Solubility and Ksp
First,
let’s define a few terms that will be useful:
- solubility product constant (Ksp) – the equilibrium constant for the equilibrium between an ionic solid (salt) and its saturated solution
- solubility – the quantity of substance that dissolves to form a saturated solution (g/L)
- molar solubility – the number of moles of solute that dissolves in 1 L to form a saturated solution (mol/L) - in the past, we have called this "molar concentration"
We
must be able to convert between all three.
Solubility Product Constant, Ksp
We
can only study the solution equilibrium for a system in which there is a saturated
solution, ie, one in which the solution is in contact with undissolved solute.
For
example,
Ca3(PO4)2(s) ↔
3Ca2+(aq) + 2PO43-(aq)
In
the forward direction, there is dissolution as the solid dissolves in
solution to produce aqueous ions.
In
the reverse direction, there is precipitation as the ions come together to
form a solid and fall out of solution.
This
is a dynamic equilibrium (Ca2+ and PO43- are
continually moving between the solution and undissolved solid).
The
equilibrium law associated with this system is:
Ksp
= [Ca2+]3[PO43-]2 = 2.0 x
10-29 at 25°C
Ksp
(like all equilibrium constants) is the ratio of products to reactants or
[P]/[R] or, in this situation, [dissolved]/[undissolved]. Since the Ksp value is so
tiny, it indicates that calcium phosphate has very low water solubility.
Calculating the Solubility Product
Constant
Example
Solid magnesium fluoride is added to water at 25°C. At equilibrium, [Mg2+] = 2.10 x
10-3 M. What is the Ksp
for magnesium fluoride?
Answer:
First, set up the balanced equilibrium equation:
MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
Then set up the equilibrium expression and sub in the data.
We are told the eq [Mg2+] which we
plug in.
To determine the eq [F-],
use the stoichiometric relationship in the balanced equation: for every 1 Mg2+
that is dissolved in solution, there will be 2F- ions, so the eq [F-]
= 2 x 2.10 x 10-3 M = 4.20 x 10-3
M.
Ksp = [Mg2+][F-]2
= (2.10 x
10-3)( 4.20 x 10-3)2
= 3.70 x
10-8
Calculating the Solubility
Example
The
Ksp for calcium fluoride is 3.9 x
10-11 at 25°C.
What is the solubility of calcium fluoride?
Answer
First
set up an ICE table. We cross out the CaF2
section because it is a solid and thus, is not included in the equilibrium
expression. The initial [ ] for the two
ions is 0 M. The change for each ion is
+x and +2x respectively because they are in a 1:2 ratio. Finally, to arrive at eq [ ], simply add in [
] to Δ [
] in each column.
CaF2(s) ↔ Ca2+(aq) + 2F-(aq)
in
[ ] ---- 0 0
Δ [
] ---- +x +2x
eq
[ ] ---- x 2x
Set
up the eq expression and fill in the Ksp (if this value was not provided in the question, we could look it up in the Ksp
table provided below). Then, sub in the two eq [ ].
Solve for x.
Ksp = [Ca2+][F-]2
3.9 x 10-11 = x(2x)2
3.9 x 10-11 = 4x3
x = 2.1 x
10-4
Notice
that in the balanced equation above, for every 1 CaF2(s) that
dissolves, 1 Ca2+(aq) shows up in solution. So, solving for x tells
us how much solid has dissolved in solution, in mol/L (which is the molar
solubility).
However, we were asked to
determine solubility – to convert molar solubility into solubility, simply
multiply molar
solubility by the molar mass of the solid (CaF2):
∴ 2.1 x
10-4 mol/L(78.1 g/mol) = 1.6 x 10-2 g/L
Homework #36, 37
I take you through the answer to HW #36 in this video.
KspTable
