AP Chemistry: Solubility Equilibria
The Solubility Product Constant (Ksp)
The constant for the solubility equilibrium for slightly soluble or nearly insoluble ionic compounds
When an excess of a slightly soluble salt is mixed with water, equilibrium is established between the solid and the dissolved ions...
Saturated Solution!
Dynamic Equilibrium
So, of course, we can write
an equilibrium expression...Ksp!
The Solubility Product Constant (Ksp)
calcium oxalate...
CaC2O4 (s) Ca+2 (aq) + C2O4­2 (aq)
Ksp = [Ca+2] [C2O4­2]
lead (II) iodide...
PbI2 (s) Pb+2 (aq) + 2 I­1 (aq)
Ksp = [Pb+2] [I­1]2
*As always, Ksp is temperature dependent!
17.1­17.3
AP Chemistry: Solubility Equilibria
Calculating Ksp
The solubility of silver chloride is 1.9 x 10­3 g/L at
25 C. What is the Ksp of silver chloride at this temperature?
0.0019
143.4
I
C
E
= 1.33 x 10­5 M
AgCl (s) Ag+1 (aq) + Cl­1 (aq)
0
0
+x
+x
1.33 x 10­5 1.33 x 10­5 Ksp = [Ag+1] [Cl­1] = (1.33 x 10­5)2 =
1.7 x 10­10 Calculating Ksp
The solubility of lead (II) arsenate, Pb3(AsO4)2, a compound found in some insecticides, is 3.0 x 10­5 g/L at 25 C. What is its Ksp at this temperature?
3.0 x 10­5
= 3.3 x 10­8 M
899.4
Pb3(AsO4)2 (s) 3 Pb+2 (aq) + 2 AsO4­3 (aq)
I
C
E
0
+3x
9.9 x 10­8 0
+2x
6.6 x 10­8 Ksp = [Pb+2]3 [AsO4­3]2 =
(9.9 x 10­8)3 (6.6 x 10­8)2 =
4.2 x 10­36 17.1­17.3
AP Chemistry: Solubility Equilibria
Table 17.1, pg. 703, has Ksp values
Check out how small some of the Ksp values are!
Which compounds will we NOT see on this table?
Which ionic compounds are ALWAYS soluble?
All sodium, potassium, ammonium and nitrate salts!
When we know Ksp, we can calculate a salt's solubility and also use it to determine if precipitation will occur under given conditions
Calculations with Ksp
What is the solubility of "fluorite" (calcium fluoride, CaF2) at 25 C if its Ksp is 3.4 x 10­11 at this temp?
CaF2 (s) Ca+2 (aq) + 2 F­1 (aq)
0
0
I
+x
+2x
C
x
2x
E
Ksp = 3.4 x 10­11 = [Ca+2] [F­1]2 = (x) (2x)2
3.4 x 10­11 = 4x3
x = 2.0 x 10­4 M
2.0 x 10­4 mol/L x 78.1 g/mol =
1.6 x 10­2 g/L
17.1­17.3
AP Chemistry: Solubility Equilibria
Solubility and the Common Ion Effect
PbCrO4 (s) Pb+2 (aq) + CrO4­2 (aq)
Lead (II) chromate is slightly soluble in water
(Ksp = 1.8 x 10­14)
When very soluble lead (II) nitrate is added...
COMMON­ION EFFECT!
Pb+2 is being added...
Reaction shifts left (LeChatelier!)
PbCrO4 precipitates out
"Ionization is repressed!"
Solubility and the Common Ion Effect
Calculate the molar solubility of BaF2 in water at 25 C (Ksp is 1.0 x 10­6), and then compare it to
the molar solubility in 0.15 M NaF.
BaF2 (s) Ba+2 (aq) + 2 F­1 (aq)
0
0
I
+x
+2x
C
x
2x
E
Ksp = 1.0 x 10­6 = [Ba+2] [F­1]2 = (x) (2x)2
1.0 x 10­6 = 4x3
x = 6.3 x 10­3 M
17.1­17.3
AP Chemistry: Solubility Equilibria
*compare it to the molar solubility in 0.15 M NaF
BaF2 (s) Ba+2 (aq) + 2 F­1 (aq)
0.15 M
0
I
+2x
+x
C
0.15 + 2x
x
E
Ksp = 1.0 x 10­6 = [Ba+2] [F­1]2 = (x) (.15 +2x)2
1.0 x 10­6 = (x) (.15)2 = 0.225x
x = 4.4 x 10­5 M
*compare molar solubilities:
6.3 x 10­3 M = 140 times greater 4.4 x 10­5 M without F­ present!
