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Solubility Products
Consider the equilibrium that exists in a saturated
solution of BaSO4 in water:
Aqueous equilibria II
Solubility Products
BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)
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Solubility Products
The equilibrium constant expression for this
equilibrium is
Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called the
solubility product.
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1
Which Ksp expression is correct for AgCl?
A
[Ag+]/[Cl-]
B
[Ag+][Cl-]
C
[Ag2+]2[Cl2-]2
D
[Ag+]2[Cl-]2
E
None of the above.
There is never any denominator in Ksp expressions
because pure solids are not included in any
equilibrium expressions.
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2
Given the reaction at equilibrium:
Zn(OH)2 (s) <--> Zn2+ (aq) + 2OH- (aq)
what is the expression for the solubility product
constant, Ksp, for this reaction?
A Ksp= [Zn2+][OH-]2 / [Zn(OH)2]
B
Ksp= [Zn(OH)2] / [Zn2+][2OH-]
C Ksp= [Zn2+][2OH-]
D
Ksp= [Zn2+][OH-]2
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Solubility Products
· Ksp is not the same as solubility.
· Solubility is generally expressed as the mass of solute
dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in
mol/L (M).
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Solubility
Solubility
The term "solubility" represents the maximum amount
of solute that can be dissolved in a certain volume
before any precipitate is observed.
The solubility of a substance can be given in terms of
grams per liter
g/L
or in terms of
moles per liter
mol/L
The latter is sometimes referred to as "molar
solubility."
Example #1
Consider the slightly soluble compound barium oxalate,
BaC2O4.
The solubility of BaC2O4 is 1.3 x 10-3 mol/L.
The ratio of cations to anions is 1:1.
· This means that 1.3 x 10-3 moles of Ba2+ can dissolve
in one liter.
· Also, 1.3 x 10-3 moles of C2O42- can dissolve in one
liter.
· What is the maximum amount (in grams) of BaC2O4
that could dissolve in 2.5 L (before a solid precipitate
or solid settlement occurs)?
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Solubility
Solubility
What is the maximum amount (in grams) of
BaC2O4 that could dissolve in 2.5 L (before a
precipitate occurs)?
Example #2
Consider the slightly soluble compound lead (II)
chloride, PbCl2.
The solubility of BaC2O4 is 1.3 x 10-3 mol/L.
BaC2O4 --> Ba2+ + C2O42-
1 mol BaC2O4
-------------------- =
1.3 x 10-3 mol BaC2O4
The solubility of PbCl2 is 0.016 mol/L.
225.3g
----------xg
The ratio of cations to anions is 1:2.
= 0.293g BaC2O4 x 2.5L = 0.73g
1L
· This means that 0.016 moles of Pb2+ can dissolve in
one liter. Molar solubility always refers to the ion
with the lower molar ratio.
This is the maximum amount that could dissolve in
2.5 L before a precipitate occurs.
2.5 L
1
· Twice as much, or 2(0.016) = 0.032 moles of Cl- can
dissolve in one liter.
-3
1.3 x 10 mol BaC2O4
x
1 L
x
225.3 g BaC2O4
1 mol BaC2O4
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Solubility
Example #3
Consider the slightly soluble compound silver sulfate,
Ag2SO4.
The solubility of Ag2SO4 is 0.015 mol/L.
The ratio of cations to anions is 2:1.
2-
· This means that 0.015 moles of SO4 can dissolve
in one liter.
· Twice as much, or 2(0.015) = 0.030 moles of Ag+ can
dissolve in one liter.
Solubility
Remember that molar solubility refers to the ion with
the lower mole ratio. It does not always refer to the
cation, although in most cases it does.
Molar
Compound Solubility of [Cation]
Compound
1.3 x 10-3
mol
[Anion]
BaC2O4
1.3 x 10-3
mol
1.3 x 10-3
mol
PbCl2
0.016 mol/L 0.016 mol/L 0.032 mol/L
Ag2SO4
0.015 mol/L 0.030 mol/L 0.015 mol/L
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If the solubility of barium carbonate, BaCO3 is 7.1 x 10-5 M,
this means that a maximum of _______barium ions, Ba2+
ions can be dissolved per liter of solution.
3
4
If the solubility of barium carbonate, BaCO3 is 7.1 x 10-5 M,
this means that a maximum of _______carbonate ions,
CO32- ions can be dissolved per liter of solution.
A
7.1 x 10-5 moles
A
7.1 x 10 -5 moles
B
half of that
B
half of that
C twice as much
C
twice as much
D one-third as much
D
one-third as much
E
E
one-fourth as much
one-fourth as much
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5
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If the solubility of Ag2CrO4 is 6.5 x 10-5 M, this means
that a maximum of _______chromate ions, CrO42-, can
be dissolved per liter of solution.
