Chemistry 12 UNIT 3
SATURATION and PRECIPITATION
SATURATED:
-the state of a solvent in which no more solute will dissolve.
The solubility of a solid will be limited:
The solid will dissolve until the concentration of the solute reaches a maximum in the solution.
This is SATURATION.
IMPORTANT NOTE:
A substance forms a saturated solution if the dissolved
substance is in Equilibrium with some of the undissolved substance.
IN OTHER WORDS:
To translate and relate this to UNIT 2:
SATURATED in Unit 3 is the same as EQUILIBRIUM in Unit 2.
CONSIDER:
A
I
+
2.0 M
B
1.0 M
⇄
C
1.0 M
+
D
1.0 M
C
E
Given that Keq for this equilibrium reaction is 1.0, determine whether the inital data given is at
equilibrium.
Ktrial = (1.0)(1.0) / (2.0) (1.0) = 0.50
If it is not, determine which way the equilibrium must shift in order for this system to reach
equilibrium.
Ksp > Ktrial
1.0 > 0.50
Therefore this system must shift to the RIGHT to
produce more products in order to achieve
equilibrium
NOW CONSIDER:
AlPO4 (s)
I
X
⇄
Al3+ (aq)
0.50 M
+
PO43- (aq)
0.50 M
C
E
Given that Ksp for this equilibrium reaction (i.e. soluble salt in solution) is 6.0 X 10 -13
Calculate K trial and determine in which direction the above system would shift to achieve
equilibrium.
Ktrial = (0.50) (0.50) = 0.25
Ksp < Ktrial
6.0 X 10 -13 <0.25
THEREFORE, this would shift to the LEFT to achieve equilibrium
Determine whether a Precipitate forms…or in other words, determine whether this solution, at its
INITIAL state, is UNSATURATED, SATURATED, or SUPERSATURATED.
This system is SUPERSATURATED initially since at I, Ksp < Ktrial
The system must shift to the left to achieve equilibrium.
Ksp versus K trial:
Ksp > Ktrial
Ksp = Ktrial
Determining Precipitation and Saturation
UNSATURATED SOLUTION:
-contains less than the maximum
amount of a substance which can
dissolve
-therefore more solute may be dissolved
in the solution
-no undissolved solid present
-not at equilibrium
-a shift to the right is still possible
SATURATED SOLUTION:
-maximum amount of substance
dissolved in solution
-some undissolved material present
-equilibrium exists between dissolved
and undissolved material
rate of ions forming (dissociation) is
equal to
rate of crystallization (or precipitation)
NO PRECIPITATE
NO PRECIPITATE
OR
The minimum possible
precipitate forms
(in equilibrium with the
amount of dissolved ions
in solution)
EQUILIBRIUM =
SATURATION
Ksp < Ktrial
SUPERSATURATED SOLUTION:
-excess salt (more than the maximum
amount of dissolved salt in saturated
solution) crystallizes out of solution
-ppt continues to form until all excess
ions have been removed from solution
and the value of K trial has been
reduced to the value of Ksp
(Equilibrium has shifted to the left)
PRECIPITATE
FORMS
SOLUBLE vs. LOW SOLUBLE
SOLUBLE
LOW SOLUBLE
High Ksp
- this can be a dangerous
generalization
- best to base the level of solubility
on "s"
Low Ksp
- this can be a dangerous
generalization
- best to base the level of solubility
on "s”
Solubility > 0.1 M at RTP
Solubility < 0.1 M at RTP
Easily dissolved in water
(i.e. easily forms ions in water)
Relatively insoluble in water
(does not readily form ions in water)
Ions compatible as
separate particles
Ions incompatible
as separate particles
Low bonding drive
High bonding drive
CONSIDER:
A molecule, AB, with a Ksp = 25
Another molecule, A3B2, with a Ksp = 25
For AB, Ksp = s2
25 = s2
s=5
For A3B2, Ksp = 108s
25 = 108s5
s = 0.75
AHA!
The molecule AB is
5
more soluble!
Using your DATA BOOKLET
USE THE SOLUBILITY TABLE IN YOUR DATA BOOKLET TO
DETERMINE WHICH ARE INSOLUBLE IN WATER:
NaOH
CaSO4
PbCl2
KCl
CaBr2
USING THE SOLUBILITY TABLE:
mostly
soluble
SULPHATE
SULPHIDE
mostly low
solubility
PLEASE NOTE:
Compounds containing alkali metals, or H+, NH4+, NO3will be soluble in water.
