Name:_______________________________________ Date:_____________________________
OPTIONAL Homework - do NOT turn in – for practice only!!
1.) Lead (II) sulfate is a key component in car batteries. Its solubility in water at 25oC is 4.25 x 10-3
g/100 mL of solution. What is the Ksp of lead (II) sulfate?
We need a balanced equation: PbSO4 (s)  Pb+2 (aq) + SO4-2 (aq)
Ksp = [Pb+2] [SO4-2]
But what are the concentrations of Pb+2 and SO4-2?? We need to determine them using the solubility
of PbSO4
However, we need concentrations of species so we need to MOLAR SOLUBILITY of PbSO4
Right now, it is given in terms of gram solubility!
4.25 x 10 3 grams 1000 mL
1 mole PbSO 4
x
x
= 1.40 x 10-4 M PbSO4
100 mL
1L
303.3grams PbSO 4
Using mole to mole ratios, we can now figure out the M of Pb+2 and SO4-2 that would be in solution!
1.40 x 10-4 M PbSO4 x
1 mole Pb 2
= 1.40 x 10-4 M Pb+2
1 mole PbSO 4
1.40 x 10-4 M PbSO4 x
1 mole SO 4
= 1.40 x 10-4 M SO4-2
1 mole PbSO 4
2
Now that we have the concentrations of species, we can plug them in and chug out an answer!
Ksp = [Pb+2] [SO4-2]
Ksp = [1.40 x 10-4] [ 1.40 x 10-4] =
1.96 x 10-8
Remember that K values cannot be directly compared to the text values because of the assumptions
that we are making – that any species that dissolves, turns into an ion, but we should be close. Let’s
compare: text value for PbSO4 = 1.6 x 10-8 pretty close 
2.) When powdered calcium fluoride is shaken in water at 18oC, only 1.5 x 10-4 grams dissolved for
every 10.0 mL of water. Calculate the Ksp for calcium fluoride. (HINT: write the correct formula
for calcium fluoride!!)
CaF2 (s) ↔ Ca+2 (aq) + 2F-1 (aq)
Ksp = [Ca+2] [F-1]2
Unfortunately, we don’t know ANYTHING about the concentrations of Ca+2 or F-1
But we do have information about the solubility of CaF2!
1.5 x 10 4 grams CaF2 1000 mL
1 mole CaF2
=
x
x
10.0 mL
1L
78.08 grams CaF2
1.9 x 10-4
mole
CaF2
L
Using the molar solubility of CaF2 we can figure out the concentration of Ca+2 and F-1 ions
1.9 x 10-4
1.9 x 10-4
mole
1 mole Ca 2
CaF2 x
= 1.9 x 10-4 M Ca+2 ions
L
1 mole CaF2
mole
2 mole F 1
CaF2 x
= 3.8 x 10-4 M F-1 ions
L
1 mole CaF2
Ksp = [1.9 x 10-4][3.8 x 10-4]2
Ksp = 2.7 x 10-11
3.) Calculate the molar solubility of Ca(IO3)2 in a 0.060 M Ca(NO3)2 solution (Ksp = 7.1 x 10-7) And
calculate the equilibrium concentrations of Ca+2 and IO3-1 in the above solution
in the Ca(NO3)2 solution, it is the Ca+2 ion that is the common ion.
0.060 M Ca(NO3)2 x
Ca(IO3)2
(s)
1 mole Ca 2
= 0.060 M Ca+2
1 mole Ca(NO3 ) 2

Ca+2 (aq)
0.060
+x
0.060 + x
X
Ksp = [Ca+2] [IO3-1]2
7.1 x 10-7 = [0.060 + x][x]2 assume x is small!!!
7.1 x 10-7 = [0.060][2x]2
7.1 x 10-7 = 0.060[4x2]
x = 1.7 x 10-3 M
2IO3-1 (aq)
0
+2x
2x
change
1.7 x 10 -3
x 100 
x 100 = 2.9 % < 5% 
initial
0.060
check assumption:
Now that is the concentration of IO3-1 ions but NOT the concentration of Ca+2
[Ca+2] = 0.060 M
[IO3-1] = 2(1.7 x 10-3 M) = 3.4 x 10-3 M
Again, in order to figure out the molar solubility of the Ca(IO3)2 we need to use the ion that is ONLY
coming from the parent solid. We “played with” the concentration of Ca+2, it is coming from 2
sources, the calcium nitrate that was added and from the calcium iodate that is dissociating. We
cannot do any calculating using a species that we screwed around with!!
We MUST use the IO3-1 concentration since it is ONLY coming from the Ca(IO3)2 that would be
dissociating!