6
If the solubility of Ag2CrO4 is 6.5 x 10-5 M, this means
that a maximum of _______ Ag+ ions can be dissolved
per liter of solution.
A
6.5 x 10-5 moles
A
6.5 x 10-5 moles
B
twice 6.5 x 10-5 moles
B
twice 6.5 x 10-5 moles
C
half 6.5 x 10-5 moles
C
half 6.5 x 10-5 moles
D
-5
D
one-fourth 6.5 x 10-5 moles
E
four times 6.5 x 10-5 moles
E
one-fourth 6.5 x 10 moles
-5
four times 6.5 x 10 moles
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7
Calculating Ksp from the Solubility
Sample Problem
The molar solubility of lead (II) bromide, PbBr2 is 1.0 x
10-2 at 25oC. Calculate the solubility product, Ksp, for
this compound.
[Pb2+] = 1.0 x 10-2 mol/L
[Br-] = 2.0 x 10-2 mol/L
= (1.0 x 10-2)(2.0 x 10-2)2
B 1/2 (3 x 10-8)
C (3 x 10-8)^1/2
D 2 (3 x 10-8)
Substitute the molar concentrations into the Ksp expression
and solve.
Ksp = [Pb2+][Br-]2
For the slightly soluble compound, AB, the molar
solubility is 3 x 10-8 moles per liter. Calculate the
Ksp for this compound. No calculator.
AB <--> A + + BA 3 x 10-8
E
(3 x 10-8)^2
= 4.0 x 10-6
AgCl < - - > Ag+ + Cl-
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8
For the slightly soluble compound, XY, the molar
solubility is 5 x 10-5 M. Calculate the Ksp for this
compound. No calculator.
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9
For the slightly soluble compound, MN, the molar
solubility is 4 x 10-6 M. Calculate the Ksp for this
compound. No calculator.
MN <--> M+ + N-
XY <--> X + + Y-
A 4 x 10-6
A 5 x 10-5
B 10 x 10-5
C 25 x 10
B 16 x 10-6
-5
C 16 x 10-12
D 5 x 10-10
E
D 16 x 10-36
25 x 10-10
BaCO3 < - - > Ba2+ + CO32-
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10
For the slightly soluble compound, AB2, the molar
solubility is 3 x 10-4 M. Calculate the solubilityproduct constant for this compound.
No calculator.
AB2 <--> A2+ + 2BA 9 x 10-4
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11
For the slightly soluble compound, X3Y, the molar
solubility is 1 x 10-4 M. Calculate the solubility
product for this compound.
No calculator.
X3Y <-> 3X+ + Y3A 3 x 10-4
B 9 x 10-8
B 3 x 10-8
C 18 x 10-8
C 27 x 10-12
D 36 x 10-8
D 27 x 10-16
E
108 x 10-12
Fe(OH)3 < - - > Fe3+ + 3(OH)Na3P <--> 3Na+ + P3-
PbCl2 < - - > Pb2+ + 2Cl-
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Calculating Solubility from the Ksp
Sample Problem
Calculate the solubility of CaF2 in grams per liter in
a) pure water
b) a 0.15 M KF solution
c) a 0.080 M Ca(NO3)2 solution
The solubility product for calcium fluoride, CaF2 is 3.9 x 10-11
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Calculating Solubility from the Ksp
a) pure water
CaF2 < - - > Ca2+ + 2FIf we assume x as the dissociation then,
Ca2+ ions = x
and [F-] = 2x
Ksp = [Ca2+] [F-]2
= (x)(2x)2
Ksp = 3.9 x 10-11 = 4x3
So x = 2.13 x 10-4 mol/L x (78 g/mol CaF2)
Solubility is 0.0167 g/L
2.13 x10-4 Ca 2+ mol/L x 1mol/L CaF2
78g/L
--------------------- x ------------1mol/L Ca2+ ions
1 mol CaF2
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Calculating Solubility from the Ksp
Calculating Solubility from the Ksp
b) a 0.15 M KF solution
remember KF is a strong electrolyte, is completely ionized.
the major source of F- ions, then [F-] =0.15M
The solubility product for calcium fluoride, CaF2 is 3.9 x 10-11
The solubility product for calcium fluoride,CaF2 is 3.9 x 10-11
[ F-] = 0.15M
2+
Calculate the solubility of CaF2 in grams per liter in
c) a 0.080 M Ca(NO3)2 solution
[Ca2+ ] = 0.08M
- 2
Ksp = [Ca ] [F ]
Ksp = 3.9 x 10
-11
= (x)(0.15)
2
Ca2+
CaF2 (s) <--->
(aq)
+
2 F- (aq)
= 0.0225x
Ksp = [Ca2+] [F-]2
So x = ______ mol/L
Solubility is = ______ x (78 g/mol CaF2) = ______ g/L
1.73 x10-9 Ca 2+ mol/L x 1mol/L CaF2
78g/L
--------------------- x ------------1mol/L Ca2+ ions
1 mol CaF2
= (0.080)(x)2
Ksp = 3.9 x 10-11 = 0.080x2
So x = 2.2 x 10-5 mol/L * (78 g/mol CaF2)/ 2
Solubility is 0.000858 g/L
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Calculating Solubility from the Ksp
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12
Recall from the Common-Ion Effect that adding a
strong electrolyte to a weakly soluble solution with a
common ion will decrease the solubility of the weak
electrolyte.
