AP SOLUBILITY EQUILIBRIA
A. THE EFFECT OF DIFFERING HEATS OF SOLUTION ON THE SOLUBILITY
Definition: The ENTHALPY OF SOLUTION (or Heat of Solution) of a substance is the energy change
associated with dissolving one mole of the substance in a particular solvent.
The solubility of a substance increases as temperature increases if the heat of solution is
ENDOTHERMIC. Now consider the effect that changing the heat of solution has on the solubility.
EXERCISES:
1. If I2(s) + 6.7 kJ
I2(in alcohol) and I2(s) + 24 kJ
in which solvent is I2(s) more soluble? Explain.
2. If O2(g)
O2(aq) + 12 kJ and N2O(g)
which gas is more soluble in water? Explain.
I2(in CCl4)
N2O(aq) + 21 kJ
3. When a solid evaporates directly (without melting), the process is called SUBLIMATION. Given
FCN(s)
FCN(g) ; ∆H = 24 kJ and ICN(s)
ICN(g) ; ∆H = 59 kJ,
(a) how do the tendencies to maximum randomness compare between FCN(s) and ICN(s)?
(b) how do the tendencies to minimum energy compare?
(c) would you expect ICN(s) to have a lower or higher vapour pressure than FCN(s) at 201 K?
The equilibrium vapour pressure of FCN(s) is 101.3 kPa at 201 K.
4. Liquid chloroform, CHCl3 , and liquid acetone, CH3COCH3 , are miscible (dissolve in each other in
all proportions).
(a) When pure CHCl3(l) is mixed with pure CH3COCH3(l), is randomness increased or
decreased?
(b) Given
CHCl3(l) + CH3COCH3(l)
1:1 solution; ∆H = –2.1 kJ
does the tendency to
minimum energy favour reactants or products?
(c) Why is it no surprise that chloroform and acetone are miscible? What would have to be true
for them to be IMMISCIBLE? (eg. like water and gasoline)
B. CRYSTAL LATTICE ENERGY AND HYDRATION ENERGY
Definitions: The CRYSTAL LATTICE ENERGY, ∆HCL , is the energy required to separate the ions in a
crystal lattice. ∆HCL is always ENDOTHERMIC.
The HYDRATION ENERGY, ∆HHYD , is the energy change involved when water molecules
surround and attach themselves to a species. ∆HHYD is always EXOTHERMIC because
energy is given off when bonds form.
The HEAT OF SOLUTION for the dissolving of an ionic species in water is the sum of the
crystal lattice and hydration energies.
∆HSOLN = ∆HCL + ∆HHYD
EXAMPLE:
NOTE:
The dissolving of NaCl(s) to form Na+(aq) and Cl–(aq) is given as follows.
NaCl(s)
Na+(g) + Cl–(g)
; ∆HCL
= 789 kJ
+
+
Na (g) + H2O(l)
Na (aq)
; ∆HHYD = –418 kJ
Cl–(g) + H2O(l)
Cl–(aq)
; ∆HHYD = –367 kJ
————————————————————————————————––
Overall:
NaCl(s)
Na+(aq) + Cl–(aq) ; ∆HSOLN =
+4 kJ
The hydration of sodium ions can be represented as
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H
H
O
H
Na+ + 4 H2O
H
Na+
O
O
+ 418 kJ
H
H
O
H
+
Na(H2O)4
H
The hydration of chloride ions can be represented as
O
H
H
Cl– + 4 H2O
H
H
Cl–
O
O
+ 367 kJ
H
H
H
H
O
–
Cl(H2O)4
The EXOTHERMIC nature of the hydration reaction is evident if one looks at the reaction in
reverse: it requires an addition of energy to remove the 4 water molecules bonded to the
central ion.
EXERCISE:
5. (Fill in the blanks) The above theory shows that there are two factors involved in determining the
solubility of an ionic compound.
(a) The Crystal Lattice Energy: (endothermic) – the greater the value of ∆HCL , the
soluble the ionic compound.
(b) The Hydration Energy: (exothermic) – the greater the value of ∆HHYD , the
soluble
the ionic compound.
—————————————————————————————————————————————––
The energies involved in the dissolving process can be represented by means of a BORN–HABER
CYCLE (see AP Thermodynamics).
IONS IN GAS PHASE
HHYD
HCL
HYDRATED IONS
HSOLN
CRYSTAL
The idea behind a Born–Haber cycle is that the energy required to form an aqueous solution
CRYSTAL + Heat
HYDRATED IONS
going counter–clockwise from CRYSTAL to HYDRATED IONS can be estimated as the sum of the crystal
lattice and hydration energies (CRYSTAL to IONS IN GAS PHASE to HYDRATED IONS, going
clockwise).
The process assumes the following sequential behaviour.
(a) The ions in the crystal are separated from one another, forming gaseous ions, and then
(b) water molecules attach themselves to the gaseous ions.
In actual fact, it is known that the dissolving process
• first involves the water molecules attaching themselves to ions in the crystal (which releases energy
due to bond formation), and then
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3
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• the energy released when water bonds to the ions is combined with the energy already available to
the ions (heat contained in the crystal and surroundings) to cause the separation of the combined
unit (ion plus attached water molecules) from the adjacent ions in the crystal.
EXERCISE:
6. Examine the Born–Haber diagram above and fill in the following blanks with the most appropriate
symbol: >, < or  .
“Highly soluble” implies
∆HHYD
∆HCL
“Moderately
soluble” implies

