KSP Solubility and Solubility Product Common Ion Effect Chemistry

Common ion effect in solubility equilibria and its a step-by step calculation

1. The Solubility Product Constant, K
sp
•Many ionic compounds are only slightly soluble in
water and equations are written to represent the
equilibrium between the compound and the ions
present in a saturated aqueous solution.
•The equilibrium system is a heterogeneous system.
•The solubility product constant, K
sp, is the product
of the concentrations of the ions involved in a
solubility equilibrium, each raised to a power equal to
the stoichiometric coefficient of that ion in the
chemical equation for the equilibrium.

2. The Solubility Equilibrium Equation And K
sp
•* Remember, solids are not in equilibrium
expressions!

3. Some Values For Solubility Product
Constants (K
sp) At 25
o
C

4. K
sp And Molar Solubility
•The solubility product constant (K
sp) is related to
the solubility of an ionic solute, but K
sp and Molar
Solubility are not the same thing.
•Molar Solubility - the molarity (concentration) of a
solute in a saturated aqueous solution
•Calculating solubility equilibria fall into two
categories:
–determining a value of K
sp from experimental data
–calculating equilibrium concentrations when K
sp is
known.

5. Calculating K
sp From Molar Solubility
It is found that 1.2x10
-3
mol of lead (II) iodide, PbI
2,
dissolves in 1.0 L of aqueous solution at 25
o
C.
What is the K
sp at this temperature?
•Ksp = [Pb
2+
][I
-
]
2
•Ksp = [1.2 x 10
-3
] [2(1.2 x 10
-3
)]
2
•2 iodide ions
form, so you
must multiply
molarity by 2!
•Ksp = 6.9 x 10
-9

6. Calculating Molar Solubility From K
sp
Calculate the molar solubility of silver chromate,
Ag
2CrO
4, in water from K
sp = 1.1x10
-12
for Ag
2CrO
4.
Ksp = [Ag
+
]
2
[CrO
4
-2
]
At equilibrium 2x x
1.1 x 10
-12
= (x)(2x)
2
1.1 x 10
-12
= 4x
3
x = 6.5 x 10
-5
M = [Ag
2
CrO
4
] = [CrO
4
2-
]
[Ag
+
] = 2(6.5 x 10
-5
)

7. The Common Ion Effect In Solubility Equilibria
•The common ion effect also affects solubility
equilibria.
•Le Châtelier’s principle is followed for the
shift in concentration of products and
reactants upon addition of either more
products or more reactants to a solution.
•The solubility of a slightly soluble ionic
compound is lowered when a second solute
that furnishes a common ion is added to the
solution.

8. Solubility Equilibrium Calculation
-The Common Ion Effect
What is the solubility of Ag
2CrO
4 in 0.10 M K
2CrO
4?
K
sp
= 1.1x10
-12
for Ag
2
CrO
4
.
Comparison of solubility of Ag
2CrO
4
In pure water: 6.5 x 10
-5
M
In 0.10 M K
2CrO
4:1.7 x 10
-6
M
The common ion effect!!
Adding more CrO
4
2-
ions shifts the equilibrium back
to the reactants, which is solid Ag
2CrO
4

9. Chapter SixteenPrentice-Hall ©2002 Slide 9 of 32
Common Ion Effect Example:
What will the molar solubility of CaF
2 (Ksp = 4.0 x 10
-11
)
in a 0.025M NaF solution
Before it dissolves
[Ca
2+
] = 0 [F
-
] = 0.025
After it dissolves
[Ca
2+
] = x [F
-
] = 0.025 + 2x

From NaFFrom CaF
2
4.0 x 10
-11
= x (0.025 – 2x)
2
Assume x is very small
due to small Ksp
4.0 x 10
-11
= x(0.025)
2
x = 6.4 x 10
-8
= molar
solubility

10. Determining Whether Precipitation Occurs
•Q
sp is the ion product reaction quotient and is
based on initial conditions of the reaction.
•Q
sp can then be compared to K
sp.
•To predict if a precipitation occurs:
- Precipitation should occur if Q
sp > K
sp.
- Precipitation cannot occur if Q
sp < K
sp.
- A solution is just saturated if Q
sp = K
sp.
Sometimes the concentrations of the ions
are not high enough to produce a
precipitate!

