Lecture 35
Precipitation Titrations
Solubility in Acid Solutions
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+]
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)
3
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)
[Ag+]/ CAg = 1/( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
4
Precipitation Reactions and
Titrations
5
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:
6
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):
7
MA(s) D M+ + AA- + H+ D HA
ksp = [M+][A-]
ka = [A-][H+] /[HA]
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+ + A8
Before equil
Solid
0
0
Equation
MA(s)
M+
A-
After Equil
Solid
s
a 1s
We can write:
Ksp = s * a1s
S = (ksp/a1)1/2
9
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
10
Before equil
Equation
After Equil
Solid
CaC2O4 (s)
Solid
0
Ca2+
s
Ksp = s * a2 s
S = (ksp/a2)1/2
Therefore, we must calculate a2
a2 = ka1ka2 / ([H+]2 + ka1[H+] + ka1ka2)
0
C2O42a 2s
a2 = (6.5x10-2 * 6.1x10-5)/( (0.001)2 + (6.5x10-2 * 0.001) + (6.5x10-2 * 6.1x10-5) )
a2 = 5.7x10-2
s = (2.6x10-9/5.7x10-2)1/2 = 2.1x10-4 M
Compare with solubility in absence of acidity!!
11
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
12
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:
13
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
14
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).
a. In pure water
AgBr(s) D Ag+ + Br-
Ksp = s * s
S = (4.0x10-13)1/2 , S = 6.3x10-7 M
15
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.
16
Precipitation Titrations
Titrations in which precipitates are formed are called
precipitation titrations. The most frequent
application of this type of titration uses silver ion to
determine chloride. Therefore, these titrations are
called argentometric titrations. According to the
indicator used, three methods can be described.
Chromate is the indicator in Mohr's method while
Fajans method makes use of adsorption indicators.
Both methods are direct methods. The third method
is an indirect method where an excess silver is
added to chloride unknown and the remaining silver
is back-titrated with a standard thiocyanate solution
in presence of Fe(III) as an indicator.
17
Titration Curves
A titration curve for a precipitation titration can be
constructed by plotting mL Ag+ against pX (X
can be Cl-, Br-, I-, or SCN-) where the sharpness
of the end point and the break is directly
proportional to:
1. ksp of the silver salt.
2. The concentration of reactants.
18
19
20
Mixtures can also be titrated provided that
enough difference in the solubilities of the
two silver salts exists (at least 103).
21
Example
Find the pCl in a 20 mL of a 0.10 M Cl- solution
after addition of 0, 10, 20, and 30 mL of 0.10 M
AgNO3. Ksp = 1.0x10-10.
Solution
1. After addition of 0 mL Ag+
[Cl-] = 0.10 M
pCl = 1.00
2. After addition of 10 mL Ag+
Initial mmol Cl- = 0.10 x 20 = 2.0
mmol Ag+ added = 0.10 x 10 = 1.0
mmol Cl- excess = 2.0 – 1.0 = 1.0
[Cl-]excess = 1.0/30 = 0.033 M
22
We should expect that this is the actual concentration
present in solution since the solubility of AgCl is
very small (as seen from the ksp) and this is
especially true since we have a common ion.
However, let us try to calculate the [Cl-]dissociation of
AgCl.
Ksp = s(1/30 + s),
Assume 1/30 >> s
1.0x10-10 = 1/30 * s , s = 3x10-9 M
Therefore, [Cl-] = 0.033 M, pCl = 1.48
23
3. After addition of 20 mL Ag+
Initial mmol Cl- = 0.10 x 20 = 2.0
mmol Ag+ added = 0.10 x 20 = 2.0
mmol Cl- excess = 2.0 – 2.0 = 0
This is the equivalence point
Ksp = s * s
S = (1.0x10-10)1/2
S = 1.0x10-5 M
24
4. After addition of 30 mL Ag+
Initial mmol Cl- = 0.10 x 20 = 2.0
mmol Ag+ added = 0.10 x 30 = 3.0
mmol Ag+ excess =3.0 – 2.0 = 1.0
[Ag+]excess = 1.0/50 = 0.02 M
Therefore, once again we have a common ion situation.
