Transcript الشريحة 1

Fractions of Dissociating Species in Polyligand Complexes
When polyligand complexes are dissociated in solution,
metal ions, ligand, and intermediates are obtained in
equilibrium with the complex. For example, look at the
following equilibria
Ag+ + NH3 D Ag(NH3)+
Ag(NH3)+ + NH3 D Ag(NH3)2+
kf1 = [Ag(NH3)+]/[Ag+][NH3]
kf2 = [Ag(NH3)2+]/[Ag(NH3)+][NH3]
We have Ag+, NH3, Ag(NH3)+, and Ag(NH3)2+ all present in
solution at equilibrium where
CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]
The fraction of each Ag+ species can be defined as:
b0 = [Ag+]/ CAg
b1 = [Ag(NH3)+]/ CAg
b2 = [Ag(NH3)2+]/ CAg
As seen for fractions of a polyprotic acid dissociating
species, one can look at the b values as b0 for the
fraction with zero ligand (free metal ion, Ag+), b1 as
the fraction of the species having one ligand
(Ag(NH3)+) while b2 as the fraction containing two
ligands (Ag(NH3)2+).
The sum of all fractions will necessarily add up to unity
(b0 + b1 + b2 = 1)
For the case of b0, we make all terms as a function of Ag+
since b0 is a function of Ag+. We use the equilibrium
constants of each step:
CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]
kf1 = [Ag(NH3)+]/[Ag+][NH3]
[Ag(NH3)+] = kf1 [Ag+][NH3]
Kf1 x kf2 = [Ag(NH3)2+]/[Ag+][NH3]2
[Ag(NH3)2+] = Kf1 x kf2 [Ag+][NH3]2
Substitution in the CAg relation gives:
CAg = [Ag+] + kf1 [Ag+][NH3] + Kf1 x kf2 [Ag+][NH3]2
CAg = [Ag+]( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
CAg /[Ag+] = ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
The inverse of this equation gives:
b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
If we use the same procedure for the derivation of
relations for other fractions we will get the same
denominator but the nominator will change
according to the species of interest:
b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
b1 = kf1 [NH3] / ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
b2 = Kf1 kf2 [NH3]2/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
Example
Calculate the concentration of the different ion
species of silver for 0.010 M Ag+ in a 0.10 M NH3
solution. Assume that ammonia concentration
will not change. Kf1 = 2.5x103, kf2 = 1.0x104
Solution
The concentration of silver ion species can be
obtained as a function of ammonia concentration
and formation constants from the relations
above:
b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
Substitution in the above equation yields:
b0 = 1/ ( 1 + 2.5x103 * 0.1 + 2.5x103 * 1.0x104 *(
0.10)2)
b0 = 4.0x10-6
b0 = [Ag+]/ CAg
4.0x10-6 = [Ag+]/0.010
[Ag+] = 4.0x10-8 M
In the same manner calculations give:
b1 = 1.0x10-3
b1 = [Ag(NH3)+]/ CAg
1.0x10-3 = [Ag(NH3)+]/ 0.010
[Ag(NH3)+] = 1.0x10-5 M
•
b2 = 1.0
b2 = [Ag(NH3)2+]/ CAg
1.0 = [Ag(NH3)2+]/ 0.010
[Ag(NH3)2+] = 0.010 M
Therefore, it is clear that most Ag+ will be in the complex
form Ag(NH3)2+ since the formation constant is large for
the overall reaction:
Kf = kf1*kf2
Kf = 2.5x103 * 1.0x104 = 2.5x107
The actual concentration of ammonia is 0.08 M not 0.10 M.
The value of 0.08 M NH3 should have been used for the
calculation of the fractions.
