precipitate.
1.carbonate ion: CO322.chromate ion: CrO423.phosphate ion: PO434.sulfide ion: S2- (Exceptions: CaS, SrS, and BaS are soluble)
5.hydroxide ion: OH- (Exceptions: Sr(OH)2 and Ba(OH)2 are soluble;
Ca(OH)2 is slightly soluble
Section 16.1
Solubility Equilibria and the Solubility Product
Low solubility salts
 Salts that have extremely low solubilities
dissociate in water according to the principles of
equilibrium. For example, the reaction for the
dissociation of the s alt AgCl is:
Saturated solution - Contains the maximum concentration of
ions that can exist in equilibrium with the solid salt at a given
temperature
Section 16.1
Solubility Equilibria and the Solubility Product
 The reverse reaction for the dissolving of the salt
would be the precipitation of the ions to form a
solid:
 The system has reached equilibrium when the
rate at which AgCl dissolves is equal to the rate at
which AgCl precipitates.
Section 16.1
Solubility Equilibria and the Solubility Product
Solubility Equilibria
 Solubility product (Ksp) – equilibrium constant;
has only one value for a given solid at a given
temperature.
 Solubility – an equilibrium position.
Section 16.1
Solubility Equilibria and the Solubility Product
Difference between liquid equilibrium and liquid solid
equilibrium
 NOTE: There is no denominator in the solubility
product equilibrium constant. The key word to
remember is PRODUCT which can remind you
that you should have a multiplication (or product)
of the concentrations of the ions. The reason that
the solid reactant is not written is because its
concentration effectively remains constant.
Section 16.1
Solubility Equilibria and the Solubility Product
Solubility Equilibria- Effect of Stoichiometry
Coefficients from the balanced equations become
exponents
Bi2S3(s)
2Bi3+(aq) + 3S2–(aq)
2
2
K sp = Bi  S 
3+
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Section 16.1
Solubility Equilibria and the Solubility Product
CONCEPT CHECK!
In comparing several salts at a given temperature,
does a higher Ksp value always mean a higher
solubility?
Explain. If yes, explain and verify. If no, provide a
counter-example.
Unlike Ka and Kb for acids and bases, the relative
values of Ksp cannot be used to predict the relative
solubilities of salts if the salts being compared
produce a different number of ions.
Unlike Ka and Kb for acids and bases, the relative values of Ksp cannot be used to predict the relative
solubilities of salts if the salts being compared produce a different number of ions.
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Section 16.1
Solubility Equilibria and the Solubility Product
Solubility of salt
 Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8)
 First, write the BALANCED REACTION:
 Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:
 In the above equation, however, we have two unknowns, [Ca2+] and [F-]2. So,
we have to write one in terms of the other using mole ratios. According to the
balanced equation, for every one mole of Ca2+ formed, 2 moles of F- are
formed. To simplify things a little, let's assign the the variable X for the
solubility of the Ca2+:
Section 16.1
Solubility Equilibria and the Solubility Product
If we SUBSTITUTE these values into the equilibrium expression,
we now only have one variable to worry about, X:
We can now SOLVE for X:
Section 16.1
Solubility Equilibria and the Solubility Product
EXERCISE!
Calculate the solubility of silver chloride in water.
Ksp = 1.6 × 10–10
[Ag+]=[Cl-] so you solubility = √Ksp
1.3×10-5 M
Calculate the solubility of silver phosphate in water.
Ksp = 1.8 × 10–18 =[Ag+]*[PO4+3]3
[Ag+]=3x [PO4+3]=x
1.6×10-5 M
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Section 16.1
Solubility Equilibria and the Solubility Product
We assigned X as the solubility of the Ca2+ which is equal
to the solubility of the salt, CaF2.
However, our units right now are in molarity (mol/L),
so we have to convert to grams:
Section 16.1
Solubility Equilibria and the Solubility Product
Calculate Ksp from Solubility
The solubility of AgCl in pure water is 1.3 x 10-5 M. Calculate the value of Ksp.
 First, write the BALANCED REACTION:
 Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:
 It is given in the problem that the solubility of AgCl is 1.3 x 10-5. Since the mole
ratio of AgCl to both Ag+ and Cl- is 1:1, the solubility of each of the ions is equal
to the solubility of AgCl. SUBSTITUTE the solubility given into the equilibrium
