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AP Chemistry
Aqueous Equilibria II:
Ksp & Solubility Products
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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
Introduction to
Solubility Equilibria
· Factors Affecting Solubility
Return to the
Table of Contents
· Precipitation Reactions and Separation of Ions
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Introduction to Solubility Equilibria
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Introduction to Solubility Equilibria
A saturated solution of CaCO3(s)
Ca2+
Many shells are made of relatively insoluble calcium carbonate, so
the shells are not at huge risk of dissolving in the ocean.
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
Introduction to Solubility Equilibria
Consider the equilibrium that exists in a saturated
solution of CaCO3 in water:
The equilibrium constant expression for this
equilibrium is
CaCO3 (s) ↔ Ca2+ (aq) + CO3 2- (aq)
Ksp = [Ca2+ ] [CO 3 2− ]
where the equilibrium constant, Ksp , is called the
solubility product.
Unlike acid-base equilibria which are homogenous,
solubility equilibria are heterogeneous, there is
always a solid in the reaction.
There is never any denominator in Ksp expressions
because pure solids are not included in any
equilibrium expressions.
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Solubility Equilibrium
1 Which Ksp expression is correct for AgCl?
CaCO3(s) --> Ca (aq) + CO3 (aq)
Ksp @ 25 C = 5.0 x 10
MgCO3(s) --> Mg2+(aq) + CO32-(aq)
Ksp @ 25 C = 6.8 x 10-6
2+
2-
-9
Answer
The degree to which an ionic compound dissociates in water can be
determined by measuring it's "Ksp" or solubility product equilibrium
constant.
A
[Ag+]/[Cl-]
B
[Ag+][Cl-]
C
[Ag2+]2[Cl2-]2
D
[Ag+]2[Cl-]2
E
None of the above.
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?
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1 Which Ksp expression is correct for AgCl?
[Ag+]/[Cl-]
+
-
B
[Ag ][Cl ]
C
[Ag2+]2[Cl2-]2
D
[Ag+]2[Cl-]2
E
B
Answer
A
None of the above.
AgCl(s) # Ag+(aq) + Cl-(aq)
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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?
Ksp = [Ag+][Cl-]
A Ksp= [Zn2+][OH-]2 / [Zn(OH)2]
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B
Ksp= [Zn(OH)2] / [Zn2+][2OH-]
C
Ksp= [Zn2+][2OH-]
D
Ksp = [Zn2+ ][OH-]2
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Answer
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?
D
Ksp= [Zn2+][OH-]2
A Ksp= [Zn2+][OH-]2 / [Zn(OH)2]
Ksp= [Zn(OH)2] / [Zn2+][2OH-]
B
C
Ksp= [Zn2+][2OH-]
D
Ksp = [Zn2+ ][OH-]2
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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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3 Which Ksp expression is correct for Fe3(PO4)2?
A
[Fe2+ ]3[PO 4 3- ]2
B
3- 2
[Fe2+]3/[POFe
4 ]3(PO4)2(s) # 3Fe2+(aq) + 2PO43-(aq)
C
[Fe3+]2 [PO43-]2 Ksp = [Fe2+]3[PO43-]2
D
[Fe2+]2 /[PO43-]2
E
None of the above.
Answer
A
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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.
True
False
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True
False
Answer
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.
False
The solution in this case is
unsaturated. It has the
capacity to dissolve more
salt.
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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.
True
False
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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.
Answer
True
False
False
For solutions of the same
concentration, Ag2CO3 would
dissociate into more ions so
therefore it would have a greater
conductivity
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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
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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
Solubility
Example #1
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)?
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.
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Solubility
Solubility
Example #2
Example #3
Consider the slightly soluble compound lead chloride,
PbCl2 .
Consider the slightly soluble compound silver sulfate,
Ag2 SO4 .
The solubility of PbCl2 is 0.016 mol/L.
The solubility of Ag 2 SO4 is 0.015 mol/L.