11. Determining Whether Precipitation Occurs
– An Example
The concentration of calcium ion in blood plasma is
0.0025 M. If the concentration of oxalate ion is
1.0x10
-7
M, do you expect calcium oxalate to
precipitate? K
sp
= 2.3x10
-9
.
Two steps:
(1)Determine the initial concentrations of ions.
(2)Compare Q
sp
with K
sp
.

12. Chapter SixteenPrentice-Hall ©2002 Slide 12 of 32
The concentration of calcium ion in blood plasma is
0.0025 M. If the concentration of oxalate ion is
1.0x10
-7
M, do you expect calcium oxalate,
CaC
2
O
4
, to precipitate? K
sp
= 2.3x10
-9
.
Determining Whether Precipitation Occurs
Q
sp = [Ca
2+
][C
2O
4
2-
] = (0.0025)(1.0 x 10
-7
)
Q
sp
= 2.5 x 10
-10
•Qsp < Ksp therefore, no precipitate will form!!!

13. Determining Whether Precipitation Occurs
In applying the precipitation criteria, the effect of dilution when
solutions are mixed must be considered.
Example: A 250.0 mL sample of 0.0012 M Pb(NO
3
)
2
(aq) is
mixed with 150.0 mL of 0.0640 M NaI (aq). Should
precipitation of PbI
2
(s), K
sp
= 7.1x10
-9
, occur?
Calculate new concentrations in total volume of 400mls = 0.4L
[Pb
2+
] = (0.250L)(0.0012M)/(0.400L) = 7.5 x 10
-4
M
[I-] = (0.150L)(0.0640M)/(0.400L) = 0.024 M
Qsp = [Pb
2+
][I
-
]
2
= (7.5 x 10
-4
)(0.024)
2
= 4.32 x 10
-7
Qsp > Ksp therefore a precipitate will form!

14. Chapter SixteenPrentice-Hall ©2002 Slide 14 of 32
Comparing Solubilities
•Which salt will be the most soluble in water.
1.CuS Ksp = 8.5 x 10
-45
2.Ag
2S Ksp = 1.6 x 10
-49
3. Bi
2S
3 Ksp = 1.1 x 10
-73
Ksp = x
2
Ksp = (2x)
2
x = 4x
3
Ksp = (2x)
2
(3x)
3
= 108x
5
Molar solubility
x = 9.2 x 10
-23
x = 3.4 x 10
-17
x = 1.0 x 10
-15
•Most Soluble = highest molar solubility!!!! Do not use Ksp!!

15. Effect of pH on Solubility
•The solubility of an ionic solute may be greatly
affected by pH if an acid-base reaction also occurs
as the solute dissolves.
•In other words, some salts will not dissolve well in
pure water, but will dissolve in an acid or a base.
•If the anion (A-) of the salt/precipitate is that of a
weak acid, the salt/precipitate will dissolve more
when in a strong acid (H+ ions will form HA with
A-)
•However, if the anion of the precipitate is that of a
strong acid, adding a strong acid will have no
effect on the precipitate dissolving more.

16. Chapter SixteenPrentice-Hall ©2002 Slide 16 of 32
•How would the addition of HCl affect the solubility of
PbCl
2?
–Cl- is the conjugate base of a strong acid, thus it is a weak
base.
–It will not react with H+ ions, so there is no effect.
•How would the addition of HCl affect the solubility of
FeS?
–S
2-
is a strong base (conjugate base of weak acid)
–Thus, it will react with H+ ions to form H
2S,
–This will shift the equilibrium to make more FeS dissolve!
Effect of pH on Solubility

17. Chapter SixteenPrentice-Hall ©2002 Slide 17 of 32
Summary
•The solubility product constant, K
sp,
represents equilibrium between a slightly
soluble ionic compound and its ions in a
saturated aqueous solution.
•The common ion effect is responsible for the
reduction in solubility of a slightly soluble
ionic compound.
•The solubilities of some slightly soluble
compounds depends strongly on pH.