We should expect that this is the actual Ag+
concentration present in solution since the solubility
of AgCl is very small (as seen from the ksp) and this
is especially true since we have a common ion.
However, we can calculate the Cl- concentration from
dissociation of AgCl in presence of excess Ag+ since
this is the only source of Cl-.
25
Ksp = (0.02 + s) * s
Assume that 0.02 >> s
1.0x10-10 = 0.02 S
S = 5.0X10-9 M = [Cl-]
pCl = 8.30
26
27
Methods for Chloride Determination
a. Mohr Method
This method utilizes chromate as an indicator.
Chromate forms a precipitate with Ag+ but
this precipitate has a greater solubility than
that of AgCl, for example. Therefore, AgCl is
formed first and after all Cl- is consumed, the
first drop of Ag+ in excess will react with the
chromate indicator giving a reddish
precipitate.
2 Ag+ + CrO42- (Yellow) = Ag2CrO4 (Red)
28
In this method, neutral medium should be used
since, in alkaline solutions, silver will react
with the hydroxide ions forming AgOH. In
acidic solutions, chromate will be converted to
dichromate. Therefore, the pH of solution
should be kept at about 7. There is always
some error in this method because a dilute
chromate solution is used due to the intense
color of the indicator. This will require
additional amount of Ag+ for the Ag2CrO4 to
form.
29
The amount of chromate added to the solution is
critical for the success of the titration. As a first
estimate, consider the calculation of the initial
concentration of CrO42– that must be present so that
Ag2CrO4 just begins to form at the equivalence point
in the titration of 0.1 M Cl– with 0.1 M Ag+. The value
of the solubility product for AgCl is 1.82 × 10–10, so
that at the equivalence point:
[Ag+] = [Cl–] = (1.82 × 10–10)1/2 = 1.35 × 10–5 M
The value of [CrO42–] at the equivalence point required
for Ag2CrO4 to just begin forming is given by the
following:
Ksp = [Ag+]2 [CrO42–] = 1.29 × 10–12
30
[CrO42–] = Ksp/[Ag+]2 = 1.29 × 10–12/(1.35 × 10–5)2
[CrO42–] = 7.08 × 10–3 M
However, since a volume of titrant equal to the volume
of analyte solution was added, the original
concentration of CrO42– was halved, so that its
concentration of CrO42– was:
Original [CrO42–] = 2 × 7.08 × 10–3 M = 0.0142 M
In practice, a concentration of the order of 7 × 10–3 M
CrO42– at the equivalence point is not practical
because the yellow color of the chromate ions
masks the color change at the end point. A
concentration of CrO42– at the end point of about 2.5
× 10–3 M is generally used for the Mohr titration.
31
This introduces a positive titration error
because a concentration of Ag+ higher
than 1.35 × 10-5 M must be reached, and
a finite amount of Ag2CrO4 must be
precipitated before the color of the
silver chromate can be observed.
Compensation may be made for the Mohr
chloride titration error by running an
indicator blank.
32
Example
Find the concentration of chloride in a 25 mL
solution to which few drops of K2CrO4 were
added, if the end point required 20 mL of 0.10
M AgNO3.
Solution
Ag+ + Cl- D AgCl(s)
The reaction between silver ions and chloride
is 1:1
mmol Ag+ = mmol Cl0.10 x 20 = MCl- * 25
MCl- = 0.08 M
33
b. Volhard Method
This is an indirect method for chloride
determination where an excess amount of
standard Ag+ is added to the chloride solution
containing Fe3+ as an indicator. The excess Ag+
is then titrated with standard SCN- solution untill
a red color is obtained which results from the
reaction:
Fe3+ (Yellow) + SCN- = Fe(SCN)2+ (Red)
The indicator system is very sensitive and usually
good results are obtained. The medium should
be acidic to avoid the formation of Fe(OH)3 .
