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AP Chemistry
Aqueous Equilibria II:
Ksp & Solubility Products
Table of Contents: Ksp & Solubility Products
Click on the topic to go to that section
· Introduction to Solubility Equilibria
· Calculating Ksp from the Solubility
· Calculating Solubility from Ksp
· Factors Affecting Solubility
· Precipitation Reactions and Separation of Ions
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Introduction to
Solubility Equilibria
Return to the
Table of Contents
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Introduction to Solubility Equilibria
Many shells are made of relatively insoluble calcium carbonate, so
the shells are not at huge risk of dissolving in the ocean.
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Introduction to Solubility Equilibria
A saturated solution of CaCO3(s)
Ca2+
CO32-
Ca2+
CO32-
CaCO3(s)
Calcium carbonate is a relatively insoluble ionic salt. Would the
picture look different for a soluble ionic salt such as Na2CO3?
Which solution would be the better electrolyte?
Answer
Ionic compounds dissociate into their ions to different degrees when
placed in water and reach equilibrium with the non-dissociated solid
phase when the solution is saturated.
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Introduction to Solubility Equilibria
Consider the equilibrium that exists in a saturated
solution of CaCO3 in water:
CaCO3 (s) ↔ Ca2+ (aq) + CO3 2- (aq)
Unlike acid-base equilibria which are homogenous,
solubility equilibria are heterogeneous, there is
always a solid in the reaction.
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Introduction to Solubility Equilibria
The equilibrium constant expression for this
equilibrium is
Ksp = [Ca2+ ] [CO 3 2− ]
where the equilibrium constant, Ksp , is called the
solubility product.
There is never any denominator in Ksp expressions
because pure solids are not included in any
equilibrium expressions.
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Solubility Equilibrium
CaCO3(s) --> Ca2+(aq) + CO32-(aq)
Ksp @ 25 C = 5.0 x 10-9
MgCO3(s) --> Mg2+(aq) + CO32-(aq)
Ksp @ 25 C = 6.8 x 10-6
In both cases above, the equilibrium lies far to the left, meaning
relatively few aqueous ions would be present in solution.
Which saturated solution above would have the higher
conductivity and why?
Answer
The degree to which an ionic compound dissociates in water can be
determined by measuring it's "Ksp" or solubility product equilibrium
constant.
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.
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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, K sp , 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
3 Which Ksp expression is correct for Fe3(PO4)2?
A
[Fe2+ ]3[PO 4 3- ]2
B
[Fe2+]3/[PO43-]2
C
[Fe3+]2 [PO43-]2
D
[Fe2+]2 /[PO43-]2
E
None of the above.
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4 When 30 grams of NaCl are mixed into 100 mL of
distilled water all of the solid NaCl dissolves. The solution
must be saturated and the Ksp for the NaCl must be very
high.
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True
False
5 The conductivity of a saturated solution of Ag2CO3 would
be expected to be less than the conductivity of a
saturated solution of CaCO3. Justify your answer.
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True
False
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Solubility
The term solubility represents the maximum amountof 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 asmolar solubility. For any slightly
soluble salt the molar solubilityalways refers to the ion with the
lower molar ratio.
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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 BaC
2 O4 that could
dissolve in 2.5 L (before a solid precipitate or solid settlement
occurs)?
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Solubility
Example #1
What is the maximum amount (in grams) of BaC2O4 that could
dissolve in 2.5 L (before a precipitate occurs)?
The solubility of BaC2O4 is 1.3 x 10-3 mol/L.
BaC2O4 (s) --> Ba2+ (aq) + C2O42- (aq)
1.3 x 10-3 mol BaC2O4
2.5L x -------------------1 liter
3.25 x 10- 3 g x
BaC2O4
1 mole
225.3 g
=
3.25 x 10 - 3 g
BaC 2O4
= 0.73g BaC2O4
0.73g is the maximum amount of BaC2O4 that could dissolve
in 2.5 L before a precipitate forms.
Solubility
Example #2
Consider the slightly soluble compound lead chloride,
PbCl2 .
The solubility of PbCl2 is 0.016 mol/L.
The ratio of cations to anions is 1:2.
This means that 0.016 moles of Pb 2+ can dissolve
in one liter.
Twice as much, or 2(0.016) = 0.032 moles of Cl - can
dissolve in one liter.
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Solubility
Example #3
Consider the slightly soluble compound silver sulfate,
Ag2 SO4 .
The solubility of Ag 2 SO4 is 0.015 mol/L.