∆HHYD
∆HCL
“Slightly soluble” implies
∆HHYD
∆HCL
C. COMMON ION EFFECT CALCULATIONS
The basis of the common ion effect is simple and obvious: since the product of two ion concentrations is
constant, then as the concentration of one ion increases the concentration of the other ion must decrease.
An additional important point, which is not obvious at first glance, is that forcing one ion concentration to
be high effectively prevents an ionic compound from dissolving if the compound contains an ion which
already exists in high concentrations in solution.
EXAMPLE:
Solid SrSO4 is added to 1.00 L of 0.200 M Na2SO4 until no more SrSO4 can dissolve. If
KSP is 3.4 x 10–7 for SrSO4 , what is [Sr2+] in the saturated solution?
CARE! Many students mis–read problems like this and assume that KSP is for Na2SO4 .
You must properly identify the compound described by the KSP value.
Set up the problem as follows.
Sr2+
X
2 Na+
0.40

SrSO4(s)
–X
Na2SO4
–0.20
+ SO 24
X
+ SO 24
0.20
; KSP = [Sr2+][ SO 24 ]
(Note: only the “boxed” ions concern us.)

 the values for [Sr2+] and [ SO 24 ].
Since KSP refers to SrSO4 , find
[Sr2+] = X
[ SO 24 ] = X + 0.20
Now

( Note that there are two sources of SO 24 . The SO 24 is

COMMON to both the SrSO4 and the Na2SO4 .
KSP = 3.4 x 10–7 = X•(X + 0.20) , which is QUADRATIC!

STOP!!!

ANALYSIS: Since X•(X + 0.20) equals a small number (3.4 x 10–7), then one or both of
the terms “X” and “X + 0.20” must be SMALL. But 0.20 >> 3.4 x 10–7 so that 0.20 is
“large” (compared to 3.4 x 10–7) and the addition of X to give “X + 0.20” makes the term
even bigger. Therefore, the value of X MUST be VERY small in order to make the overall
term, X•(X + 0.20), equal to a small value.
Since 0.20 is large and X is very small, then
0.20 + X  0.20
that is:
X is negligible with respect to 0.20.