LOW SOLUBLE means a precipitate forms.
DOES A PRECIPITATE FORM AND WHAT IS IT?
FOR EACH OF THE FOLLOWING QUESTIONS, you should be able to:
i) Write the dissociation equations for each given formula
ii) Use solubility table to determine precipitates in the products formed
iii) Write the Formula equation (Show the double replacement reaction)
iv) Write the Complete ionic equation (with spectator ions)
v) Write the Net Ionic equation (Precipitation equation)
a)
NaOH and HCl
b)
Bi(NO3)2 and NaOH
c)
Pb(C2H3O2)2 and K2SO4
d)
CuSO4 and FeCl3
e)
FeSO4 and (NH4)2S
f)
K2CO3 and Sr(NO3)2
g)
NaCl and KOH
h)
NaI and AgNO3
i)
Al2SO3 and CaCl2
j)
Na2SO4 and CaCl2
k)
K2SO4 and Ba(C2H3O2)2
l)
MgSO4 and CaBr2
UNIT 3 SAMPLE CALCULATIONS:
1. Recall previous example:
AlPO4 (s) ⇄
Al3+ (aq) +
PO43- (aq)
I
X
0.50 M
0.50 M
R
E
a) Given that Ksp for this equilibrium reaction (i.e. soluble salt in solution)
is 6.0 X 10 -13, Calculate the K trial.
b) Considering the K trial that you have calculated, determine in which direction
the above equilbrium would shift. OR IN OTHER WORDS, determine if a
precipitate would form.
c) What would the new equilbrium concentrations of each ion be? OR IN OTHER
WORDS, what is the concentration of each ion in the saturated solution?
d) HOW MUCH PRECIPITATE FORMS if there was originally 0.25 moles of
AlPO4(s) dissolved in 500. ml of solution.
2. The ion concentrations in 2.00 L of 0.32 M K3PO4 are:
3. Which is the least soluble in water:
a) CaS
b) Fe(OH)3
c) KMnO4
d) NH4HC2O4
4. A solution contains two cations, each having a concentration of 0.20 M.
When an equal volume of 0.20 M OH- is added, these cations are removed
from the solution by precipitation. These ions are:
a) Ba2+ and K+ b) Sr2+ and Na+ c) Sr2+ and Mg2+ d) Mg2+ and Ca2+
5. The solubility of Mn(IO3)2 is 4.8 x 10-3. What is the value of the Ksp?
6. Write the net ionic equation representing the reaction that occurs when 50.0
ml of 0.20 M ZnSO4 and 50.0 ml of 0.20 M BaS are combined.
7. When 1.00 g of Mg CO3 is added to 2.0 L of water, some, but not all, will
dissolve to form a saturated solution. Calculate the mass of solid that
remains undissolved.
ANSWERS IN CLASS. Will NOT be posted on Website.
Worksheet 3.2: Ksp Calculations
1. Calculate the solubility product of the following:
a) Mg(OH)2 whose solubility is 1.4 x 10-4 M
b) La(IO3)3 whose solubility is 6.9 x 10-4 M
2. The solubility of PbSO4 in water is 3.8 x 10-2 g/L.
Calculate the Ksp of PbSO4.
3. [Ag+] = 2.2 x 10-4 M in a saturated solution of
Ag2C2O4.
Determine the solubility product of the compound.
4. SrF2 has a Ksp = 2.18 x 10-8.
Determine the concentrations of the strontium and
fluoride ions in a saturated solution at that
temperature.
5. How many grams of PbI2 will dissolve in 250. mL of
water if its solubility product equals 1.7 x 10-5?
ANSWERS
1a) Ksp = [Mg+2] [OH-]2 = (1.4 x 10-4) (2.8 x 10-4)2 = 1.1 x 10-11
1b) La(IO3)3 Ksp = 6.4 x 10-12
2 Ksp of PbSO4 = 1.7 x 10-8
3 Ag2C2O4 Ksp = 5.3 x 10-12
4 [Sr+2]=1.76x10-3 M and [F-] = 3.52x10-3 M
5 1.8 g of PbI2