The ratio of cations to anions is 2:1.
This means that 0.015 moles of SO 4 2- can dissolve
in one liter.
Twice as much, or 2(0.015) = 0.030 moles of Ag + can
dissolve in one liter.
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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
[Anion]
BaC2O4
1.3 x 10-3
mol
1.3 x 10-3
mol
PbCl 2
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
1.3 x 10-3
mol
6 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.
A
7.1 x 10 -5 moles
B
half of that
C
twice as much
D
one-third as much
E
one-fourth as much
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If the solubility of barium carbonate, BaCO3 is 7.1 x 10-5 M,
7 this means that a maximum of _______carbonate ions,
CO32- ions can be dissolved per liter of solution.
A
7.1 x 10-5 moles
B
half of that
C
twice as much
D
one-third as much
E
one-fourth as much
If the solubility of Ag2CrO4 is 6.5 x 10-5 M, this means
8 that a maximum of _______silver ions, Ag +, can be
dissolved per liter of solution.
A
6.5 x 10-5 moles
B
twice 6.5 x 10-5 moles
C
half 6.5 x 10-5 moles
D
one-fourth 6.5 x 10-5 moles
E
four times 6.5 x 10-5 moles
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Calculating Ksp from
the Solubility
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Table of Contents
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Calculating Ksp from the Solubility
Sample Problem
The molar solubility of lead (II) bromide, PbBr2 is 1.0 x 10 -2 at
25 o C. Calculate the solubility product, K sp , for this compound.
The molar solubility always refers to the ion of the lower molar
ratio, therefore [Pb2+ ] = 1.0 x 10-2 mol/L and
[Br-] = 2.0 x 10-2 mol/L
Substitute the molar concentrations into the K sp expression and solve.
Ksp = [Pb2+ ][Br -]2
= (1.0 x 10-2 )(2.0 x 10-2 )2
= 4.0 x 10 -6
9 For the slightly soluble salt, CoS, the molar
solubility is 5 x 10-5 M. Calculate the Ksp for this
compound.
A
5 x 10 -5
B
1.0 x 10 -4
C
2.5 x 10 -4
D
5 x 10 -10
E
2.5 x 10 -9
10 For the slightly soluble salt, BaF 2 , the molar solubility
is 3 x 10-4 M. Calculate the solubility-product
constant for this compound.
A
9 x 10 -4
B
9 x 10 -8
C
1.8 x 10 -7
D
3.6 x 10 -7
E
1.08 x 10 -10
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11 For the slightly soluble salt, La(IO3)3, the molar
solubility is 1 x 10-4 M. Calculate Ksp.
A
3 x 10 -12
B
3 x 10 -16
C
2.7 x 10 -11
D
2.7 x 10 -15
E
1 x 10 -12
12 For the slightly soluble compound, Ca3(PO4)2, the
molar solubility is 3 x 10-8 moles per liter. Calculate
the Ksp for this compound.
A
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9.00 x 10 -16
B 1.08 x 10-38
C 8.20 x 10-32
D 1.35 x 10-13
E 3.0 x 10-20
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13 The concentration of hydroxide ions in a saturated
solution of Al(OH)3 is 1.58x10-15. What is the Ksp of
Al(OH)3?
14 What is the Ksp of Fe(OH)3(s) if a saturated solution of it
has a pH of 11.3?
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A 2.0 x 10-12
B 1.6 x 10-15
C 2.1 x 10-46
D 1.4 x 10-8
E 5.4 x 10-16
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Calculating Solubility
from the Ksp
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Table of Contents
Calculating Solubility from the Ksp
Example: What is the molar solubility of a saturated aqueous
solution of BaCO3? (Ksp @25 C = 5.0 x 10-9)
BaCO3(s) --> Ba2+(aq) + CO32-(aq)
Ksp = 5.0 x 10-9 = [Ba2+][CO32-]
Since neither ion concentration is known, we will substitute "x"
for the [Ba2+] and "x" for the [CO32-].
5.0 x 10-9 = (x)(x) = x2
"x" = [Ba2+] = [CO32-] = 7.07 x 10-5 M
Since 1 Ba or 1 CO32- are required for 1 BaCO3, the molar
solubility of the BaCO3(s) = 7.07 x 10-5 M.
2+
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Calculating Solubility from the Ksp
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Example: What is the molar solubility of a saturated aqueous
solution of PbI2? (Ksp @25 C = 1.39 x 10-8)
PbI2(s) --> Pb2+(aq) + 2I-(aq)
Ksp = 1.39 x 10-8 = [Pb2+][I-]2
Since neither ion concentration is known, we will substitute "x"
for the [Pb2+] and "2x" for the [I-].