expression to get Ksp:
1.3 x 10-5 M X 1.3 x 10-5 M
Section 16.1
Solubility Equilibria and the Solubility Product
EXERCISE!
Calculate the solubility of AgCl in:
Ksp = 1.6 × 10–10
a) 100.0 mL of 4.00 x 10-3 M calcium chloride.
2.0×10-8 M
b) 100.0 mL of 4.00 x 10-3 M calcium nitrate.
1.3×10-5 M
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Section 16.1
Solubility Equilibria and the Solubility Product
Ion Product (Qsp)
 The product of the concentrations of the ions at
any moment in time (not necessarily at
equilibrium).
 Imagine we have a saturated solution of AgCl.
The equilibrium reaction for the dissociation of
this salt is:
Section 16.2
Precipitation and Qualitative Analysis
Precipitation (Mixing Two Solutions of Ions)
 Q > Ksp; precipitation occurs and will continue until
the concentrations are reduced to the point that
they satisfy Ksp.
 Q < Ksp; no precipitation occurs.
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Section 16.1
Solubility Equilibria and the Solubility Product
Common Ion Effect
 The decrease in the solubility of a salt that occurs
when the salt is dissolved in a solution that already
contains another source of one of its ions. For
example, if AgCl is added to a NaCl solution
(which contains the common ion, Cl-) the
solubility of the AgCl decreases.
Section 16.2
Precipitation and Qualitative Analysis
Effect of adding a common ion
 [Ag+] is equal to [Cl-] at equilibrium because the
mole ratio of Ag+ to Cl- is 1:1. What would
happen to the solution if a tiny bit of AgNO3 (a
soluble salt) were added?
 Since AgNO3 is soluble, it dissociates completely
to give Ag+ and NO3- ions. There would now be
two sources of the Ag+ ion, from the AgCl and
from the AgNO3:
Section 16.2
Precipitation and Qualitative Analysis
Adding AgNO3 increases the Ag+ concentration and the solution is
no longer at equilibrium. The ion product (Qsp) at that moment is
bigger than the solubility product (Ksp). The reaction will
eventually return to equilibrium but when it does, the [Ag+] is no
longer equal to the [Cl-]. Instead, the [Ag+] will be larger than the
[Cl-].
Section 16.2
Precipitation and Qualitative Analysis
 Let's go back to the saturated AgCl solution. What would happen this time if a
tiny bit of NaCl (a soluble salt) were added? Since NaCl is soluble, it dissociates
completely to give Na+ and Cl- ions. There would now be two sources of the Clion, from AgCl and from NaCl:
 Adding NaCl increases the Cl- concentration and the solution is no longer at
equilibrium. The ion product (Qsp) at that moment is bigger than the solubility
product (Ksp). The reaction will eventually return to equilibrium but when it
does, the [Ag+] is no longer equal to the [Cl-]. Instead, the [Cl-] will be larger
than the [Ag+].
Section 16.1
Solubility Equilibria and the Solubility Product
CONCEPT CHECK!
How does the solubility of silver chloride in water
compare to that of silver chloride in an acidic
solution (made by adding nitric acid to the
solution)?
Explain.
The solubilities are the same.
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Section 16.1
Solubility Equilibria and the Solubility Product
CONCEPT CHECK!
How does the solubility of silver phosphate in water
compare to that of silver phosphate in an acidic
solution (made by adding nitric acid to the
solution)?
Explain.
The silver phosphate is more soluble in an acidic
solution.
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Section 16.1
Solubility Equilibria and the Solubility Product
CONCEPT CHECK!
How does the Ksp of silver phosphate in water
compare to that of silver phosphate in an acidic
solution (made by adding nitric acid to the
solution)?
Explain.
The Ksp values are the same.
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2
Section
16.1
9
Solubility Equilibria and the Solubility Product
Effect of pH on Solubility
 Sometimes it is necessary to account for other
reactions aqueous ions might undergo.
– For example, if the anion is the conjugate base of
a weak acid, it will react with H3O+.
– You should expect the solubility to be affected by
pH. By adding and complexing out ions you can
affect the pH of solution which could affect ppt
reactions.
3
Section
16.1
0
Solubility Equilibria and the Solubility Product
– Consider the following equilibrium.
CaC2O 4 (s )
H2O
2
2
Ca (aq)  C 2O 4 (aq)
– Because the oxalate ion is conjugate base, it will
react with H3O+ (added acid to lower pH).
2