Precipitation Calculations
Precipitation is merely another way of looking at solubility equilibrium ­ will precipitation occur for given starting ion concentrations?
AKA...with given concentrations, will the reaction: go FORWARD (further dissolving),
or in REVERSE (precipitation)...
Q!
Qc = Ksp reaction at equilibrium
(saturated solution)
Qc > Ksp reaction goes left (reactants)
PRECIPITATION!
Qc < Ksp reaction goes right (products)
(unsaturated solution)
*Qc = solubility (ion) product
17.1­17.3
AP Chemistry: Solubility Equilibria
Precipitation Calculations
Lead (II) nitrate, Pb(NO3)2, and sodium chloride, NaCl, are added to water to get a solution that is
0.050 M Pb(NO3)2 and 0.10 M NaCl.
Will lead (II) chloride, PbCl2, precipitate out?
Ksp for PbCl2 = 1.6 x 10­5
PbCl2 (s) Pb+2 (aq) + 2 Cl­1 (aq)
Qc = [Pb+2] [Cl­1]2
= (.050) (.10)2
= 5.0 x 10­4
Qc > Ksp Yes, precipitate!
...until equilibrium reached (saturated)
Precipitation Calculations
Precipitation is an important process in the lab and in chemical industry ­ we precipitate Mg+2 from seawater to manufacture Mg metal
Solubilities and precipitation of salts in the ocean and different bodies of water and underground in aquifers and caves can be a major studying area for environmental engineers
Typical problem: given two solution volumes and concentrations, will a precipitate form?
17.1­17.3
AP Chemistry: Solubility Equilibria
If 50.0 mL of .0010 M BaCl2 is added to 50.0 mL of .00010 M Na2SO4, will BaSO4 precipitate out?
Ksp for BaSO4 = 1.1 x 10­10
BaSO4 (s) Ba+2 (aq) + SO4­2 (aq)
Qc = [Ba+2] [SO4­2]
[Ba+2] [SO4­2] (.0010 M) (.050 L) =
5.0 x 10­5 mol
(.00010 M) (.050 L) =
5.0 x 10­6 mol
0.100 L total
5.0 x 10­4 M
0.100 L total
5.0 x 10­5 M
Qc = [Ba+2] [SO4­2] = (5.0 x 10­4) (5.0 x 10­5) =
2.5 x 10­8 > Ksp Yes! Precipitate!
Fractional Precipitation
A technique of separating two or more ions from a solution by adding reactant that precipitates one ion first, then another, and so on.
A solution is made with .10 M BaCl2 and .10 M SrCl2. By adding K2CrO4, we can get BaCrO4 and SrCrO4 to precipitate out. Which will precipitate first,
and at which [K2CrO4]?
Ksp for BaCrO4 = 1.2 x 10­10
Ksp for SrCrO4 = 3.5 x 10­5
BaCrO4: 1.2 x 10­10 = [Ba+2] [CrO4­2] = (.10) [CrO4­2]
[CrO4­2] = 1.2 x 10­9 M
SrCrO4: 3.5 x 10­5 = [Sr+2] [CrO4­2] = (.10) [CrO4­2]
[CrO4­2] = 3.5 x 10­4 M
K2CrO4 is added slowly...
At [CrO4­2] = 1.2 x 10­9 M, BaCrO4 starts to precip At [CrO4­2] = 3.5 x 10­4 M, SrCrO4 starts to precip When SrCrO4 starts to precip, what % Ba+2 left?
1.2 x 10­10 = [Ba+2] [CrO4­2] = [Ba+2] (3.5 x 10­4)
[Ba+2] =
17.1­17.3
3.4 x 10­7 M
= .0000034 (.00034%)
.10 M