The ratio of cations to anions is 1:2.
The ratio of cations to anions is 2:1.
This means that 0.016 moles of Pb 2+ can dissolve
in one liter.
This means that 0.015 moles of SO 4 2- can dissolve
in one liter.
Twice as much, or 2(0.016) = 0.032 moles of Cl - 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
mol
1.3 x 10
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
-3
1.3 x 10
mol
-3
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.
-3
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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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 The ratio of ions is 1:1 the
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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.
Answer
A
7.1 x 10-5 moles
B
half of that
C
maximum amount of Ba2+ is
twice as much
7.1 x 10-5 moles per 1 liter.
C
twice as much
D
one-third as much
D
one-third as much
E
one-fourth 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.
7.1 x 10-5 moles
B
half of that
A
Answer
A
C
A
The ratio of ions is 1:1 the
maximum amount of CO32- is
twice as much
7.1 x 10-5 moles per 1 liter.
D
one-third as much
E
one-fourth as much
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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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If the solubility of Ag2CrO4 is 6.5 x 10-5 M, this means
8 that a maximum of _______silver ions, Ag +, can be
B
dissolved per liter of solution.
The ratio of ions is 2:1. The
6.5 x 10-5 moles molar solubility always refers
D
to the ion of the lowest mole
ratio, in this case CrO42-.
Therefore, the maximum
half 6.5 x 10-5 moles
amount of Ag+ is twice
-5
one-fourth 6.5 x 10
6.5 xmoles
10-5 moles per 1 liter.
E
four times 6.5 x 10-5 moles
B
C
Calculating Ksp from
the Solubility
Answer
A
twice 6.5 x 10-5 moles
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Table of Contents
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9 For the slightly soluble salt, CoS, the molar
solubility is 5 x 10-5 M. Calculate the Ksp for this
compound.
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 = [Pb ][Br ]
2+
- 2
= (1.0 x 10 )(2.0 x 10 )
-2
= 4.0 x 10
-2 2
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
-6
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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
Answer
9 For the slightly soluble salt, CoS, the molar
solubility is 5 x 10-5 M. Calculate the Ksp for this
compound.
E
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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
Ksp = (5 x 10-5 M)2
B
9 x 10 -8
Ksp = 2.5 x 10
C
1.8 x 10 -7
D
3.6 x 10 -7
E
1.08 x 10 -10
Ksp = [Co2+][S2-]
-9
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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
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11 For the slightly soluble salt, La(IO3)3, the molar
solubility is 1 x 10-4 M. Calculate Ksp.
E
9 x 10 -4
Ksp = [Ba ][F-]2
2+
A
3 x 10 -12
B
3 x 10 -16
C
2.7 x 10 -11
9 x 10 -8
C
1.8 x 10 -7
D
3.6 x 10 -7 Ksp = 1.08 x 10-10
D
2.7 x 10 -15
E
1.08 x 10 -10
E
1 x 10 -12
Answer
B
[Ba2+] = 3 x 10-4, [F-] = 6 x 10-4
Ksp = (3 x 10 )(6 x 10 )
-4
-4 2
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11 For the slightly soluble salt, La(IO3)3, the molar
solubility is 1 x 10-4 M. Calculate Ksp.
3 x 10 -12
B
3 x 10 -16
Ksp = [La3+][IO3-]3
C
2.7 x 10 -11
D
2.7 x 10Ksp = 3 x 10
E
1 x 10 -12
Answer
A
B
[La3+] = 1 x 10-4, [IO3-] = 3(1x 10-4)
Ksp = (1 x 10-4)(3 x 10-4)3
-15
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C
Ksp = [Ca ] [PO43-]2
Answer
B 1.08 x 10
-38
C 8.20 x 10-32
2+ 3
[PO43-] = 3 x 10-8
The ratio of Ca2+ to PO43- 3:2
[Ca 2+] = 3 x 10-8 x 3/2 = 4.5 x 10-6
D 1.35 x 10-13 Ksp = (4.5 x 10-6)3(3 x 10-8)2
E 3.0 x 10-20
A
9.00 x 10 -16
B 1.08 x 10-38
C 8.20 x 10-32
D 1.35 x 10-13
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.
