Transcript Salt Solubility

Salt Solubility
Chapter 18
1
Solubility product constant
• Ksp
• Unitless
• CaF2(s)  Ca2+(aq) + 2F-(aq)
• Ksp = [Ca2+][F-]2
• [Ca2+] = molar solubility (in M)
• Get Ksp values from Appendix J, pages A24-25
2
We have problems
• Ksp for BaSO4 = 1.1 x 10-10 @ 25°C
– Calculate this salt’s solubility in water in g/L.
3
Solution
BaSO4( s )  Ba 2 ( aq )  SO4 2 ( aq )
I
C
E
0
+x
x
K sp  1.1 10
10
0
+x
x
2
2
 [ Ba ][ SO4 ]  x
2
x  1.0  105 M
g
g
m
ol
1.0  10
 233
 0.0024
L
mol
L
5
4
Problem
• If [Ca2+] = 2.4 x 10-4 M, what is the Ksp
value of CaF2?
5
Solution
CaF2(s)  Ca
I
C
E
----
2+
-
(aq)
+ 2F (aq)
0
+x
x
0
+2x
2x
(2.4  10 )
-4
(2 x 2.4  10 )
-4
K sp = [Ca 2+ ][F- ]2  [2.4 10-4 ][4.8 10-4 ]2  5.5 1011
6
Solubility and the Common Ion
Effect
• Problem: How much AgCl (g/L) would
dissolve in water given Ksp= 1.8 x 10-10?
7
AgCl( s )  Ag
I
C
E

( aq )
 Cl
0
+x
x
K sp =1.8 10
-10

( aq )
0
+x
x


 [ Ag ][Cl ]  x
2
x  1.3 10 M
-5
g
1.3 10 M 143
-5
mol
g
 0.0019
L
8
Solubility and the Common Ion
Effect
• If solid AgCl is placed in 1.00L of 0.55M
NaCl, what mass of AgCl (g/L) would
dissolve?
• Before solving this, would you expect it to
be lesser or greater than in pure water?
– Think back to Le Châtelier’s Principle and the
pH of buffers
9
AgCl( s )  Ag
I
C
E

( aq )
 Cl
0
+x
x
K sp =1.8 10
-10

( aq )
0.55
+x
x+0.55


 [ Ag ][Cl ]  [ x][ x  0.55]  [ x][0.55]
(We can approximate since K is very small.)

