•
CaC
2
O
4
, calcium oxalate, are 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.
CHEM1622013 CHAPTER 17
37
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
)
log(ab) = 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) = 119.7147
log(K) = 119 + 0.7147
log(a) + log(b) = log(ab)
(???)
K = 10
0.7147
x 10
119
K = 5.18 x 10
119
38
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.
39
SOLUBILITY EQUILIBRIA
p.5.CHEM1622010 8th WEEK RECITATION
ET Note: Given the solubility of a salt, find KspZ&Z 75 Use the following data to calculate the Kspvalue for each solid. a. The solubility* of CaC2O4is 4.8 x 105mol/L. CaC2O4(s) ←→Ca2+(aq) + C2O42(aq) CaC2O4(s)←→Ca2+(aq) + C2O42(aq) Initial Y 0 Change X +X Equilibrium Y  X +X CaC2O4(s)←→Ca2+(aq) + C2O42(aq) Initial Y 0 Change 4.8 x 105+4.8 x 105+4.8 x 105
0
+X
X
0
b. The solubility of BiI3is 1.32 x 105mol/L. BiI
0
+3X
3X
0
)
)
CHEM1622013 CHAPTER 17
40
From Kimmel’s notes
ET Note: Given the solubility of a salt, find Ksp; the solubility of the salt is based on its pH.
A saturated calcium hydroxide solution has a pH of 12.14.
Calculate the K
sp
of
calcium hydroxide.
pH = 12.14
pOH = 14.00  12.14 = 1.86
[OH

] = 10
pOH
= 10
1.86
= 1.38 x 10
2
Ca(OH)
2
(solid)
←
→
Ca
2+
(aq) + 2OH

(aq)
Ca(OH)
2
(s)
←
→
Ca
2+
(aq)
+
2OH

(aq)
Initial
Y
0
0
Change
Equilibrium
1.38 x 10
2
Ca(OH)
2
(s)
←
→
Ca
2+
(aq)
+
2OH

(aq)
Initial
Y
0
0
Change
X
+X
+2X
Equilibrium
YX
+X
1.38 x 10
2
0 + 2X = 1.38 x 10
2
X = 0.0069
Ca(OH)
2
(s)
←
→
Ca
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 Chemistry, pH, Solubility, CHEM1622013 CHAPTER