Chem 12 Ksp
Worksheet
1)
Find the solubility of;
a)
Silver iodide
Dissociation eq’n
Ksp
y =
b)
=
y
8.5 x 10-17
c)
y
Ksp
Ksp
Fe(OH)3 (s)
Dissociation eq’n
z =
I-(aq)
+
=
=
y
[Ag+][I-] = (y)(y)
8.5 x 10-17 = y2
= y2
9.2 x 10-9 M = solubility of Silver Iodide
Iron (III) hydroxide
Ksp
Ag+(aq)
AgI (s)
=
z
2.6 x 10-39
Fe+3(aq)
+
3 OH-(aq)
z
Ksp
Ksp
3z
= [Fe3+][OH-]3 = (z)(3z)3
= 2.6 x 10-39 = 27z4
= 9.63 x 10-41 = z4
=
27z4
9.91 x 10 -11 M = solubility of Iron (III) hydroxide
Lead (II) bromide
PbBr2 (s)
Dissociation eq’n
w
6.6 x 10-6
Ksp
=
w =
1.2 x 10 -2 M =
Pb+2(aq)
+
w
Ksp
Ksp
= [Pb2+][Br-]2 = (w)(2w)2
= 6.6 x 10-6 = 4w3
= 1.65 x 10-6 = w3
= 1.18 x 10 -2 = w
solubility of Lead (II) bromide
2 Br -(aq)
2w
=
4w3
d)
Aluminum hydroxide (not on your table)
Al+3(aq)
Al(OH)3 (s)
Dissociation eq’n
2)
ø
+
ø
3.0 x 10-34
Ksp
=
w =
1.8 x 10-9 M
Ksp
Ksp
=
3 OH -(aq)
3ø
= [Al3+][OH-]3 = (ø)(3ø)3
= 3.0 x 10-34 = 27w4
= 1.11 x 10-35 = w4
= 1.83 x 10-9
=
27ø4
solubility of Aluminum hydroxide.
Silver ions are slowly added to the a solution containing both chromate ions and bromide
ions of equal concentrations (0.200 M). What compound will precipitate first and why?
AgBr (s)
Ag+(aq)
x
+
Br-(aq)
0.200 M
Ksp = [Ag+][Br-] = [Ag+](0.20) = 5.4 x 10-13
[Ag+] = 2.7 x 10 -12 M
The maximum [Ag+] that can exist in this solution is 2.7 x 10 -12 M
Ag2CrO4 (s)
2 Ag+(aq)
2y
+
CrO4-2(aq)
0.200 M
Ksp = [Ag+]2[CrO4-2] = [Ag+]2(0.20) = 1.1 x 10-12
[Ag+]2 = 5.5 x 10 -12
[Ag+] = 2.3 x 10-6 M
The maximum [Ag+] that can exist in this solution is [Ag+] = 2.3 x 10-6 M
Since we are adding the silver ion starting at zero concentration and increasing slowly,
we will achieve 2.7 x 10 -12 M before the 2.3 x 10-6 M, that is, we will exceed the Ksp of AgBr
before we exceed the Ksp or Ag2CrO4, hence the AgBr will ppt first.
3)
What is the greatest concentration of sulphate ion that can be added to a 0.600 M solution
of strontium ions without forming a precipitate?
Sr2+(aq)
SrSO4 (s)
+
0.600 M
Ksp = [Sr2+][SO4-2]

= (0.600)() = 3.4 x 10-7
=

SO4-2(aq)
[SO4-2]
= 5.7 x 10-7 M
The maximum concentration of sulphate ion that can exist is 5.7 x 10 -7 M
4)
Will a precipitate form if you mix 27.00 mL of 0.000258 M lead (II) nitrate with
13.00 mL of 0.000784 M magnesium iodide? Show all work.