1.39 x 10-8 = (x)(2x)2 = 4x3
"x" = [Pb2+] = 1.51 x 10-3 M
Since 1 Pb required 1 PbI2, the molar solubility of the PbI2(s) =
1.51 x 10-3 M.
2+
15 Calculate the concentration of silver ion when the
solubility product constant of AgI is 1 x 10-16 .
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A 0.5 (1 x 10-16)
B 2 (1 x 10-16)
C
(1 x 10-16)2
D
(1 x 10-16)
16 Calculate the molar solubility of PbF2 that has a Ksp at
25℃ = 3.6 x 10-6.
Students type their answers here
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17 The Ksp of a compound of formula AB3 is 1.8 x 10 -18.
What is the molar solubility of the compound?
18 The Ksp of a compound of formula AB3 is 1.8 x 10 -18.
The molar mass is 280g/mol. What is the solubility?
19 Which of the following ionic salts would have the highest
molar solubility?
A NiCO3(s) Ksp = 6.61 x 10-9
B MnCO3(s) Ksp = 1.82 x 10-11
C ZnCO3(s) Ksp = 1.45 x 10-11
D Ag2CrO4(s) Ksp = 9.00 x 10-12
E All have the same molar solubility
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Factors Affecting Solubility
Return to the Table
of Contents
Common Ion Effect
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Consider a saturate solution of barium sulfate:
BaSO4 (s)
Ba2+ (aq) + SO4 2- (aq)
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.
So adding any soluble salt containing either Ba2+ or
SO4 2- ions will decrease the solubility of barium sulfate.
Common Ion Effect
Sample Problem
Calculate the solubility of CaF 2 in grams per liter in
a) pure water
b) a 0.15 M KF solution
c) a 0.080 M Ca(NO 3 )2 solution
The solubility product for calcium fluoride, CaF2 is 3.9 x 10-11
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Common Ion Effect
a) pure water
Ca 2+ (aq) + 2F - (aq)
If we assume x as the dissociation then,
Ca2+ ions = x and [F-] = 2x
Ksp = [Ca2+ ] [F- ]2
= (x)(2x)2
Note
CaF2(s)
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
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Common Ion Effect
b) a 0.15 M KF solution
Remember KF, a strong electrolyte, is completely ionized and the major
source of F- ions. [F- ] =0.15M
Note
The solubility product for calcium fluoride, CaF2 is 3.9 x 10-11
[ F- ] = 0.15M
Ksp = [Ca2+ ] [F- ]2
= (x)(0.15)2
Ksp = 3.9 x 10-11 = 0.0225x
So x = ______ mol/L
Solubility is = ______ x (78 g/mol CaF2) = ______ g/L
Common Ion Effect
Calculate the solubility of CaF2 in grams per liter in
c) a 0.080 M Ca(NO3 )2 solution
[Ca2+ ] = 0.08M
The solubility product for calcium fluoride,CaF2 is 3.9 x 10 -11
CaF 2 (s)
Ca2+
(aq)
+
2 F - (aq)
Ksp = [Ca2+ ] [F-] 2 = (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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Common Ion Effect
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Recall from the Common-Ion Effect that adding a strong electrolyte
to a weakly soluble solution with acommon ion will decrease the
solubility of the weakelectrolyte.
Compare the solubilities from the previous Sample Problem
CaF2 (s)
Ca2+ (aq)
+ 2 F- (aq)
CaF2 dissolved with:
Solubility of CaF2
pure water
0.016 g/L
0.015 M KF
1.35x10-7 g/L
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.
20 What is the molar solubility of a saturated solution of
Ag2CrO4? Ksp at 25℃ is = 1.2 x 10-12.
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#
A 1.1 x 10-4
B 6.7 x 10-5
C 8.4 x 10-5
D 5.5 x 10-7
E 2.2 x 10-8
21 What is the molar solubility of a saturated solution of
Ag2CrO4 in 0.100M K2CrO4? Ksp at 25℃ is = 1.2 x 10-12.
A 3.0 x 10-12
B 6.3 x 10-5
C 5.1 x 10-8
D 3.5 x 10-7
E 1.7 x 10-6
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22 What is the molar solubility of a saturated solution of
Ag2CrO4 in 0.200M AgCl? Ksp at 25℃ is = 1.2 x 10-12.