3.4 x 10-3 M IO3 -1 x
1 mole Ca(IO3 ) 2
2 mole IO 3
1
=
1.7 x 10-3 M Ca(IO3)2
4.) Predict the effect on equilibrium for each change. Make sure to indicate the direction the reaction
will go for each change. Using your knowledge of chemistry – EXPLAIN what happens for each
reaction (you must write more than left, right, or stays the same!!). Talk about the effect on the
solubility of the parent species. Each addition is happening independently (into a NEW beaker
of solution of BaSO4 or CaCO3). These additions are not happening all into the same beaker one
after the other!
a. BaSO4 (s)  Ba+2 (aq) + SO4-2 (aq)
i. You add Ba(NO3)2 to the solution
If Ba(NO3)2 is added to the solution, then we are increasing the number of Ba+2
ions in the solution (it is the common ion). This will drive the equilibrium to the
left. What does that mean chemically? It means that we form MORE solid, so we
have fewer ions in solution. In effect, the addition of a common ion suppresses or
prevents the ionization of the parent BaSO4 species which means BaSO4 is less
soluble.
ii. You add more BaSO4 to the solution
Adding more BaSO4 would cause the equilibrium to shift to the right. What does
this mean chemically? It means that more BaSO4 would be able to dissolve, thus
we would increase the concentration of ions in solution
iii. You add Ba3(PO4)2 to the solution
Absolutely nothing!! Ba3(PO4)2 is insoluble! There would be no appreciable
addition of ions into the solution, so it really would not affect the ionization of
the parent BaSO4 species! Tricky tricky . . .
b. CaCO3 (s)  Ca+2 (aq) + CO3-2 (aq)
i. You add more CaCO3 to the solution
Adding more CaCO3 to the solution would cause the equilibrium to shift to the
right. What does this mean chemically? It means that more CaCO 3 can dissolve
which means more ions will be put into solution.
ii. You add Na2CO3 to the solution
Na2CO3 is a soluble salt. It will put Na+1 and CO3-2 ions into solutions. However,
only the CO3-2 ions are shared between the species, so this makes the CO3-2 the
common ion. Addition of CO3-2 to the reaction will cause the equilibrium to shift
left (away from the added species) back to the parent CaCO 3. This will result in
the suppression of ionization of CaCO3 making it less soluble (less able to
dissociate/ turn into ions).
5.) Write the balanced equation to explain whether the addition of H 3O+1 will affect the solubility of
these ionic compounds: THINGS TO THINK ABOUT: Each solution is initially dissolved in
water. Then, after equilibrium is reached, acid is added. The question is – will any of the anions
in solution interact with the acid added? If the ion reacts with the added acid, how, using your
knowledge of Le’Chatelier’s Principle, will that affect the solubility of the parent compound?
a. Copper (II) hydroxide
Cu(OH)2 (s)  Cu+2 (aq) + 2OH-1 (aq)
Addition of H3O+1 ions WILL have an effect on the solubility of the Cu(OH)2 compound.
H3O+1 ions will react with OH-1 ions floating in solution. In fact, these two species will
neutralize one another:
H3O+1 (aq) + OH-1
(aq)
→ H2O (l)
This means that we are essentially removing the OH-1 ions from solution. If the [OH-1] is
decreasing, that will encourage more Cu(OH)2 to dissociate so it will INCREASE the
solubility of the compound.
b. calcium fluoride
CaF2 (s)  Ca+2 (aq)  2F-1 (aq)
F-1 is the strong conjugate base of a weak acid. This means, that in the presence of H 3O+1
ions F-1 will abstract the proton and since HF is a weak acid, it will not dissociate (as would
HBr in part a).
F-1 (aq) + H3O+1 (aq) → H2O (l) + HF (aq)
This will cause the concentration of F-1 ions to decrease which in turn, will INCREASE the
amount of CaF2 that can be dissolved. So more CaF2 will dissolve, the solubility
INCREASES.
6.) To improve the quality of x-ray photos used in the diagnosis of intestinal disorders, patients
drink aqueous suspensions (that doesn’t taste very good!!) of BaSO4 before the x-ray procedure.
The Ba+2 suspension is opaque to x-rays. What is the molar solubility of BaSO4 (Ksp = 1.1 x 10-10)
in pure water? What would it be in a 0.10 M solution of Na2SO4?
Question 3a: What is the molar solubility in water: First we will need an ICE table, since the solid
BaSO4 does not appear in the K expression, we will have to calculate information about the solid
from its daughter ions.