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Now solve the simplified equation.
X • 0.20 = 3.4 x 10–7
X = 1.7 x 10–6 M = [Sr2+]
(Note that the assumption, 0.20 + X = 0.20 + 1.7 x 10–6

0.20 is correct.)
IMPORTANT: Whenever you make a simplifying assumption, you MUST state the assumption or you will
be marked as if you made a serious math mistake.
In our example above, it would be

sufficient to show
0.20 + X  0.20 (since X is negligible with respect to 0.20)
The common ion effect is most useful in situations where one ion in a solution is deliberately raised to
relatively high concentrations to 
prevent significant amounts of a second ion from dissolving. For
example, if we produced a precipitate of SrSO4 and were concerned about the amount of SrSO4 which
dissolves when distilled water is used to “wash” the precipitate, we might wash the SrSO4 with a solution
of 1 M Na2SO4. The high [ SO 24 ] in the wash solution prevents the SrSO4 from dissolving to a significant
extent.
EXAMPLE:
If Sr(NO3)2(s) is added to 2.00 L of 4.00 x 10–6 M SrSO4 , how many moles of Sr(NO3)2

must be added to just start forming a precipitate of SrSO 4? KSP = 3.4 x 10–7 for SrSO4 .
Again, set up the problem as before.
Sr2+
SrSO4(s)
–4.00 x 10–6
Sr(NO3)2
–X
4.00 x 10–6
Sr2+
X 
+ SO 24
; KSP = [Sr2+][ SO 24 ]
4.00 x 10–6
+ 2 NO 3Р
2X

Since KSP refers to SrSO4 , find the values for [Sr2+] and [ SO 24 ].

[Sr2+] = X + 4.00 x 10–6
(Note that there are two sources of Sr2+)
[ SO 24 ] = 4.00 x 10–6

and KSP = 3.4 x 10–7 = (X + 4.00 x 10–6)(4.00 x 10–6)
 Again, make a simplifying assumption – in this particular example, this assumption does
NOT have to be made but the math is a bit easier if we do. Since [ SO 24 ] is already small,
the term “X + 4.00 x 10–6” must be relatively large so as to prevent the product
(X + 4.00 x 10–6)(4.00 x 10–6) from becoming TOO SMALL. This implies
X >> 4.00 x 10–6, so that

X + 4.00 x 10–6  X
The KSP equation then becomes
X• (4.00 x 10–6) = 3.4 x 10–7

so that
and
moles Sr(NO3)2 required = 0.085

X = 0.085 M = [Sr(NO3)2]
mol
x 2.00 L = 0.17 mol
L
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GENERAL SIMPLIFYING RULE
Recall that in the first example: [Sr2+] = X
[ SO 24 ] = 0.20 + X

0.20
[Sr2+] = X + 4.00 x 10–6  X
[ SO 24 ] = 
4.00 x 10–6

In each example two different compounds are involved: one compound is insoluble and contains both

of the ions involved in the KSP expression. The other compound is soluble and contains only one of
 KSP expression (this compound has no KSP expression given for itself). A
the ions involved in the
simplifying rule is shown below.
while in the second example:
The soluble compound contributes one ion to the solution. Make the approximation so that the
insoluble compound contributes only the other ion.
Another way to look at this is to point out that we always ignore the “corner box” when we lay out the
concentrations (as shown in the previous examples).
SrSO4(s)
–X
Sr2+
X
+ SO 24
X
; KSP = [Sr2+][ SO 24 ]
Na2SO4
–0.20
2 Na+
0.40 
+ SO 24
0.20
Ignore the BOLD box
Sr2+
+ SO 24
; KSP = [Sr2+][ SO 24 ]
SrSO4(s)
–4.00 x 10–6
Sr(NO3)2
–X
GENERAL COMMENTS
4.00 x 10–6
Sr2+
X 