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A 3.0 x 10 -12
B 6.3 x 10-5
C 3.11 x 10-11
D 3.5 x 10-7
E 6.7 x 10-6
Changes in pH
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The solubility of almost any ionic compound is affected by
changes in pH.
Consider dissociation equation for magnesium hydroxide:
Mg(OH)(s) # Mg2+(aq) + 2OH-(aq)
What do you expect will happen to the equilibrium if
the pH of this system is lowered by adding a strong acid?
Will one of the substances in the equilibrium interact with the
strong acid?
Would the Mg(OH)2 be more or less soluble?
(Think Le Châtelier’s Principle.)
Changes in pH
Changes in pH can also affect the solubility of salts that contain
the conjugate base of a weak acid.
Consider the dissociation of the salt calcium fluoride:
CaF2 (s)
Ca 2+(aq) + 2F -(aq)
What do you expect will happen to the equilibrium if
the pH of this system is lowered by adding a strong acid?
Will one of the substances in the equilibrium interact with the
strong acid?
Would the CaF 2 be more or less soluble?
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Changes in pH
Sample Problem
Calculate the molar solubility of Mn(OH) 2 in
a) in a solution buffered at pH=9.5
b) in a solution buffered at pH=8.0
c) pure water
The solubility product for Mn(OH)2 at 25℃ is 1.6 x 10-13 .
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Changes in pH
Sample Problem
Calculate the molar solubility of Mn(OH)2 in
a) in a solution buffered at pH = 9.5
In a solution buffered at pH=9.5, the [H+] = 3.2 x 10-10,
the [OH-] = 3.2x 10-5.
The solubility product for Mn(OH)2, is 1.6 x 10-13
[ OH- ] = 3.2 x 10-5M
1.6 x 10 -13 = [Mn2+ ] [OH- ]2 = (x)(3.2 x 10-5)2
x = 1.6 x 10-13 /(3.2 x 10-5)2 = 1.56 x 10 -4 mol/L
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Changes in pH
Sample Problem
Calculate the molar solubility of Mn(OH)2 in
b) in a solution buffered at pH = 8.0
In a solution buffered at pH=8.0, the [H+] = 1 x 10-8, the [OH-] = 1x 10-6.
[ OH- ] = 1 x 10-6M
1.6 x 10 -13 = [Mn2+ ] [OH- ]2 = (x)(1 x 10 -6)2
x = 1.6 x 10-13 / (1 x 10 -6)2 =
So x = 0.16 mol/L
Note
The solubility product for Mn(OH)2, is 1.6 x 10-13
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Changes in pH
Sample Problem
Calculate the molar solubility of Mn(OH)2 in
c) in pure water
In pure water the pH=7.0, the [H+] = 1 x 10-7, the [OH-] = 1x 10-7.
Note
The solubility product for Mn(OH)2, is 1.6 x 10-13
[ OH- ] = 1 x 10-7M
1.6 x 10 -13 = [Mn2+ ] [OH- ]2 = (x)(1 x 10-7)2
x = 1.6 x 10-13 / (1 x 10-7)2 =
So x = 16 mol/liter
Changes in pH
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If a substance has a basic anion, it will be more soluble in an
acidic solution.
If a substance has an acidic cation, it will be more soluble in
basic solutions.
We will discuss in a little while the affect of pH changes on
substances that are amphoteric.
Do you remember what it means when a substance is
amphoteric?
23 Given the system at equilibrium
AgCl (s)
Ag+ (aq) + Cl- (aq)
When 0.01 M 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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24 Which of the following substances are more soluble in
acidic solution than in basic solution? Select all that
apply.
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A PbCl2
B BaCO3
C AgI
D Fe(OH)3
E MgF2
25 What is the solubility of Zn(OH)2 in a solution that is
buffered at pH = 8.5? Ksp = 3.0 x 10-16
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Students type their answers here
26 Will the solubility of Zn(OH)2 in a solution that is buffered
at pH = 11.0 be greater than in a solution buffered at
8.5? Explain.
A Yes
B No
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27 The molar solubility of NH4Cl increases as pH
_________ .
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A increases
B decreases
C is unaffected by changes in pH
28 The molar solubility of Na2CO3 increases as pH
_________ .
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A increases
B decreases
C is unaffected by changes in pH
Complex Ions
Metal ions can act as Lewis acids and form complex ions
with Lewis bases in the solvent. The formation of
complex ions particularly with transitional metals can
dramatically affect the solubility of a metal salt.