9.00 x 10 -16
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.
-16
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A
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Ksp = 8.20 x 10-32
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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?
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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?
Al(OH)3(s) #
3+
14 What is the Ksp of Fe(OH)3(s) if a saturated solution of it
has a pH of 11.3?
Al3+ (aq) + 3OH - (aq)
- 3
Ksp = [Al ][OH ]
Answer
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The ratio of Al3+ to OH - is 1 to 3.
[Al3+ ] = (1/3) 1.58x x 10-15= 5.3 x 10 -16
Ksp= (5.3 x 10 -16)(1.58 x 10-15)3
A 2.0 x 10-12
B 1.6 x 10-15
C 2.1 x 10-46
Ksp = 2.1 x 10 -60
D 1.4 x 10-8
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E 5.4 x 10-16
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14 What is the Ksp of Fe(OH)3(s) if a saturated solution of it
has a pH of 11.3?
E
Fe(OH)3 (s) # Fe3+ (aq) + 3OH - (aq)
B 1.6 x 10
-15
Ksp = [Fe3+][OH-]3
pH = 11.3, pOH = 3.7
Answer
A 2.0 x 10-12
[OH-] = 10 -3.7 = 2.0 x 10-4
Calculating Solubility
from the Ksp
The ratio of [Fe3+] to [OH-] to = 1:3
[Fe3+] = 1/3 x 2.0 x 10-4 = 6.7 x 10-5
C 2.1 x 10-46
Ksp = (6.7 x 10-5)(2.0 x 10-4)3
Ksp = 5.4 x 10 -16
D 1.4 x 10-8
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Table of Contents
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E 5.4 x 10-16
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Calculating Solubility from the Ksp
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)
Example: What is the molar solubility of a saturated aqueous
solution of PbI2? (Ksp @25 C = 1.39 x 10-8)
BaCO3(s) --> Ba2+(aq) + CO32-(aq)
PbI2(s) --> Pb2+(aq) + 2I-(aq)
Ksp = 5.0 x 10-9 = [Ba2+][CO32-]
Ksp = 1.39 x 10-8 = [Pb2+][I-]2
Since neither ion concentration is known, we will substitute "x"
for the [Ba2+] and "x" for the [CO32-].
Since neither ion concentration is known, we will substitute "x"
for the [Pb2+] and "2x" for the [I-].
5.0 x 10-9 = (x)(x) = x2
1.39 x 10-8 = (x)(2x)2 = 4x3
"x" = [Ba2+] = [CO32-] = 7.07 x 10-5 M
"x" = [Pb2+] = 1.51 x 10-3 M
Since 1 Ba or 1 CO3 are required for 1 BaCO3, the molar
solubility of the BaCO3(s) = 7.07 x 10-5 M.
Since 1 Pb required 1 PbI2, the molar solubility of the PbI2(s) =
1.51 x 10-3 M.
2+
2-
2+
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15 Calculate the concentration of silver ion when the
solubility product constant of AgI is 1 x 10-16 .
A 0.5 (1 x 10-16)
A 0.5 (1 x 10-16)
B 2 (1 x 10-16)
B 2 (1 x 10-16)
C
(1 x 10-16)2
C
(1 x 10-16)2
D
(1 x 10-16)
D
(1 x 10-16)
Answer
15 Calculate the concentration of silver ion when the
solubility product constant of AgI is 1 x 10-16 .
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D
Ksp = [Ag+][I-]
1 x 10-16 = x2
x = √1x10-16
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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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16 Calculate the molar solubility of PbF2 that has a Ksp at
25℃ = 3.6 x 10-6.