A 0.5 (1 x 10-16)
B
Solubility of CaF2
pure water
0.016 g/L
2 (1 x 10-16)
C (1 x 10-16)2
Compare the solubilities from the previous Sample Problem
CaF2 (s) <---> Ca2+ (aq)
+ 2 F- (aq)
CaF2 dissolved with:
Calculate the concentration of silver ion when the
solubility product constant of AgI is 10-16.
D
(1 x 10-16)
0.015 M KF
0.080 M Ca(NO3)2
0.0017 g/L
These results support Le Chatelier's Principle that increasing
a product concentration will shift equilibrium to the left.
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13
The Ksp of a compound of formula AB3 is 1.8 x 10-18.
The molar solubility of the compound is ----
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14
The Ksp of a compound of formula AB3 is 1.8 x 10-18.
The solubility of the compound is ---The molar mass is 210g/mol
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15
The concentration of hydroxide ions in a saturated
solution of Al(OH)3 is 1.58 x10-15. The concentration of
Al 3+ is -------
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16
The concentration of hydroxide ions in a saturated
solution of Al(OH)3 is 1.58 x10-15. The ksp of Al (OH)3
is-----
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Factors Affecting Solubility
· Recall The Common-Ion Effect
· If one of the ions in a solution equilibrium is already
dissolved in the solution, the equilibrium will shift to
the left and the solubility of the salt will decrease.
BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)
· So adding any soluble salt containing either
Ba2+ or SO42- ions will decrease the solubility
of barium sulfate.
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Factors Affecting Solubility
Write the dissociation equation for magnesium hydroxide:
Suppose you have a saturated solution of magnesium
hydroxide in a flask. What will happen if you add a small
amount of strong acid to it? (Think Le Châtelier’s
Principle.)
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Factors Affecting Solubility
· The Common-Ion Effect applies when a basic anion
is involved, such as conjugate base of a weak acid.
· Consider the dissociation of the salt calcium fluoride:
CaF2
<-->
______ + ______
What do you expect will happen to equilibrium point if
the pH of this system is lowered by adding a strong
acid?
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Factors Affecting Solubility
Main idea: Adding an acid, or otherwise lowering the
pH of a solution will _______________________ the
solubility of a salt containing a
___________________________________.
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Factors Affecting Solubility
· pH
· If a substance has a
basic anion, it will be
more soluble in an acidic
solution.
· Substances with acidic
cations are more soluble
in basic solutions.
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Factors Affecting Solubility
· Complex Ions
· Metal ions can act as Lewis acids and form
complex ions with Lewis bases in the solvent.
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Factors Affecting Solubility
· Amphoterism
· Amphoteric metal oxides
and hydroxides are soluble
in strong acid or base,
because they can act
either as acids or bases.
· Examples of such
cations are Al3+, Zn2+, and
Sn2+.
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17
Given the system at equilibrium
AgCl (s) <--> Ag+ (aq) + Cl- (aq)
When 0.01M HCl is added to the sytem, the point of
equilibrium will shift to the ________.
A right and the concentration of Ag+ will decrese
B
right and the concentration of Ag+ will increase
C left and the concentration of Ag+ will decrease
D left and the concentration of Ag+ will increase
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Factors Affecting Solubility
· Complex Ions
· The formation of
these complex ions
increases the
solubility of these
salts.
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Precipitation Problems
Solubility rules to memmorize by this time. Will help
you to identify which will precipitate as you mix two
solutions.
· Any salt made with a Group I metal is soluble.
· All salts containing nitrate ion are soluble.
· All salts containing ammonium ion are soluble.
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18
What is the name of the solid precipitate that is formed
when a solution of sodium chloride is mixed with a
solution of silver nitrate?
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19
A potassium bromide
A sodium silver
B
sodium nitrate
B
calcium carbonate
C chloride nitrate
C potassium calcium
D silver chloride
D carbonate bromide
E
Not enough information
E
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20
What is the name of the solid precipitate that is formed
when a solution of potassium carbonate is mixed with a
solution of calcium bromide?