Precipitation Reactions and
Titrations
We had previously looked at precipitation equilibria in
solution and we are familiar with calculations of solubility
of sparingly soluble salts in pure water, in presence of a
common ion, and in presence of diverse ions. However,
we should remember that a salt is formed from a metal
ion (a very weak conjugate Lewis acid) that will not react
with water, and a conjugate base that may be weak and
will not react with water (like Cl-, Br-, I-) or a strong
conjugate base that will react with water (all conjugate
bases of weak acids). Calculations presented so far deal
with the first situation where the conjugate base is a very
weak base. Let us now look at sparingly soluble salts of
weak acids:
Effect of Acidity on the Solubility of Precipitates
The conjugate base of a weak acid is strong
enough to react with water upon dissociation
of the precipitate. At low pH values, excessive
amounts of the conjugate base will be
converted to the weak acid and thus forces the
precipitate to further dissociate to produce
more anions to satisfy the equilibrium constant
(ksp). Look at the general example assuming Ais the conjugate base of a weak acid and in
presence of acid solution (low pH):
MA(s) D M+ + AA- + H+ D HA
ksp = [M+][A-]
ka = [HA]/[A-][H+]
Since some of the formed A- is converted to HA, [M+] no
longer equals [A-]. However, we can write:
CT = [A-] + [HA], where CT = [M+]
[A-] = a1 CT
Recall EDTA equilibria in solution when its chelate
dissociates in water. Therefore, for the equilibrium
reaction:
MA(s) D M+ + A-
Before equil
Solid
0
0
Equation
MA(s)
M+
A-
After Equil
Solid
s
a1s
We can write:
Ksp = s * a1s
S = (ksp/a1)1/2
Example
Calculate the solubility of CaC2O4 (ka1 = 6.5x10-2,
ka2 = 6.1x10-5, ksp = 2.6x10-9) in a 0.001 M HCl
solution.
Solution
CaC2O4 D Ca2+ + C2O42The released oxalate will form H2C2O4, HC2O4-, and
some will remain as C2O42- where:
CC2O42- = s = [Ca2+] = [H2C2O4] + [HC2O4-] + [C2O42-]
and:
[C2O42-] = a2 s
Before equil
Equation
After Equil
Solid
CaC2O4 (s)
Solid
0
Ca2+
s
0
C2O42a2s
Ksp = s * a2 s
S = (ksp/a2)1/2
Therefore, we must calculate a2
a2 = ka1ka2 / ([H+]2 + ka1[H+] + ka1ka2)
a2 = (6.5x10-2 * 6.1x10-5)/( (0.001)2 + (6.5x10-2 * 0.001) + (6.5x10-2 * 6.1x105) )
a2 = 5.7x10-2
s = (2.6x10-9/5.7x10-2)1/2 = 2.1x10-4 M
Compare with solubility in absence of acidity!!
Example
Find the solubility of Ag3PO4 at pH 3. a3 at pH
3=3.3x10-14, ksp= 1.3x10-20
Solution
Ag3PO4 D 3 Ag+ + PO43Solid
3s
a3s
Ksp = (3 s)3 * a3 s
S = (ksp/27 a 3)1/4
S = (1.3x10-20/27x3.3x10-14)1/4
S = 1.1x10-2 M
Effect of Complexation on Solubility
We have seen earlier that acidity affects the
solubility of precipitates that contains the
conjugate base of a weak acid. In the same
manner, if a sparingly soluble precipitate was
placed in a solution that contains a ligand which
can form a complex with the metal ion of the
precipitate, solubility of the precipitate will
increase due to complex formation. For
example, if solid AgCl is placed in an ammonia
solution, the following equilibria will take place:
AgCl(s) D Ag+ + ClAg+ + NH3 D Ag(NH3)+
Ag(NH3)+ + NH3 D Ag(NH3)2+
The dissociated Ag+ ions will form complexes with
ammonia, forcing AgCl to further dissolve. Now, the
solubility of AgCl will increase and will equal [Cl-].
However, CAg = [Cl-] = s
CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]
Therefore, [Ag+] = bo s
One can then write:
AgCl(s) D Ag+ + ClKsp = [Ag+][Cl-]
Ksp = bo s * s
S = (ksp/bo)1/2
Example
Find the solubility of AgBr (ksp = 4x10-13) in pure water and
in 0.10 M NH3 (bo in 0.10 M ammonia = 4.0x10-6).
Solution
a. In pure water
AgBr(s) D Ag+ + Br-
Ksp = s * s
S = (4.0x10-13)1/2 , S = 6.3x10-7 M
b. In presence of 0.10 M NH3
Ksp = bo s * s
S = (ksp/bo)1/2
S = (4.0x10-13/4.0x10-6)1/2
S = 3.2x10-4 M
% increase in solubility = {(3.2x10-4 – 6.3x10-7)/6.3x10-7} x 100 =
5.1X104 %
Huge increase in solubility is expected as calculation
above suggests.