4.00 x 10–6
+ 2 NO 3Р
2X
Ignore the BOLD box


1. Read Exercises 19 and 20. As you see, they look very similar and yet one is a simple KSP problem
and the other is a common ion effect problem. Here is how to tell the difference.
• A common ion effect problem only mentions two different compounds (each compound supplies
one of the ions used in the calculation).
• A simple KSP problem mentions either ONE or THREE compounds. If one compound is
mentioned, the compound dissolves to form an equilibrium, for example:
AgCl(s)
Ag+ + Cl–.
If three compounds are mentioned, one ion is taken from each of two different compounds to
create a precipitate (the third compound). For example:
AgNO3(aq) + NaCl(aq)
AgCl(s) + …
(There is no mention of the NaNO3 remaining in solution as the fourth compound involved in
this double replacement.)
2. Common ion effect problems are tricky and lead to frequent errors by students. The following
procedure works well in ensuring a good problem layout and a lower likelihood of making errors. (If
you look at the way the example problems are laid out, you will see that this procedure has been
applied every time.)
1. Identify the compound belonging to the KSP. Write the insoluble compound in
equilibrium with its ions and write the KSP expression after the equilibrium equation.
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2. Identify the soluble compound and show it dissolving into its ions, under the
equilibrium for the insoluble compound. The soluble compound is the compound for
which no KSP is given.
3. Determine which substance (the insoluble compound or the soluble compound) has
an unknown concentration and which has a known concentration. Put either the
given concentration or an unknown under each compound. Write the concentration
of ions produced when the compound ionizes (the concentrations are treated like the
∆–line of a “START, ∆, EQUILIBRIUM” calculation).
4. For each ion used in the KSP expression, add up the contributions from each
substance used.
5. Make an approximation such that the soluble compound contributes one ion and the
insoluble compound contributes the other.
6. Solve the KSP expression for the unknown.
EXERCISES:
7. How many moles of AgBr dissolve in 1.0 L of 0.020 M KBr solution?
8. How many moles of AgCl dissolve in 5.0 L of 0.10 M NaCl solution?
9. If 5.0 x 10–6 mol of PbSO4 dissolves in 1.0 L of a K2SO4 solution, what is the [K2SO4 ]?
10.
If 2.0 x 10–5 mol of Cu(IO3)2 dissolves in 2.0 L of a NaIO3 solution, what is the [NaIO3]?
11. Powdered AgOH is slowly added to 2.0 L of a 0.020 M solution of NaOH until no more AgOH
dissolves. If KSP = 1.4 x 10–8 for AgOH,
(a) what is [Ag+] in the saturated solution?
(b) what mass of AgOH is added to the solution to produce this concentration of Ag +?
12. When excess Ag2CO3 is shaken with 1.00 L of 0.0100 M K2CO3 , 4.01 mg of Ag2CO3 dissolves.
Calculate the solubility product of Ag2CO3.
13. If BaSO4 is added to a solution of K2SO4 , the solubility of the BaSO4 is 2.0 x 10–7 M. What is the
concentration of the K2SO4 solution?
14. A solution contains 0.020 mol of (NH4)2S. How many grams of MnS must be added to 1.0 L of the
solution to just start forming a precipitate of MnS? KSP = 1.4 x 10–15 for MnS.
15. Silver oxide dissolves according to ½ Ag2O + ½ H2O
Ag+ + OH–. How many moles of Ag2O
dissolve in 2.0 L of solution with [OH–] = 1.0 x 10–3 M, if KSP = [Ag+][OH–] = 2.0 x 10–8 for AgOH?
16. KSP = 3.7 x 10–15 for Al(OH)3. A precipitate just begins to form when 50.0 mL of 2.0 x 10–6 M Al(OH)3
is mixed with 50.0 mL of NaOH solution. What is the ORIGINAL concentration of the NaOH solution,
before being mixed with the Al(OH)3?
17. What is the solubility (expressed in grams per litre) of Mg(OH) 2 in 0.050 M MgSO4.
18. A solution contains 1.00 x 10–4 mol of NaOH in 1.00 L of water. If 1.56 x 10–9 g of Cu(OH)2(s) is able
to dissolve in the solution, what is KSP for Cu(OH)2?
19. A solution contains Na3AsO4. If 9.92 x 10–5 g of Ag2SO4 dissolves in 1.00 L of the solution, what
mass of Na3AsO4 is present in 1.00 L of the solution? KSP = 1.00 x 10–23 for Ag3AsO4
20. A solution contains K3AsO4. If 6.50 x 10–5 g of Ag3AsO4 dissolves in 1.00 L of the solution, what
mass of K3AsO4 is present in 1.00 L of the solution? KSP = 1.00 x 10–23 for Ag3AsO4