For example, the addition of excess ammonia to AgCl will cause
the AgCl to dissolve. This process is the sum of two reactions
resulting in:
AgCl(s) + 2NH3(aq)
Ag(NH3)2+(aq) + Cl- (aq)
Added NH3 reacts with Ag+ forming
Ag(NH3)2+. Adding enough NH3
results in the complete dissolution
of AgCl.
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Amphoterism
Some metal oxides and hydroxides are soluble in strongly
acidic and in strongly basic solutions because they can act
either as acids or bases. These substances are said to be
amphotheric.
Examples of such these substances are oxides and
hydroxides of Al3+ , Zn 2+ , and Sn2+ .
They dissolve in acidic solutions because their anion is
protonated by the added H + and is pulled from solution
causing a shift in the equilibrium to the right. For example:
Al(OH)3(s) #
Al3+(aq) + 3 OH-(aq)
#
Amphoterism
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However these oxides and hydroxides also dissolve in
strongly basic solutions. This is because they form complex
ions containing several typically four hydroxides bound to
the metal ion.
Aluminum hydroxide reacts with OH
in the following reaction:
-
to form a complex ion
Al(OH) 3(s) + OH - (aq) # Al(OH)4- (aq)
As a result of the formation of the complex ion, Al(OH)4- , aluminum
hydroxide is more soluble.
Many metal hydroxides only react in strongly acidic solutions.
Ca(OH)2, Fe(OH)2 and Fe(OH)3 are only more soluble in acidic
solution they are not amphoteric.
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29 Which of the following factors affect solubility?
A
pH
B
Formation of Complex Ions
C
Common-Ion Effect
D
A and C
E
A, B, and C
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Precipitation Reactions
and Separation of Ions
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Table of Contents
Precipitation Reactions and Separation of Ions
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Do you remember the solubility rules?
They were useful before when we were trying to qualitatively
determine if a given reaction would produce a precipitate. They will
be useful now for the same reason however now we are going to
add a quantitative component that we will discuss soon.
In general, soluble salts were:
· Any salt made with a Group I metal is soluble.
· All salts containing nitrate ion are soluble.
· All salts containing ammonium ion are soluble.
Do you remember what metal cations tended to be insoluble?
Ag+, Pb2+, and Hg 2+
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30 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?
A sodium silver
B sodium nitrate
C
chloride nitrate
D
silver chloride
E Not enough information
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31 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?
A potassium bromide
B calcium carbonate
C
potassium calcium
D
carbonate bromide
E Not enough information
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32 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 Pb(SO4)2
C
Pb2SO4
D
Mg(NO3)2
E Not enough information
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33 The K sp 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
E no, because Qsp > Ksp
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34 The Ksp for zinc carbonate is 1 x 10-10 .
If equivalent amounts 0.2M sodium carbonate and
0.1M zinc nitrate are mixed, what happens?
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.
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Separation of Ions
When metals are found in natural they are usually found as metal
ores. The metal contained in these ores are in the form of
insoluble salts. To make extraction even more difficult the ores
often contain several metal salts. In order to separate out the
metals, one can use differences in solubilities of salts to separate
ions in a mixture.
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Separation of Ions
Imagine, you have a test tube that contains Ag+,
Pb2+ and Cu2+ ions and you want to selectively remove
each ion and place them into separate test tubes.
What reagent could you add to the test tube that will
form a precipitate with one or move of the cations and
leave the others in solution? You can use your
knowledge of the solubility rules or Ksp values for
various metal salts to help you accomplish this goal.
Separation of Ions
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You should remember that Ag+ and Pb2+ readily form
insoluble salts and that Cu2+ does not form
insoluble salts as readily.
Looking at some solubility product values, you will
find the following:
Salt
Ksp
Ag2S
6 x 10-51
PbS
3 x 10-28
CuS
6 x 10-37
AgCl
1.8 x 10-10
PbCl2
1.7 x 10-5
You will notice that CuCl2 is not to be found.
This means CuCl2 is a soluble salt!
Separation of Ions
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Adding Cl- should precipitate the Ag+ and Pb2+ ions but
not the Cu2+ ions. We can remove Ag+ and Pb2+ from
the test tube. Now, how can we separate the Ag+ and Pb2+ ions?
Salt
Ksp
Ag2S
6 x 10
PbS
3 x 10-28
CuS
6 x 10-37
AgCl
1.8 x 10-10
PbCl2
1.7 x 10-5
-51
Do you notice the
significant difference
between the Ksp values for
Ag2S and PbS?