What is the name of the solid precipitate that is formed
when a solution of lead (IV) nitrate is mixed with a
solution of magnesium sulfate?
A PbSO4
B
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Precipitation Problems
Most of the problems in this section ask you whether a
precipitate will form after mixing certain solutions together.
General Problem-Solving Strategy
Step 1 - Determine which of the products is the precipitate.
Write the Ksp expression for this compound.
Pb(SO4)2
C Pb2SO4
Step 2 - Calculate the cation concentration of this slightly
soluble compound.
D Mg(NO3)2
E
Not enough information
Not enough information
Step 3 - Calculate the anion concentration of this slightly
soluble compound.
Step 4 - Substitute the values into the reaction quotient (Q)
expression. Recall that this is the same expression as K.
Step 5 - Compare Q to K to determine whether a precipitate
will form.
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Precipitation Problems
Analyzing quantitative results:
If Q = Ksp
If Q > Ksp
If Q < Ksp
then you have an
exactly perfect
saturated solution with
not one speck of
undissolved solid.
then YES you will
observe a precipitate;
the number of cations
and anions exceeds
the solubility
then NO precipitate
will form; there are so
few cations and
anions that they all
remain dissolved
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21
The Ksp for Zn(OH)2 is 5.0 x10-17. Will a precipitate
form in a solution whose solubility is 8.0x10-2
mol/L Zn(OH)2?
A yes, because Qsp < Ksp
B yes, because Qsp > Ksp
C no, because Qsp = Ksp
D no, because Qsp < Ksp
In a solution,
· If Q = Ksp, the system is at equilibrium and the
solution is saturated.
· If Q > Ksp, the salt will precipitate until Q = Ksp.
· If Q < Ksp, more solid can dissolve until Q = Ksp.
E no, because Qsp > Ksp
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Precipitation Problems
Sample Problem
Will a precipitate form if you mix 50.0 mL of 0.20 M
barium chloride, BaCl2, and 50.0 mL of 0.30 M sodium
sulfate, Na2SO4?
· Qualitatively, you know that ____________ will be
the solid that would theoretically form.
· Quantitatively, however, the question is, are there
enough _________________ ions and
____________________ ions to produce a solid
precipitate, or will there be so few of them that they will
all dissolve?
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Sample Problem - Answers
Will a precipitate form if you mix 50.0 mL of 0.20 M barium
chloride, BaCl2, and 50.0 mL of 0.30 M sodium sulfate, Na2SO4?
Step 1 - Determine which of the products is the precipitate. Write
the Ksp expression for this compound.
BaSO4 (s) <--> Ba2+ (aq) + SO42- (aq)
Step 2 - Calculate the cation concentration of this slightly soluble
compound.
M1V1 =M2V2
Step 3 - Calculate the anion concentration of this slightly soluble
compound.
M1V1 =M2V2
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M2 = (M1V1) / V2
M2= (0.20M*50.0mL) / 100 mL
M2 = 0.10 M BaCl2
[Ba2+] = 0.10 M
M2 = (M1V1) / V2
M2= (0.30M*50.0mL) / 100 mL
M2 = 0.15 M Na2SO4
[SO42-] = 0.15 M
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The Ksp for zinc carbonate is 1 x 10-10.
If equivalent amounts of 0.1M sodium carbonate and
0.1M zinc nitrate are mixed, what happens?
22
Sample Problem - Answers (con't)
Will a precipitate form if you mix 50.0 mL of 0.20 M barium
chloride, BaCl2, and 50.0 mL of 0.30 M sodium sulfate, Na2SO4?
A A zinc carbonate precipitate forms, since Q>K.
B
A zinc carbonate precipitate forms, since Q<K.
C A sodium nitrate precipitate forms, since Q>K.
D No precipitate forms, since Q=K.
Step 4 - Substitute the values into the reaction quotient (Q)
expression. Recall that this is the same expression as K.
Ksp = [Ba2+] [SO42-] = (0.10) (0.15) = 0.015
Step 5 - Compare Q to K to determine whether a precipitate
will form.
The Ksp for barium sulfate is 1 x 10-10.
Therefore, since Q > K, there will be a precipitate formed
when you mix equal amounts of 0.20 M BaCl2, and 0.30 M
Na2SO4.
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23
Which of the following factors affect solubility?
A
pH
B
Concentration
C
Common-Ion Effect
D
A and C
E
A, B, and C
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Selective Precipitation of Ions
One can use
differences in
solubilities of salts to
separate ions in a
mixture.
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HW question
http://www.ktf-split.hr/periodni/en/abc/kpt.html
To a solution containing 2.0 x 10-5 M Barium ions and
1.8 x 10-4 M lead ions, Na2CrO4 is added.
Which would precipitate first from this solution?
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