Solubility Equilibria And The Solubility Product Constant (Chemistry Worksheet)
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37.6 SOLUBILITY EQUILIBRIA AND THE SOLUBILITY PRODUCT CONSTANT
is the quantity of solute that can dissolve in a quantity of
solvent and a certain temperature.
is a dynamic equilibrium (reversible)
between the solute and solvent in a saturated solution.
When sodium iodide is placed in water, the charged ends of polar H
O molecules pulls the ions
2
+
-
Ag
from their crystal lattice structure. The dissolution equation i:s AgI
+ I
(s)
(aq)
(aq)
Once in solution, the ions can collide with each other to re-form the solid precipitate showing
+
-
AgI
that: Ag
+ I
(aq)
(aq)
(s)
+
-
The equilibrium between dissolution and precipitation is shown as: AgI
⇌ Ag
+ I
(s)
(aq)
(aq)
The Solubility Product Constant (K
) is the value obtained from the equilibrium law, but based
sp
on a saturated solution. K
has no units and its values can be found in Appendix B4 (text)
sp
Since solubility equilibrium involves heterogeneous equilibria and the concentration of a solid
remains constant, we write the solubility product constant equation with reference to the ions
only (whose concentrations can vary)
Example Problem 1: Write solubility constant equation for a saturated solution of Al
(SO
)
2
4
3(aq)
Balanced Equation for Equilibrium:
K
=
sp
Example Problem 2: In a saturated solution of Ca
(PO
)
, the concentration of calcium ions is
3
4
2
-7
-7
4.53 x 10
mol/L and 3.02 x 10
mol/l. Calculate the K
of Ca
(PO
)
sp
3
4
2
Balanced Equation: Ca
(PO
)
⇌
3
4
2(aq)
2+
-7
3-
-7
G: [Ca
] = 4.53 x 10
[PO
] = 3.02 x 10
R: K
,
4
sp = ?
A: K
=
S: K
=
sp
sp
Molar solubility and K
both describe the solubility of a substance in different ways. You can use
sp
the balanced dissociation equation and the molar solubility of the solid to determine the
concentration of the ions in solution. Then calculate the K
. You can determine the K
sp
sp -
solubility expressed in mass per volume by converting it to mol per volume.
-14
Example Problem 3: Calculate the K
value for Bi
S
which has a solubility of 4.68 x 10
g/100 mL.
sp
2
3(s)
3+
2-
Balanced Equation: Bi
S
⇌ 2 Bi
+ 3 S
2
3(aq)
(aq)
(aq)
-14
G: Bi
S
Solubility = 4.68 x 10
g/100 mL
2
3
3+
2-
Bi
S
2Bi
+
3S
⇌
2
3(s)
(aq)
(aq)
R: K
= ?
I
-
0
0
sp
C
-
+2x
+3x
3
3+
2
2-
]
A: K
= [Bi
]
[S
E
-
+2x
+3x
sp
(aq)
(aq)
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