(0.027 L Pb(NO3)2)(0.000258 mol Pb(NO3)2) = 6.97 x 10-6 moles of Pb(NO3)2
(1.00 L Pb(NO3)2)
= 1.02 x 10-5 mol MgI2
(0.013 L MgI2)(0.000784 mol MgI2)
(1.00 L MgI2)
[Pb(NO3)2] = (6.97 x 10-6 moles of Pb(NO3)2) = 1.74 x 10-4 M
(0.040 L)
(1.02 x 10-5 mol MgI2)
(0.040 L)
[MgI2] =
Pb(NO3)2 (aq)
Pb2+(aq)
MgI2 (aq)
Mg2+(aq)
Pb(NO3)2 (aq)
Pb2+(aq)
+
2 NO3-(aq)
[Pb2+] =
[NO3-] =
[Mg2+] =
[I-]
=
2 I-(aq)
+
MgI2 (aq)
PbI2 (s)
+
2 I-(aq)
PbI2 (s)
Ksp = [Pb2+][I-]2 = 8.5 x 10-9
Compare Trial Ion Product (TIP) to Ksp
Since Ksp > TIP
+
= 2.55 x 10-4 M
+
1.74 x 10-4 M
3.48 x 10-4 M
2.55 x 10-4 M
5.10 x 10-4 M
Mg(NO3)2 (aq)
TIP = [Pb2+][I-]2 = (1.74 x 10-4)(5.10 x 10-4)2
TIP = 4.53 x 10-11
The solution is not yet saturated and no solid forms.
5)
Give one example of solubility increasing with increasing temperature and one example
of solubility decreasing with increasing temperature.
What can be generalized from these examples?
eg #1 - Sodium acetate’s (ionic compound) solubility increases as temperature increases
eg #2 - Carbon dioxide gas’ solubility decreases as temperature increases.
When temperature increases, the randomness of the system (solute + solvent) will
increase. In the case of an ionic solid, “increased randomness” implies that the solid will
go into solution, hence increasing the solute’s solubility. In the case of any gas,
“increased randomness” would imply the gaseous form will be preferred over the
solvated form. This will decrease the solubility of gases in water as temperature
increases.
6)
Give as many ways as possible for increasing the solubility of:
“increasing the solubility” is just another way of saying “stress the system so it shifts to
the right”
a)
MgCrO4
MgCrO4 (aq)
Mg2+(aq)
+
CrO4-2(aq)
We could remove Mg2+ by adding and alkali metal compounds which contain
sulphite, carbonate, phosphate, or hydroxide. These would all form insoluble compounds
with magnesium removing it from the above equilibrium.
We could also remove CrO4-2 from the system by adding barium or silver nitrates
since both barium and silver chromates have low solubility (as evidenced by the reported
Ksp’s). Of course we could also increase solubility by heating the system.
b)
PbCO3
Add soluble forms of sulphite, phosphate, hydroxide, sulphide, sulphate or any of
the halides to remove the Lead (II).
Add any cation except for NH4+, H+, the alkali’s, or Pb+2 to remove carbonate.
Increase the heat content of the system.
c)
Fe(OH)2
Add soluble forms of sulphite, carbonate, phosphate, or sulphide to remove the
Iron (II) ion.
Add any cation except NH4+, the alkali’s, Sr+2 or Fe+2 to remove the hydroxide.
Increase the heat content of the system.
7)
a)
b)
c)
Give one example of a molecular solution in water.
Why does your chosen molecule dissolve in water?
Why does this compound remain as a molecule instead of dissociating into ions?
a)
b)
Sugar dissolved in water.
Sugar has multiple -OH groups hanging off the main ring. All of these groups are
polar and can hydrogen bond with the solvent’s -OH groups. Since like dissolves
like, they are partially soluble.
Sugar is covalently bonded, the main sugar ring is composed of perfectly covalent
C-C linkages and slightly polar C-O linkages. The H’s and OH’s that hang on the
ring are also covalently bonded. NOne of the bonds in that molecule have an
electronegativity difference of greater than 1.7, therefore there is no opportunity
for sugar to form ions.
c)