Students type their answers here
PbF2(s) # Pb2+ (aq) + 2F- (aq)
Answer
Ksp = [Pb2+ ][F- ]2
Ksp = x(2x)2
3.6 x 10-6 = 4x3
x = ∛3.6 x 10 -6 /4
x = 9.7 x 10-3 mol/liter
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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?
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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
compound?
3+
ABthe
3(s) # A (aq) + 3B (aq)
Ksp = [A3+ ][B- ]3
Answer
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Ksp = x(3x)3
1.8 x 10-18 = 27x4
x = ∜6.7 x 10 -20
x = 1.6 x 10-5 mol/liter
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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?
Answer
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?
Slide 38 (Answer) / 91
The molar solubility is
1.6 x 10-5 mol/liter x 280g/mol
= 4.6 x 10-3 g/liter
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19 Which of the following ionic salts would have the highest
molar solubility?
Slide 39 (Answer) / 91
19 Which of the following ionic salts would have the highest
molar solubility?
D
Ag2CrO4 would have
the highest
molar
MnCO3(s) Ksp = 1.82 solubility.
x 10-11 The ratio of
ions is 2:1, this results
Ksp-11= 4x3 . When
ZnCO3(s) Ksp = 1.45 xin10
you take the cube root
sp you
-12 get a molar
Ag2CrO4(s) Ksp = 9.00ofxK10
solubility that is larger
than the other salts
All have the same molar
solubility
listed.
A NiCO3(s) Ksp = 6.61 x 10-9
B MnCO3(s) Ksp = 1.82 x 10-11
B
C ZnCO3(s) Ksp = 1.45 x 10-11
C
D Ag2CrO4(s) Ksp = 9.00 x 10-12
D
E All have the same molar solubility
E
Answer
A NiCO3(s) Ksp = 6.61 x 10-9
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Common Ion Effect
Consider a saturate solution of barium sulfate:
Factors Affecting Solubility
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.
Return to the Table
of Contents
So adding any soluble salt containing either Ba2+ or
SO4 2- ions will decrease the solubility of barium sulfate.
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Common Ion Effect
Common Ion Effect
a) pure water
CaF2(s)
Ca 2+ (aq) + 2F - (aq)
If we assume x as the dissociation then,
Ca2+ ions = x and [F-] = 2x
Ksp = [Ca2+ ] [F- ]2
The solubility product for calcium fluoride, CaF2 is 3.9 x 10-11
= (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
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Common Ion Effect
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
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
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
Ksp = [Ca2+ ] [F- ]2
CaF 2 (s)
Note
[ F- ] = 0.15M
Ca2+
(aq)
+
2 F - (aq)
Ksp = [Ca2+ ] [F-] 2 = (0.080)(x)2
= (x)(0.15)2
Ksp = 3.9 x 10 -11 = 0.080x2
Ksp = 3.9 x 10-11 = 0.0225x
So x = 2.2 x 10 -5 mol/L * (78 g/mol CaF2 )/ 2
So x = ______ mol/L
Solubility is 0.000858 g/L
Solubility is = ______ x (78 g/mol CaF2) = ______ g/L
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Common Ion Effect
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.
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20 What is the molar solubility of a saturated solution of
Ag2CrO4? Ksp at 25℃ is = 1.2 x 10-12.
#
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
Note
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
Slide 47 (Answer) / 91
20 What is the molar solubility of a saturated solution of
Ag2CrO4? Ksp at 25℃ is = 1.2 x 10-12.