Maybe we can precipitate
one of the salts out before
the other if we control the
concentration of S2- added.
Which salt Ag2S and PbS should precipitate first
when we begin to add S2-?
Separation of Ions
If we have 0.100M concentrations of Ag+ and Pb2+
and we begin to add 0.200M K2S the Ag2S should
precipitate first.
For Ag2S:
Ksp = 6 x 10 -51 = [Ag+]2[S2-] = (0.100)2(x)
x = [S2-] = 6 x 10-49M. If this concentration of
S2- is added Ag2S will precipitate.
For PbS:
Ksp = 3 x 10 -28 = [Pb2+][S2-] = 0.100(x)
x = [S2-] = 3 x 10-27M. A greater amount of S2is needed to precipitate the PbS.
Therefore, Ag2S will precipitate first.
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Separation of Ions Problems
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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.
Step 2 - Calculate the cation concentration of this slightly
soluble compound.
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.
Separation of Ions Problems
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In order for a precipitate to form the equilibrium that
exists between the solution and the insoluble salt must
reside on the left. We can determine to which side the
equilibrium will shift using Q, the Reaction Quotient.
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
If Q < Ksp
then NO precipitate
will form; there are
so
few cations and
anions that they all
remain dissolved
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.
Separation of Ions 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?
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 M2 = (M1V1) / V2
M2= (0.20M*50.0mL) / 100 mL
M2 = 0.10 M BaCl2
[Ba2+] = 0.10 M
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Separation of Ions Problems
Slide 82 / 91
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?
Step 3 - Calculate the anion concentration of this slightly
soluble
compound.
M1V1 =M2V2 M2 = (M1V1) / V2
M2= (0.30M*50.0mL) / 100 mL
M2 = 0.15 M Na2SO4
[SO42-] = 0.15 M
Step 4 - Substitute the values into the reaction quotient (Q)
expression. Recall that this is the same expression as K.
Q = [Ba2+] [SO42-] = (0.10) (0.15) = 0.015
Separation of Ions Problems
Slide 83 / 91
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?
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.
Separation of Ions Problems
In summary, to selectively precipitate metal ions from a
solution that contains a number of metal ions you should
use the solubility rules and Ksp values to determine an
experimental strategy.
The solubility rules may lead you to the identity of an
anion that will result in separation of certain metal ions
however, at other times the quantity of the added anion
will be instrumental in the separation given that metal
salts have different degrees of solubility as seen in their
Ksp values.
Slide 84 / 91
35 A solution contains 2.0 x 10-5 M barium ions and
1.8 x 10-4 M lead (II) ions. If Na2CrO4 is added,
which will precipitate first from solution? The Ksp
for BaCrO4 is 2.1 x 10 -10 and the Ksp for PbCrO4 is
2.8 x 10-13.
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A BaCrO4
B PbCrO4
C They will precipitate at the same time.
D It's impossible to determine
with the information provided.
36 A solution contains 2.0 x 10-4 M Ag+ and 2.0 x 10-4 M Pb2+.
If NaCl is added, will AgCl (Ksp = 1.8 x 10 -10) or PbCI2
(Ksp = 1.7 x 10-5) precipitate first?
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A AgCl
B PbCl2
C They will precipitate at the same time.
D It is impossible to determine with
the information provided.
37 A solution contains 2.0 x 10-4 M Ag+ and 1.7 x 10-3 M Pb2+.
If NaCl is added. What concentration of Cl- is needed to
Students type their answers here
begin
precipitation. AgCl (Ksp = 1.8 x 10 -10) and PbCI2
(Ksp = 1.7 x 10-5)
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38 A solution contains 2.0 x 10-4 M Ag+ and 1.7 x 10-3 M Pb2+.
If NaCl is added. What will be the concentration of the
their answers here
first Students
ion totype
precipitate
when the second ion begins to
precipitate? AgCl (Ksp = 1.8 x 10 -10) and PbCI2 (Ksp =
1.7 x 10-5)
39 Will Co(OH)2 precipitate from solution if the pH of a
0.002M solution of Co(NO3)2 is adjusted to 8.4? Ksp for
Co(OH)2 is 2.5 x 10 -14.
Slide 88 / 91
Slide 89 / 91
A Yes
B No
40 Will a precipitate form if you mix 25.0 mL of 0.250 M
calcium chloride, and 50.0 mL of 0.155 M lithium
chromate? The Ksp for calcium chromate is 4.5 x 10-9.
Students type their answers here
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