C 2O 4 (aq )  H 3O (aq)
H2O

HC 2O 4 (aq)  H 2O(l)
– According to Le Chatelier’s principle, as C2O42- ion is
removed by the reaction with H3O+, more calcium oxalate
dissolves (increase solubility).
– Therefore, you expect calcium oxalate to be more soluble in acidic
solution (lower pH) than in pure water. The acidity will react with
the oxalate and shift the equil toward the right and allow more
calcium oxalate to dissolve.
Section 16.2
Precipitation and Qualitative Analysis
Selective Precipitation (Mixtures of Metal Ions)
 Use a reagent whose anion forms a precipitate with only
one or a few of the metal ions in the mixture.
 Example:
 Solution contains Ba2+ and Ag+ ions.
 Adding NaCl will form a precipitate with Ag+ (AgCl),
while still leaving Ba2+ in solution.
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Section 16.2
Precipitation and Qualitative Analysis
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
 At a low pH, [S2–] is relatively low and only the very
insoluble HgS and CuS precipitate.
 When OH– is added to lower [H+], the value of [S2–]
increases, and MnS and NiS precipitate.
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Section 16.2
Precipitation and Qualitative Analysis
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
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Section 16.2
Precipitation and Qualitative Analysis
Separating the Common
Cations by Selective
Precipitation
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Section 16.3
Equilibria Involving Complex Ions
Complex Ion Equilibria
 Charged species consisting of a metal ion surrounded by
ligands.
 Ligand: Lewis base
 Formation (stability) constant.
 Equilibrium constant for each step of the formation
of a complex ion by the addition of an individual
ligand to a metal ion or complex ion in aqueous
solution.
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3
Section
16.2
6
Precipitation and Qualitative Analysis
Complex-Ion Equilibria
– For example, the silver ion, Ag+, can react with
ammonia to form the Ag(NH3)2+ ion.

Ag  2(: NH 3 )  ( H 3 N : Ag : NH 3 )

3
Section
16.2
7
Precipitation and Qualitative Analysis
– A complex is defined as a compound
containing complex ions.
– A ligand is a Lewis base (makes electron
pair available) that bonds to a metal ion to
form a complex ion. Lewis Acid is the
cation.
3
Section
16.2
8
Precipitation and Qualitative Analysis
Complex-Ion Formation
 The aqueous silver ion forms a complex ion with
ammonia in steps.


Ag (aq )  NH 3 (aq)
Ag( NH 3 ) (aq )

Ag( NH 3 ) (aq )  NH 3 (aq)

Ag( NH 3 )2 (aq )
– When you add these equations, you get the overall
equation for the formation of Ag(NH3)2+.

Ag (aq )  2NH 3 (aq)

Ag( NH 3 )2 (aq )
3
Section
16.2
9
Precipitation and Qualitative Analysis


Ag (aq )  2NH 3 (aq)
Ag( NH 3 )2 (aq )
 The formation constant, Kf, is the equilibrium
constant for the formation of a complex ion from
the aqueous metal ion and the ligands.
– The formation constant for Ag(NH3)2+ is:

[ Ag( NH 3 )2 ]
Kf 

2
[ Ag ][NH 3 ]
– The value of Kf for Ag(NH3)2+ is 1.7 x 107.
Section 16.2
Precipitation and Qualitative Analysis
Complex-Ion Formation
– The large value means that the complex ion is
quite stable.
– When a large amount of NH3 is added to a solution of
Ag+, you expect most of the Ag+ ion to react to form
the complex ion (large Kf - equil lies far to right).
– Handle calculations same way as any other K
M
4
Section
16.2
1
a
tPrecipitation and Qualitative Analysis
e
r
Complex-Ion Formation
i
The dissociation constant, Kd , is the
a
or
inverse,
value
of
K
.
f
l
w
a
s
d
e
v
e
l
reciprocal,
The equation for the dissociation of Ag(NH3)2+ is