x = 3.3 10 M  [ Ag ]
g
-10
-8 g
3.3 10 M 143
 4.7  10
mol
L
-10
10
Basic anions and salt solubility
• Do we recall what a conjugate acid/conjugate
•
base is?
If salt has conjugate base of weak acid, then
salt more soluble than value given by Ksp
– Why?
PbS(s)  Pb2+(aq) + S2-(aq)
S2-(aq) + H2O(l)  HS-(aq) + OH-(aq);
Where net = Ksp  Kb (>Ksp)
11
Effect of pH on solubility
• Mg(OH)2 (s)  Mg2+(aq) + 2OH-(aq)
• If one adds base, OH-, which way will the
equilibrium shift?
– Will more solid form or dissolve?
• If one adds acid, H3O+, which way will the
equilibrium shift?
– Will more solid form or dissolve?
12
Effect of pH on solubility
• Mg3(PO4)2 (s)  3Mg2+(aq) + 2PO4-3(aq)
• If one adds base, OH-, which way will the
equilibrium shift?
– Will more solid form or dissolve?
• If one adds acid, H3O+, which way will the
equilibrium shift?
– Will more solid form or dissolve?
13
Effect of pH on solubility
• MgCl2 (s)  Mg2+(aq) + 2Cl-(aq)
• If one adds base, OH-, which way will the
equilibrium shift?
– Will more solid form or dissolve?
• If one adds acid, H3O+, which way will the
equilibrium shift?
– Will more solid form or dissolve?
14
So what can we say about the
conjugate base and solubility?
15
The reaction quotient, Q
• Q tells us:
– Whether it’s at equilibrium
• If not, which way it’ll shift
• CaF2(s)  Ca2+(aq)+ 2F-(aq)
• Ksp = [Ca2+][F-]2
• Q = [Ca2+][F-]2
• If Q = Ksp, then @ eq. &
•
•
soln is saturated
If Q < Ksp, then will shift
to right (dissolve
more) & soln is
unsaturated
If Q > Ksp, then will shift
to left (not dissolve
anymore, ppt out) &
soln is supersaturated
(will ppt out)
16
Check out the heating pads & video
• http://genchem.chem.wisc.edu/demonstra
tions/Gen_Chem_Pages/11solutionspage/c
rystallization_from_super.htm
17
Problem
• AgCl placed in water. After a certain
amount of time, concentration= 1.2 x 10-5
M.
• Has the system reached eq.?
• If not, will more solid dissolve?
• How much more?
• (Ksp = 1.8 x 10-10)
18
Q  [ Ag  ][Cl  ]  [1.2 105 ][1.2 105 ]  1.4 1010
1.8 1010  1.4 1010
K sp  Q
Thus, the system is not yet at equilibrium and will proceed to the right,
dissolving more solid.
At equilibrium, 1.3 105 mol of each ion are present.
Therefore, 1.3 105 mol  1.2 105 mol  0.110 5 mol  moles required to dissolve to reach eq.
0.1105 mol 143 g
 1104 g of additional AgCl will dissolve
mol
19
Selective precipitation
• Will a precipitate form, given the ion
concentrations?
• What concentrations of ions are needed to
bring about precipitation?
• Basically, compare Q to Ksp
20
Problem
• If [Mg2+] = 1.5x10-6 M, and enough
hydroxide ions are added to make the
solution 1.0 x 10-4 M of hydroxide ions, will
Mg(OH)2 precipitate?
• If not, will it occur if the concentration of
hydroxide ions is increased to 1.0 x 10-2
M?
• Ksp = 5.6 x 10-12
21
Mg(OH) 2(s)  Mg
2+
-
(aq)
+2OH (aq)
Q  [1.5 106 ][1.0 104 ]2  1.5 1014
Q(1.5 1014 )  K sp (5.6 1012 )
 Unsaturated solution and precipitation doesn't occur yet.
When [OH - ]=1.0 10-2 M :
Q  [1.5 106 ][1.0 10-2 ]2  1.5 1010
Q(1.5 1010 )  K sp (5.6 1012 )
 Precipitation will occur.
22
Complex ion equilibria
• Complex ion = a central
•
•
metal ion (e- acceptor =
Lewis acid) bound to 1 or
more ligands
AgCl(s) + 2NH3(aq)  [Ag(NH3 )2 ]+(aq)  Cl-(aq)
Ligand = neutral
molecule or ion that acts
as Lewis base (e- donor)
with the central metal ion
Kf values on page A-26,
Appendix K
23
Example
• What is Knet for the reaction below?
AgCl(s) + 2NH3(aq)  [Ag(NH3 )2 ]
+
 Cl (aq)
-
(aq)
24
AgCl(s) + 2NH 3(aq)  [Ag(NH 3 ) 2 ]+ (aq)  Cl -(aq)
AgCl(s)  Ag + (aq)  Cl-(aq) ; K sp  1.8 10-10
Ag + (aq)  2NH 3(aq)  [Ag(NH 3 ) 2 ]+ (aq) ; K f = 1.6 107
K net
[Ag(NH 3 ) 2 ]+ [Cl- ]
[Ag(NH 3 ) 2 ]+
+

 [Ag ][Cl ] 
 K sp  K f  (1.8 10-10 )  (1.6 107 )  2.9 10 3
2
+
2
[NH 3 ]
[Ag ][NH 3 ]
25
Problem
• What is the net rxn and value of Knet for
dissolving AgBr in a solution containing
S2O32-?
• Ksp = 5.4 x 10-13
• Kf = 2.0 x 1013
26
AgBr(s)  Ag + (aq)  Br - (aq) ; K sp  5.4 10-13
Ag
+
(aq)
 2S2O3
2-
 [Ag(S2O3 ) 2 ]
3-
(aq)
; K f = 2.0 10
13
(aq)
net rxn: AgBr(s) + 2S2O32-(aq)  [Ag(S2O3 ) 2 ]3-(aq)  Br -(aq)
K net  K sp  K f  (5.4 10-13 )  (2.0 1013 )  10.
27