where there are only free ions, this free ion concentration is referred to as
activity
. Ksp is like the ideal gas law, which is only for gases not attracted to other
gases.
Ksp works for “ideal” solutes only.
In a sense, there are two concentrations of ions, the stoichiometric concentration and the activity.
K
sp
is based on
activity, the effective concentrations of ions regarding solubility.)
Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
41
IMPORTANCE OF SOLUBILITY IN INDUSTRY AND MEDICINE
BaSO
4
, an
insoluble
salt, shows pathway blockage of digestive tract.
CaC
2
O
4
, calcium oxalate, is main component of kidney stones.
Ca
2+
is in the blood, which reacts with oxalates in certain fruits and
vegetables, e.g., rhubarb and spinach.
Treatment: adjust diet.
Ca
5
(PO
4
)
3
OH, hydroxyapatite, is tooth enamel.
Ca
5
(PO
4
)
3
OH + H
3
O
+
→ Ca
5
(PO
4
)
3
+
+ 2H
2
O
Tooth enamel is soluble in acid.
Ca
5
(PO
4
)
3
OH
+ F

→ Ca
5
(PO
4
)
3
F (fluoroapatite) + OH

Ca
5
(PO
4
)
3
F + H
3
O
+
Ca
5
(PO
4
)
3
+
+ HF + H
2
O
Fluoroapatite less soluble in acid.
Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
42
HOW TO WORK WITH LARGE NUMBERS
Some calculations are too big (or too small) for your calculator; i.e.,
when K is very large or very small, your calculator may not be able to
handle it.
So work with logs.
e.g.:
K = (1.2 x 10
52
)
2
(3.6x10
15
)
a = (1.2 x 10
52
)
2
b = (3.6 x 10
15
)
K = (a)(b)
log(K) = log(a) + log(b)
log(K) = log((1.2x10
52
)
2
) + log(3.6x10
15
)
log(K) = 2 x log(1.2x10
52
) + log(3.6x10
15
)
log(K) = 104.1580 + 15.5563 = 119.7143
log(K) = 119.7143
log(K) = 119 + 0.7143
log(a) = b + c
a = 10
b
x 10
c
K = 10
0.7143
x 10
119
K = 5.18 x 10
119
Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
43
Solubility Product Constants*
AgCl(s)
Ag
+
(aq) + Cl

(aq)
This is the equilibrium between a saturated solution of AgCl and solid
AgCl.
All of the AgCl that dissolves is dissociated into Ag
+
and Cl

ions.
Equilibrium expression: K
sp
= [Ag
+
][Cl

]
Pure liquids and pure solids don’t appear in equilibrium constant
expressions.
K
sp
, the solubility product constant, is the product of the concentrations
of the ions involved in the solubility equilibrium, each raised to a power
equal to the stoichiometeric coefficient of that ion in the chemical
equation for the equilibrium.
A(s)
3B(aq) + 2C(aq)
A(s)
B(aq) + B(aq) + B(aq)
+ C(aq) + C(aq)
([B]
3
[C]
2
) = K
sp
H&P Example 16.1 mod.:
Write the equilibrium equations and equilibrium constant expressions for
(a) iron(III) phosphate and (b) chromium(III) hydroxide.
(a)
FePO
4
(s)
Fe
3+
(aq) + PO
4
3
(aq)
[Fe
3+
][PO
4
3
] = K
sp
(b)
Cr(OH)
3
(s)
Cr
3+
(aq) + 3OH

(aq)
[Cr
3+
][OH

]
3
= K
sp
*The product of the equilibrium concentration of the ions in the solution is equal to K, the equilibrium
constant (= the solubility product constant = K
sp
).
Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
44
SOLUBILITY EQUILIBRIA
p.5.CHEM1622010 8th WEEK RECITATION
ET Note: Given the solubility of a salt, find K
sp
Z&Z 75
Use the following data to calculate the K
sp
value for each solid.
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 pH, solubility equilibria lecture, equilibria lecture notes