#
A 1.1 x 10-4
Ag2CrO4(s) # 2Ag+ (aq) + CrO42-(aq)
Answer
C 8.4 x 10-5
D 5.5 x 10-7
Ksp = [Ag ] [CrO4 ] = (2x) (x)
+ 2
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
B 6.7 x 10-5
Slide 48 / 91
2-
B 6.3 x 10-5
2
1.2 x 10 -12 = 4x3
x = ∛ 1.2 x 10-12/4
C 5.1 x 10-8
D 3.5 x 10-7
x = 6.7 x 10-5
E 2.2 x 10-8
E 1.7 x 10-6
[This object is a pull
tab]
Slide 48 (Answer) / 91
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
Ag2CrO4(s) # 2Ag+ (aq) + CrO42-(aq)
Answer
C 5.1 x 10-8
D 3.5 x 10-7
Ksp = [Ag+]2[CrO42-] = (2x)2(0.100M)
1.2 x 10 -12 = 0.400 x2
D 3.5 x 10-7
E 6.7 x 10-6
[This object is a pull
tab]
Slide 49 (Answer) / 91
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.
C
Ag2CrO4(s) # 2Ag
D 3.5 x 10
-7
Answer
C 3.11 x 10-11
Slide 50 / 91
Changes in pH
The solubility of almost any ionic compound is affected by
changes in pH.
A 3.0 x 10 -12
B 6.3 x 10-5
B 6.3 x 10-5
C 3.11 x 10-11
x = √1.2 x 10-12/0.400
x = 1.7 x 10-6
E 1.7 x 10-6
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.
A 3.0 x 10 -12
E
B 6.3 x 10-5
Slide 49 / 91
+
(aq)
+ CrO4
2-
(aq)
Consider dissociation equation for magnesium hydroxide:
Mg(OH)(s) # Mg2+(aq) + 2OH-(aq)
Ksp = [Ag+]2[CrO42-] = (0.200)2(x)
1.2 x 10 -12 = 0.040 x
x = 3.11 x 10 -11
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?
E 6.7 x 10-6
Would the Mg(OH)2 be more or less soluble?
[This object is a pull
tab]
(Think Le Châtelier’s Principle.)
Slide 50 (Answer) / 91
Slide 51 / 91
Changes in pH
Answer
The solubility of almost any ionic compound is affected by
changes in pH.
The added H+ will react with the
Consider dissociation OH
equation
for magnesium
hydroxide:
and take
it out of solution.
Mg(OH)(s) # Mg2+(aq) + 2OH-(aq)
The equilibrium will shift to the
right and the salt, Mg(OH)2, will be
more soluble.
What do you expect will happen to the equilibrium if
the pH of this system is lowered by adding a strong acid?
[This
object is a pull
Will one of the substances in the
equilibrium
interact with the
tab]
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)
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?
Slide 51 (Answer) / 91
Slide 52 / 91
Changes in pH
Answer
Changes in pH can also affect the solubility of salts that contain
the conjugate base of a weak The
acid.F- will react with H+
forming hydrofluoric acid
and some of it will leave
Consider the dissociation of the salt calcium fluoride:
from solution. According
to Le Châtelier’s Principle
2+ equilibrium
CaF2 (s)
Cathe
(aq) + 2F (aq) will shift to
the right and CaF2 will
become
more soluble.
[This object is a pull
What do you expect will happen
tab]to the equilibrium if
the pH of this system is lowered by adding a strong acid?
Ca 2+(aq) + 2F -(aq)
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 .
Will one of the substances in the equilibrium interact with the
strong acid?
Would the CaF 2 be more or less soluble?
Slide 53 / 91
Slide 54 / 91
Changes in pH
Changes in pH
Sample Problem
Calculate the molar solubility of Mn(OH)2 in
a) in a solution buffered at pH = 9.5
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=9.5, the [H+] = 3.2 x 10-10,
the [OH-] = 3.2x 10-5.
In a solution buffered at pH=8.0, the [H+] = 1 x 10-8, the [OH-] = 1x 10-6.
[ 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
The solubility product for Mn(OH)2, is 1.6 x 10-13
[ 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
Slide 55 / 91
Slide 56 / 91
Changes in pH
Changes in pH
Sample Problem
Calculate the molar solubility of Mn(OH)2 in
c) in pure water
If a substance has a basic anion, it will be more soluble in an
acidic solution.
