CHEM 2400/2480
Lecture 13
3a
Separation Based on Sulfide Precipitation
H2S º H+ + HS-
K1 = 9.5 x 10-8
HS- º H+ + S2-
K2 = 1.3 x 10-14
In a saturated solution, [H2S] = 0.10 M
and
ˆ by adjusting the pH, we can regulate [S2-].
Week 7-1
CHEM 2400/2480
e.g. Find the conditions under which Pb2+ and Tl+ can be separated quantitatively
by sulfide precipitation from a solution that is 0.10 M in each cation.
Ksp (PbS) = 7.0 x 10-28
Ksp (Tl2S) = 1.0 x 10-22
PbS will precipitate at a lower S2- concentration than the Tl2S.
Assume we wish to lower the Pb2+ concentration to 10-5 M, then
This is the [S2-] required to lower the [Pb2+] to 10-5 M.
This value should be compared to the [S2-] needed to initiate the
precipitation of Tl2S from a 0.10 M Tl+ solution i.e.
Thus to achieve a separation we need,
[S2-] > 7 x 10-23 M but < 1 x 10-20 M
Week 7-2
CHEM 2400/2480
We can adjust [H+] to hold [S2-] in the range:
Note: Read text Box 12-2 page 241 about negative pH.
Thus by maintaining [H+] in this range, we can precipitate PbS
without precipitating Tl2S.
Week 7-3
CHEM 2400/2480
3b
Effect of pH
-The solubility of the salt of a weak acid depends on the pH
of the solution
- i.e. fluorides, sulphides, carbonates, phosphates etc.
MA (s) º M+ (aq) + A- (aq)
Ksp
A- (aq) + H2O (l) º HA (aq) + OH-
Kb
- these two equilibria determine the solubility of MA (s)
- if the pH is lowered, then
A- (aq) + H3O+ (aq) 6 HA (aq) + H2O (l)
and more MA (s) will dissolve to restore equilibrium.
Calculate the solubility of CaF2 in water
i.
Ksp = 3.9 x 10-11
CaF2 (s) º Ca2+ (aq) + 2F- (aq)
F- (aq) + H2O (l) º HF (aq) + OH- (aq) Kb = 1.5 x 10-11
H2O (l) º H+ + OH- (aq)
ii.
Kw = 1.0 x 10-14
Charge balance
[H+] + 2[Ca2+] = [OH-] + [F -]
iii.
Mass balance
If F- did not react with water, then [F-] = 2[Ca2+]
but we know about Kb
ˆ [F -] + [HF] = 2[Ca2+] = [F -]Total
Week 7-4
CHEM 2400/2480
iv.
from the mass balance:
Week 7-5
CHEM 2400/2480
But
- This is a difficult equation to solve because we do not know what the pH is
i.e.
- the solubility is pH dependent.
Week 7-6
CHEM 2400/2480
If we know the pH, the problem is simplified.
e.g. what is the solubility at pH = 3.00?
Ksp = 3.9 x 10-11
Kb = 1.5 x 10-11
In this case, the charge balance is not valid because we
had to add the components of a buffer to fix
the pH at 3.00.
At pH 3.00, [H+] = 1.00 x 10-3 M
and
[OH-] = 1.00 x 10-11 M
The solubility of CaF2 was derived as
In water alone, s = 2.1 x 10-4 M and pH = 7.108
Week 7-7
CHEM 2400/2480
Lecture 14
OK. But what if the pH is not fixed by external means
i.e. a buffer?
How do we proceed with this problem?
We can use a spreadsheet without employing any approximation.
STRATEGY: Find the pH at which the charge balance is satisfied
- this is the pH that results from the compound
dissolving (no external buffer has been added
to fix the pH)
- calculate the solubility at this pH.
Let us use CaCO3 as a second example since this is important wrt
deterioration of statues, art work and buildings.
PROCEDURES:
1.
Derive the equation for the solubility of CaCO3 . Use either
Ka's or Kb's.
2.
Calculate the concentrations of all charged species
- use equilibrium constants to do this.
3.
Find the pH at which the charged balance is satisfied.
Week 7-8
CHEM 2400/2480
3a. Solvent Effect on Anion i.e. pH
e.g. find the molar solubility of a saturated solution of CaCO3
CaCO3 (s) º Ca2+ + CO 23
CO 23
reacts with water to form CO 23 ,
HCO -3
and H2CO3
from mass balance:
[Ca2+] = [ CO 23 ]Total
= s
= [ CO 23 ] + [ HCO 3 ] + [H2CO3]
Ksp = [Ca2+] [ CO 23 ] , for which
we can obtain [ CO 23 ] from
α CO2−
3
i.e.
α CO2−
3
=
[CO 23 ]
[CO32 − ]Total
from mass balance
2[ CO 23 ] = α CO 2− [ CO 3 ]Total =
3
α CO2−
where
3
2-
then [ CO 3 ]
=
=
α CO 2− [Ca2+]
3
K1 K 2
[H + ]2 + K 1 [H+ ] + K 1 K 2
[H + ]2
K1 K 2
+ K 1 [H + ] + K 1 K 2
Week 7-9
x [Ca2+]
CHEM 2400/2480
now
Ksp = [Ca2+] [ CO 23 ]
= [Ca2+] x
=
2+ 2
[Ca ]
[H + ]2
[Ca2+]2 x
=
K1 K 2
+ K 1 [H + ] + K 1 K 2
[H + ]2
x [Ca2+]
K1 K 2
+ K 1 [H + ] + K 1 K 2
⎛ [H + ]2 + K 1 [H + ] + K 1 K 2 ⎞
K sp ⎜
⎟
K1 K 2
⎝
⎠
⎛
⎛ [H ] + K 1[H ] + K 1K 2 ⎞ ⎞
s = [Ca 2+ ] = ⎜ K sp ⎜
⎟⎟
K 1K 2
⎝
⎠⎠
⎝
+ 2
Week 7-10
+
1/2
CHEM 2400/2480
Concentration of All Charged Species
!
H+, OH-, Ca2+, CO23 , HCO3 ,
- since we input pH we can determine [H+] and [OH-]
i.e.
[OH-] = Kw / [H+]
from the equilibria equations,
[CO ] =
23
K sp
[Ca
2+
]
and
[CO 23 − ][H + ]
[H 3CO ] =
K2
−
3
also for completeness
[H + ][ HCO −3 ]
[H 2CO 3 ] =
K1
Use of Charge Balance
!
2[Ca2+] + [H+] = 2[CO23 ] + [HCO3 ] + [OH ]
and
!
2[Ca2+] + [H+] - 2[CO23 ] - [HCO3 ] - [OH ] = 0
thus, we determine the pH when the equation = 0 is satisified.
Week 7-11
CHEM 2400/2480
Notes:
i.
At the pH where charge = 0
[HCO!3 ] = [OH-]
[HCO!3 ] >> [H2CO3]
[OH-] >> [H+]
ii.
s = 1.2 x 10-4 in pure water giving a solution pH of 9.94
iii.
In acid rain where the pH is less than 7 e.g.
pH 6, s = 0.018 M
In general, we try not to make any approximations. BUT for very very small Ksp's,
we can assume pH - 7 because of the low concentration of ions i.e. low solubility.
e.g. Ag2S ( Ksp = 8 x 10-51 ); HgS ( Ksp = 5 x 10-54 )
Now try this with Ag3PO4
Find the pH of a saturated solution of silver phosphate.
Ag3PO4 (s) º 3 Ag+ + PO34
Week 7-12