Catalyst
 Complete the exam reflection
End
The Month Ahead! Final Unit!
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Friday
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Lab Report
Due!
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Electro.
Lab
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Electro.
Lab
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Saturday
School
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Review
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Review
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Review
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Unit 8
Exam
22
Unit 8
Exam
SPRING
BREAK!
Teeth in Soda/Coffee
Justify – TPS
 Why does tooth enamel, Ca10(PO4)6(OH)2, decay in acidic
solutions, but not in basic solutions?
Lecture 8.1 – Common Ion Effect and Ksp
Today’s Learning Targets
 LT 8.1 – I can explain the common ion effect and how it impacts the
overall ionization of a compound that is in equilibrium.
 LT 8.2 – I can write a Ksp for slightly soluble salts
 LT 8.3 – I can calculate the Ksp and equilibrium concentrations from
initial concentration and/or solubility data.
 LT 8.5 – I can explain and calculate, using the Ksp, how the solubility of
a compound is influenced by both the common ion effect and pH.
Precipitation Reaction
 Many reactions results in the production of a solid compound
know as a precipitate. These reactions are known as
precipitation reactions.
 We can determine whether or not
a solid will form based on a few
rules.
PbNO3 reaction
Solubility Rules
 The solubility of a compound is determined by the ions a compound
contains:
1. All group I and ammonium ions are soluble
2. All nitrate, acetate, and chlorate compounds are soluble
3. All binary compounds of halogens (other than F) with metals are
soluble, except those of Ag, Hg (I) and Pb.
4. All sulfate containing compounds are soluble, except those of barium,
strontium, calcium, lead, silver, and mercury
5. Except for rule 1, all carbonate, hydroxide, oxide, silicate, and phosphate
containing compounds are insoluble
6. Sulfide containing compounds are insoluble, except for those with
calcium, barium, strontium, magnesium, sodium, potassium, and NH4+
Soluble Ionic Compounds
Compounds Containing
Exceptions
NO3-
None
CH3COO-
None
Cl-
Compounds of Ag+, Hg2+, Pb2+
Br-
Compounds of Ag+, Hg2+, Pb2+
I-
Compounds of Ag+, Hg2+, Pb2+
SO42-
Compounds of Sr2+, Ag+, Hg2+, Pb2+
Insoluble Ionic Compounds
Compounds Containing
Exceptions
S2-
Compounds of NH4+, the alkali metal
cations, Ca2+, Sr2+, Ba2+
CO32-
Compounds of NH4+ and the alkali metal
cations.
PO42-
Compounds of NH4+ and the alkali metal
cations.
OH-
Compounds of NH4+, the alkali metal
cations, Ca2+, Sr2+, Ba2+
Class Example
 Identify the compounds that soluble and the ones that are
insoluble:
 Na2CO3
 PbSO4
Table Talk
 Identify the compounds that soluble and the ones that are
insoluble:
 Co(OH)2
 Ba(NO3)2
Solubility Product Constant (Ksp)
 A saturated solution is one that has dissolved as much solid as
possible.
 In a saturated solution, equilibrium is established:
BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq)
 The equilibrium constant for slightly soluble salts is known as the
solubility – product constants or Ksp
Ksp = [Ba2+]1[SO42-]1
Class Example
 Write the Ksp expression for calcium fluoride
Table Talk
 Write the solubility – product constant (Ksp) for silver sulfate
Solubility and Ksp
 The Ksp is NOT the solubility
 Solubility describes how many grams can dissolve in a liter of
solution
 The Ksp describes how much dissolves to form a saturated solution
 Solubility can change, but Ksp CANNOT
Ksp
Solubility of
compound
(g/L)
Solubility
Equilibrium
Molar Mass
Molar Solubility
of compound
(mol/L)
Compound
Formula
Ion
Concentration
(mol/L)
Class Example
 Solid silver chromate is added to pure water at 25 oC, and some of
the solid remains undissolved. The mixture is stirred for several
days to ensure that equilibrium is achieved between the
undissolved Ag2CrO4 (s) and the solution. Analysis of the
equilibrated solution. Analysis of the equilibrated solution shows
that its silver ion concentration is 1.3 x 10-4 M. Calculate the Ksp
for the solution
Table Talk
 A saturated solution of Mg(OH)2 in contact with undissolved
Mg(OH)2 is prepared at 25 oC. The pH of the solution is found to
be 10.17. Assuming that Mg(OH)2 dissociates completely in
water, calculate Ksp for this equilibrium. HINT – Think Kw…
Class Example
 The Ksp for CaF2 is 3.9 x 10-11 at 25 oC. Assuming that CaF2
dissociates completely upon dissolving and that there are no other
important equilibria affecting its solubility, calculate the solubility
of CaF2 in grams per liter.