In pure water the pH=7.0, the [H+] = 1 x 10-7, the [OH-] = 1x 10-7.
If a substance has an acidic cation, it will be more soluble in
basic solutions.
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 =
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?
So x = 16 mol/liter
Slide 57 / 91
23 Given the system at equilibrium
AgCl (s)
Ag+ (aq) + Cl- (aq)
Slide 57 (Answer) / 91
23 Given the system at equilibrium
AgCl (s)
Ag+ (aq) + Cl- (aq)
C
Whento
0.01M
is added,
When 0.01 M HCl is added
the HCl
sytem,
the point of
according to Le Chatelier's
equilibrium will shift to the
________.
principle the equilibrium will
Answer
When 0.01 M HCl is added to the sytem, the point of
equilibrium will shift to the ________.
shift away from the additional
Cl- to the left and the
A right and the concentration of Ag+ will decrese
+
A right and the concentration
of Agof
will
decrese
concentration
Ag+ will
B right and the concentration of Ag+ will increase
B right and the concentration of Ag+ will increase
decrease.
[This object is a pull
C
left and the concentration of Ag+ will decrease
C
left and the concentration oftab]Ag+ will decrease
D
left and the concentration of Ag+ will increase
D
left and the concentration of Ag+ will increase
Slide 58 / 91
A PbCl2
24 Which of the following substances are more soluble in
acidic solution than in basic solution? Select all that
apply.
A PbCl2
B BaCO3
B BaCO3
C AgI
C AgI
D Fe(OH)3
D Fe(OH)3
E MgF2
E MgF2
B,D,E
Answer
24 Which of the following substances are more soluble in
acidic solution than in basic solution? Select all that
apply.
Slide 58 (Answer) / 91
All these ionic compounds
contain a basic anion that
would react with the added H+
causing a shift to the right and
resulting in a more soluble salt.
[This object is a pull
tab]
Slide 59 / 91
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
Students type their answers here
Slide 59 (Answer) / 91
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
Students type their answers here
The solubility product for Zn(OH)2,
is 3.0 x 10-16
pH = 8.5, pOH = 5.5 and [ OH-] =
3.2 x 10-6M
Ksp = [Zn2+] [OH-]2 = (x)(3.2 x 10-6)2
x = 3.0 x 10-16/(3.2 x 10-8)2
x = 3.0 x 10-5 mol/L
[This object is a pull
tab]
Slide 60 / 91
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
No
8.5? Explain.
A Yes
A Yes
B No
B No
Answer
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.
Slide 60 (Answer) / 91
The solubility of Zn(OH)2
increases in more acidic
solutions. When the pH
increases, as in this case, from
8.5 to 11.0 the solubility will
decrease.
[This object is a pull
tab]
Slide 61 / 91
27 The molar solubility of NH4Cl increases as pH
_________ .
Slide 61 (Answer) / 91
27 The molar solubility of NH4Cl increases as pH
_________ .
A
NH4Cl has an acidic cation, NH4+.
As pH increases the solution
becomes more basic, the
equilibrium shifts to the right and
A increases
B decreases
B decreases
C is unaffected by changes in pH
C is unaffected by changes in pH
Answer
A increases
the salt becomes more soluble.
[This object is a pull
tab]
Slide 62 / 91
28 The molar solubility of Na2CO3 increases as pH
_________ .
Slide 62 (Answer) / 91
28 The molar solubility of Na2CO3 increases as pH
_________ .
B
A increases Na2CO3 has a basic anion, CO32-.
As pH increases the solution
B decreases becomes more basic, the
equilibrium shifts to the left and the
Answer
A increases
B decreases
C is unaffected by changes in pH
soluble.
C is unaffectedsalt
by becomes
changesless
in pH
[This object is a pull
tab]
Slide 63 / 91
Slide 64 / 91
Amphoterism
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.
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.
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:
Examples of such these substances are oxides and
hydroxides of Al3+ , Zn 2+ , and Sn2+ .