Ag( NH 3 )2 (aq )
Ag  (aq )  2NH 3 (aq)
The equilibrium constant equation is

2
1 [ Ag ][NH 3 ]
Kd 


K f [ Ag( NH 3 )2 ]
4
Section
16.2
2
Precipitation and Qualitative Analysis
Equilibrium Calculations with Kf
What is the concentration of Ag+(aq) ion in 1.00 liters of solution
that is 0.010 M AgNO3 and 1.00 M NH3? The Kf for Ag(NH3)2+ is 1.7
x 107.
Section 16.3
Equilibria Involving Complex Ions
Complex Ion Equilibria
Be2+(aq) + F–(aq)
BeF+(aq)
K1 = 7.9 × 104
BeF+(aq) + F–(aq)
BeF2(aq)
K2 = 5.8 × 103
BeF2(aq) + F–(aq)
BeF3– (aq)
K3 = 6.1 × 102
BeF3– (aq) + F–(aq)
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BeF42– (aq)
K4 = 2.7 × 101
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Section 16.3
Equilibria Involving Complex Ions
Complex Ions and Solubility
 Two strategies for dissolving a water–insoluble ionic
solid.
 If the anion of the solid is a good base, the solubility
is greatly increased by acidifying the solution.
 In cases where the anion is not sufficiently basic, the
ionic solid often can be dissolved in a solution
containing a ligand that forms stable complex ions
with its cation.
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Section 16.3
Equilibria Involving Complex Ions
 solubility and complex ions
Section 16.3
Equilibria Involving Complex Ions
Complex Ions Are Formed From Lewis Acid Metal
Interactions
Section 16.3
Equilibria Involving Complex Ions
CONCEPT CHECK!
Calculate the solubility of silver chloride in 10.0 M ammonia
given the following information:
Ksp (AgCl) = 1.6 × 10–10
Ag+ + NH3
AgNH3+ + NH3
AgNH3+
Ag(NH3)2+
K = 2.1 × 103
K = 8.2 × 103
0.48 M
Calculate the concentration of NH3 in the final equilibrium
mixture.
9.0 M
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Section 16.3
Equilibria Involving Complex Ions
Combining the two equilibria
Ag+ + NH3 <=>
AgNH3+
AgNH3+ + NH3 <=> Ag(NH3)2+
K = 2.1 × 103
K = 8.2 × 103
Ag+ + 2NH3 <=> [Ag(NH3)2]
Kf= 2.1 × 103 x 8.2 × 103 = 1.6 x 107
Section 16.3
Equilibria Involving Complex Ions
 AgCl(s) + 2 NH3(aq) <=> [Ag (NH3)2]+ + Clwhere K = [Ag (NH3)2] [Cl-] / [NH3]2 is a combination
of 3 equilibriums:
 A) AgCl(s) <=> Ag+ + Clwhose Ksp = [Ag+] [Cl-] = Ksp = 1.8 X 10-10
 B) Ag+ + 2NH3 <=> [Ag(NH3)2]
whose Kf = [Ag(NH3)2] / [Ag+] [NH3]2 = 1.6 X 10^7
Section 16.3
Equilibria Involving Complex Ions
if multiplying the ratio (A) time (B) gives your ratio:
 ([Ag+] [Cl-]) * ([Ag(NH3)2] / [Ag+] [NH3]2 )
=[Ag (NH3)2] [Cl-]/[NH3]2
then multiplying the Ksp times the Kf gives your K
 (1.8 X 10-10) (1.6 X 107) = 2.88 X 10-3
(with a constant that large, we need to use the
quadratic)
Section 16.3
Equilibria Involving Complex Ions
Set Up a Rice Table
AgCl(s)
2 NH3(aq)
Ag (NH3)2+
Cl-
X
X
10
-2X
K = [Ag (NH3)2] [Cl-] / [NH3]2
2.88 X 10-3 = [X] [X] / [10 - 2X]2