34
The color of the Fe3+ indicator is intense
yellow, which prevents observation of
the Fe(SCN)2+ red color. Therefore, a
dilute Fe3+ indicator solution should be
used. It is important to be able to
calculate the theoretical concentration
of Fe3+ indicator to appreciate the
existence of an error. This can be done
using the following arguments:
At the equivalence point, we have AgSCN
only, therefore [Ag+] = CSCN- :
35
[Ag+] = [SCN-] + [HSCN] + [Fe(SCN)2+]
[Ag+] = ksp/[SCN-]
ka = [H+][SCN-]/[HSCN]
Ka = 0.1, at pH = 2
ksp/[SCN-] = [SCN-] + [H+][SCN-]/ka + [Fe(SCN)2+]
An average observer can distinguish the red
color of Fe(SCN)2+ when the concentration of
the complex is about 6.4*10-4 M. Therefore,
one can find the concentration of SCN-.
1.1*10-12 /[SCN-] = [SCN-] + 10-2 [SCN-]/0.1 + 6.4*10-6
1.1[SCN-]2 +6.4*10-6[SCN-] – 1.1*10-12 = 0
[SCN-] = 1.77*10-7 M
36
Fe3+ + SCN- = Fe(SCN)2+
Kf = [Fe(SCN)2+]/[Fe3+][SCN-]
1.4*102 = (6.4*10-6)/[Fe3+](1.77*10-7)
[Fe3+] = 0.26 M
However, the [Fe3+] used in Volhard
method should not exceed 0.002 M,
which results in an error. Such an error
is usually much less than that obtained
when the theoretical [Fe3+] is used.
37
Another problem arises at the end of the
titration, where we have both AgSCN(s)
and AgCl(s). The solubility of AgCl (Ksp
= 1*10-10) is greater than that of AgSCN
(Ksp = 1*10-12) , therefore the following
reaction takes place when all Cl- is
over:
AgCl(s) + SCN- D AgSCN + Cl-
This is an important reason for error, if
not taken care of.
38
This problem had been overcome by two
main procedures:
The first method includes addition of
some nitrobenzene, which surrounds
the precipitate and shields it from the
aqueous medium.
The second procedure involves filtration
of the precipitate directly after
precipitation, which protects the
precipitate from coming in contact with
the added SCN- solution
39
Example
A 10 mL of a chloride sample was treated with 15 mL of
0.1182 M AgNO3. The excess silver was titrated with
0.101 M SCN- requiring 2.38 mL to reach the red
Fe(SCN)2+ end point. Find the concentration of
chloride (AtWt = 35.5) in g/L.
Solution
mmol Ag+ reacted = mmol Ag+ taken - mmol Ag+ back-titrated
mmol Ag+ reacted = mmol ClmmolAg+ back-titrated = mmol SCNmmol Cl- = 0.1182x15 - 0.101x2.38 = 1.53
MCl- = 1.53/10 = 0.153 M
g/L Cl- = 0.153 x 35.5 = 5.44g
40
C. Fajans Method
Fluorescein and its derivatives are adsorbed to the
surface of colloidal AgCl. After all chloride is
used, the first drop of Ag+ will react with
fluorescein (Fl-) forming a reddish color.
Ag+ + Fl- (Yellowish green) = AgFl (Red)
41
Principle of adsorption:
Consider the titration of
Cl- with Ag+.
1. Before the equivalence
point, Cl- is in excess
and the primary layer is
Cl-. This repulses the
indicator anions; and
the more loosely held
secondary (counter)
layer of adsorbed ions
42
Beyond the equivalence point (end point as well),
Ag+ is in excess and the surface of the
precipitate becomes positively charged, with the
1 layer being Ag+. This will now attract the
indicator anion and adsorb it in the 2 (counter)
layer.
43
The color of the adsorbed indicator is different from
that of the un-adsorbed indicator, signaling the
completion of the titration. The degree of adsorption
of the indicator can be decreased by increasing the
acidity.
Since fluorescein and its derivatives are weak
acids, the pH of the solution should be slightly
alkaline to keep the indicator in the anion form
but, at the same time, is not alkaline enough to
convert Ag+ into AgOH. Fluorescein derivatives
that are stronger acids than fluorescien (like
eosin) can be used at acidic pH without
problems.
44