BaSO4

(s)
Ba+2 (aq)
0
+x
x
X
SO4-2 (aq)
0
+x
x
Ksp = [Ba+2] [SO4-2]
1.1 x 10-10 = [x][x]
x = 1.0 x 10-5 M
But this is the molarity of the Ba+2 ion or of the SO4-2 ion. The question asks about the molar
solubility of the BaSO4
1.0 x 10-5 M Ba+2 x
1 mole BaSO 4
1 mole Ba  2
= 1.0 x 10-5 M BaSO4
Question 3b: We are going to examine what happens mathematically when we add a common ion to
the solution. In this case, we are adding Na2SO4. The common ion being added is SO4-2. We can
think about what should happen and then make sure our math confirms this. According to
Le’Chatelier’s Principle, if we add more species, the reaction will shift AWAY from the added species,
so if we add more sulfate, we should form MORE BaSO4. And the presence of the common ion which
will form more solid, means that our molar solubility should be a smaller number. Let’s see.
Step 1: We need to know how much SO4-2 we add to the solution (not of the compound as a whole,
because the Na+1 ions are just watching!)
2
1 mole SO 4
0.10 M Na2SO4 x
= 0.10 M SO4-2
1 mole Na 2 SO 4
BaSO4
(s)

Ba+2 (aq)
0
+x
x
X
Ksp = [Ba+2] [SO4-2]
1.1 x 10-10 = [x][0.10 + x] assume x is small!!!
1.1 x 10-10 = [x][0.10]
x = 1.1 x 10-9 M
SO4-2 (aq)
0.10
+x
0.10 + x
check assumption:
change
1.1 x 10 -9
x 100 
x 100 = 1.1 x 10-6 % <<< 5% 
initial
0.10
Now that is the concentration of Ba+2 ions but NOT the concentration of SO4-2
[Ba+2] = 1.1 x 10-9 M
[SO4-2] = 0.10 M
In order to figure out the molar solubility of the BaSO4 we need to use the ion that is ONLY coming
from the parent solid. We “played with” the concentration of SO4-2, it is coming from 2 sources, the
sodium sulfate that was added and from the barium sulfate that is dissociating. We cannot do any
calculating using a species that we screwed around with!!
We MUST use the Ba+2 concentration since it is ONLY coming from the BaSO4 that would be
dissociating!
1.1 x 10-9 M Ba+2 x
1 mole BaSO 4
1 mole Ba  2
= 1.1 x 10-9 M BaSO4
Notice the difference in the 2 molar solubilities:
1.0 x 10-5 M BaSO4 in pure water
1.1 x 10-9 M BaSO4 in a solution that already contains SO4-2
7.) A common laboratory method for preparing a precipitate is to mix solutions of ions together.
Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?
Plan: Decide which slightly soluble salt can form and look up the Ksp value in the table given.
Determine the correct concentration of those ions in solution and compare the Qsp value to the
known Ksp value.
We are mixing Ca(NO3)2 with NaF. So what happens:
Ca(NO3)2 (aq) + 2NaF (aq) → CaF2 (s) + 2NaNO3 (aq)
The insoluble salt that can be formed is CaF2. The question is – are those concentrations
enough to cause the species to form a ppt? Looking up the known Ksp value for CaF2 we find
Ksp = 3.4 x 10-11
Now, we need to calculate the Qsp to determine if a ppt will form or not!
CaF2 (a)  Ca+2 (aq) + 2F-1 (aq)
Qsp = [Ca+2][F-1]2
The question is – what are the concentrations of the Ca+2 and the F-1?
We can figure this out, knowing what we started with, the total volume of the solution, and
the mole:mole ratios!
0.30 moles Ca(NO3 ) 2
1 mole Ca 2
mole Ca 2
x
= 0.30
x .100 L = 0.030 moles Ca+2
1L
1 mole Ca(NO3 ) 2
1L
New [Ca+2] in the total solution =
0.030mole Ca 2
= 0.10 M Ca+2
0.300L
0.060moles NaF 1 mole F 1
mole F -1
x
= 0.060
x 0.200 L = 0.012 moles F-1
1L
1 mole NaF
1L
New [F-1] in the total solution =
0.012 mole F 1
= 0.040 M F-1
0.300 L
Plugging those numbers into the Qsp expression:
Qsp = [Ca+2][F-1]2
Qsp = [0.10][0.040]2
Qsp = 1.6 x 10-4
In order to determine if a ppt will form, we need to compare the Qsp value to the Ksp value:
Qsp = 1.6 x 10-4 Ksp = 3.4 x 10-11
Qsp >>> Ksp !!!!
Since the Qsp value is much greater than the Ksp value the solution is supersaturated and we
will have a ppt present.
8.) Define the following:
a. Chemical change: a change that is associated with a chemical reaction. The species
does not retain its identity and it does turn into something “new” or different.
b. Physical change: a change associated with a substance that retains its identity. No new
species are formed.
9.) Identify the following as physical or chemical changes:
a.
Green patina forming on copper: chemical
b.
Combustion of gasoline: chemical
c.
Steam condensing: physical
d. gallium melting: physical
e.
Sugar dissolving in coffee: physical
f.
Milk spoiling : chemical