Table Talk
 The Ksp for LaF3 is 2 x 10-19. What is the solubility of LaF3 in
moles per liter?
Talking it to the Next Level!
 Based on your current level, attempt the questions around the
classroom.
 Pink – Cake Walk Level
 Yellow – Heating Up Level
 Blue – Expert Chemist Questions
Fluoridation of Water Supply
 Tap water contains fluoride ions in order to help prevent cavities.
 Fluoridinated water replaces the OH group on tooth enamel with
a fluoride ion.
 The Ksp of normal tooth enamel (Ca10(PO4)6(OH)2) is 6.8 x 10-37
 The Ksp of tooth enamel with OH group replaced Ca10(PO4)6F2 is
5.0 x 10-51.
Justify – TPS
 The Ksp of normal tooth enamel (Ca10(PO4)6(OH)2) is 6.8 x 10-37
 The Ksp of tooth enamel with OH group replaced Ca10(PO4)6F2 is
5.0 x 10-51.
 Why do you think adding fluoride to water is crucial to lowering
the amount of cavities we have as a country?
Factors that Impact Solubility
 There are three ways in which solubility of slightly soluble salts
can be impacted:
1. Common Ion Effect
2. The pH of Solution
3. Presence of Complexing Agent
Common Ion Effect
 Recall the acetate equilibrium:
CH3COOH ⇌ H+ + CH3COO If we had both acetic acid and acetate in solution, than the acetic
acid will ionize less in order to achieve equilibrium.
 Very little movement needs to be done in order to achieve
equilibrium.
 Whenever a weak electrolyte and a strong electrolyte containing a
common ion are together, the weak electrolyte ionizes less than
when it is alone in solution
 This is known as the common ion effect
Class Example
 Calculate the molar solubility of CaF2 that is in a solution of 0.010
M Ca(NO3)2. NOTE – The Ksp for CaF2 is 3.9 x 10-11
Table Talk
 For manganese (II) hydroxide the Ksp is 1.6 x 10-13. Calculate the
molar solubility of manganese (II) hydroxide in a solution that
contains 0.020 M NaOH
pH and Solubility
 The pH of a solution impacts the solubility for any compound that
has a basic component when dissolved in solution
 Consider the equilibrium:
Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2 OH- (aq) Ksp = 1.8 x 10-11
 If we are in a basic solution, then we will have more solid due to
the presence of extra hydroxide ions
 The solubility of a compound containing a basic component
increases as the pH is lowered.
Class Example
 Which of these substances are more soluble in acidic solution than
in basic solution:
(1) Ni(OH)2
(2) CaCO3
(3) BaF2
(4) AgCl
Table Talk
 Which of these substances are more soluble in acidic solution than
in a neutral solution:
(1) ZnCO3
(2) ZnS
(3) BiI3
(4) AgCN
(5) Ba3(PO4)2
White Board Problems
White Board Problems
 The molar solubility of Ag2SO4 is 0.014 M at 25 oC. Based on this
information, what is the value of Ksp for Ag2SO4
 The value of the solubility product constant for the reaction:
BaF2 (s) ⇌ Ba2+ (aq) + 2 F- (aq)
Is 1.1 x 10-6. Calculate the concentration of F- ions in a saturated
solution of BaF2
 500 mL of a 0.0050 molar NaF solution is added to 400 mL of a
0.0050 molar Ba(NO3)2 solution. Will there be a precipitate? Use
the equilibrium from the previous problem.
Exit Slip
1. Calculate the molar solubility of a pure solution of CaF2. The Ksp
of CaF2 is 3.9 x 10-11.
2. Calculate the molar solubility of CaF2 that is in a solution of 0.010
M Ca(NO3)2. NOTE – The Ksp for CaF2 is 3.9 x 10-11. Discuss how
this answer in #1 and #2 demonstrates the impact the common
ion effect has on the dissociation of a slightly soluble salt
Rate Yourself!
 Rate yourself 1 – 4 on LTs 8.1, 8.2, 8.3, and 8.5
Closing Time
 Read 17.1, 17.4, and 17.5
 Homework: HW 8.1