Chemistry 130
Solubility equilibria
Dr. John F. C. Turner
409 Buehler Hall
jturner@ion.chem.utk.edu
Chemistry 130
Solubility and sparingly soluble salts
Solubility is an important feature of geology, medicine and biology. Solubility
is determined by the equilibrium between a solid and a solution of the solid
in a liquid.
For calcium oxalate, the equilibrium would be written as
Ca C2 O4
2+
2-
Ca aq
s
C2 O4
aq
And the equilibrium constant is therefore
Ca C2 O4
2+
s
2+
Chemistry 130
s
C2 O4
2-
K sp = [Ca aq ][C2 O4
Note that [Ca C2 O4
2-
Ca aq
aq
aq
]
] = 0 as solids have an invariant concentration
Solubility and sparingly soluble salts
K sp is the solubility product and is used for sparingly soluble salts.
We only apply the solubility product for materials that are sparingly soluble
because
we know that there is only one equilibrium between the solute and the
solid
●
the solution is dilute and therefore we can easily apply thermodynamics
to the problem
2+
2Ca C2 O4 s
Ca aq
C2 O4 aq
The solubility product is related to the molar− 1solubility by − 1
s mol L
s mol L
22
K sp = [Ca2+
][C
O
]
=
s
aq
2 4 aq
●
s=
Chemistry 130
K sp
1
2
Solubility and sparingly soluble salts
The solubility product depends only on the ions in solution and is constant
for a set of particular ions.
Le Chatelier's principle applies and so if the concentration of one of the ions
is increased, precipitation of the least soluble phase will occur.
Ca C2 O4
2+
s
2-
Ca aq
C2 O4 aq
−1
K sp
s=
Chemistry 130
−1
s mol L
s mol L
22
= [Ca2+
][C
O
]
=
s
aq
2 4 aq
K sp
1
2
Solubility and sparingly soluble salts
Q. Calculate the molar solubility of calcium oxalate ifK sp = 2.7 × 10− 9
The equlibrium is
2+
Ca C2 O4 s
Ca aq
2-
C2 O4 aq
−1
K sp
−1
s mol L
s mol L
2
= 2.7 × 10− 9 = [Ca2aq+ ][C2 O2]
=
s
4 aq
1
−9 2
s = 2.7 × 10
Chemistry 130
−5
= 5.196 × 10
−1
mol L
Solubility and sparingly soluble salts
Q. Calculate the molar solubility of calcium oxalate ifK sp = 2.7 × 10− 9
If 10 ml of a 0.1 M solution of sodium oxalate is added to this solution, what
is the final of calcium ions in solution?
Ca C2 O4
2+
Ca aq
s
2[Ca2+
]
=
[C
O
aq
2 4 aq ] =
2-
So initially, [C2 O4
−9
2-
C2 O4
K sp = 2.7 × 10
aq
2K sp and K sp = [Ca2+
][
C
O
aq
2 4 aq ]
−5
aq ]initial = 5.196 × 10
−1
mol L
−3
10 ml of a 0.1 M solution of sodum oxalate contains 1 × 10 mol of oxalate
So the new concentration of oxalate is
2−5
−3
−3
[C2 O4 aq ]new = 5.196× 10
1× 10 mol = 1.05196× 10 mol
−9
K sp
2.7 × 10
2+
−6
−1
[Ca aq ] =
=
=
2.57
×
10
mol
L
2−3
[C2 O4 aq ]new
1.05196× 10
Chemistry 130
Solubility and sparingly soluble salts
If the ion concentration product is greater than the solubility product,
precipitation will take place.
This is simply Le Chatelier's principle in action – we force the solubility
equilibrium towards the solid by increasing the concentration of one of the
ions in solution – the ions being the products of the equilibrium
Chemistry 130
Chemistry 130