AgCl(s) + 2NH3(aq)
Ag(NH3)2+(aq) + Cl- (aq)
Added NH3 reacts with Ag+ forming
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) #
Ag(NH3)2+. Adding enough NH3
Al3+(aq) + 3 OH-(aq)
results in the complete dissolution
of AgCl.
#
Slide 65 / 91
Amphoterism
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.
Slide 66 / 91
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
Slide 66 (Answer) / 91
Slide 67 / 91
29 Which of the following factors affect solubility?
pH
B
Formation of Complex Ions
C
Common-Ion Effect
D
A and C
E
A, B, and C
Answer
A
Precipitation Reactions
and Separation of Ions
E
A, B and C are correct.
[This object is a pull
tab]
Return to the
Table of Contents
Slide 68 / 91
Slide 69 / 91
Precipitation Reactions and Separation of Ions
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.
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
Do you remember what metal cations tended to be insoluble?
Ag+, Pb2+, and Hg 2+
Slide 69 (Answer) / 91
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
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
D
C
chloride nitrateNaCl(aq) + AgNO3(aq)
D
silver chloride
Answer
Slide 70 / 91
#
NaNO3(aq) + AgCl(s)
E Not enough information
C
potassium calcium
D
carbonate bromide
E Not enough information
[This object is a pull
tab]
Slide 70 (Answer) / 91
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?
Answer
A potassium bromide
Slide 71 / 91
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
K2CO3 (aq) + CaBr2 (aq) # 2KBr(aq) + CaCO3(s)
B calcium carbonate
B Pb(SO4)2
C
potassium calcium
C
Pb2SO4
D
carbonate bromide
D
Mg(NO3)2
E Not enough information
E Not enough information
[This object is a pull
tab]
Slide 71 (Answer) / 91
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
A PbSO4
Slide 72 / 91
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 ?
Answer
Pb(NO3)4(aq) + MgSO4(aq) # Mg(NO3)2 (aq) +
PbSO4(s)
B Pb(SO4)2
A yes, because Qsp < Ksp
C
Pb2SO4
B yes, because Qsp > Ksp
D
Mg(NO3)2
C
no, because Qsp = Ksp
D
no, because Qsp < Ksp
E Not enough information
[This object is a pull
tab]
E no, because Qsp > Ksp
Slide 72 (Answer) / 91
Answer
33 The K sp for Zn(OH)2 is 5.0 x10 -17B. Will a precipitate
-2
2+
form in a solution whose
Q = [Znsolubility
][OH-] is 8.0x10
mol/L Zn(OH)2 ?
2
Q = x(2x)
Q = (8.0 x 10-2)(1.6 x 10-1)2
Q =<
2.0
x 10 -3
A yes, because Q
Ksp
sp
Q>Ksp
B yes, because Qsp > Ksp
When Q>Ksp, a precipitate will
[This object is a pull
C no, because Qform.
sp = Ksp tab]
D
no, because Qsp < Ksp
E no, because Qsp > Ksp
Slide 73 / 91
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.
Slide 73 (Answer) / 91
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
Na2CO3 (aq)+ Zn(NO3)(aq) # ZnCO3 (s) + NaNO3 (aq)
A A zinc carbonate precipitate forms, since Q>K.
The salt we are interested in is
C
A
Answer
22+
ZnCO3(s) # Znprecipitate
(aq) + CO3 (aq)forms, since Q<K.
B A zinc carbonate
Q = [Zn2+][CO32-]
Q= nitrate
(0.2)(0.1)=
2 x 1-2
sodium
precipitate
Slide 74 / 91
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.
forms, since Q>K.
Q>Ksp , therefore ZnCO3 will precipitate.
D No precipitate forms, since Q=K.
[This object is a pull
tab]
Slide 75 / 91
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.
Slide 76 / 91
Separation of Ions
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!
Slide 77 / 91
Separation of Ions
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 10-51
PbS
3 x 10-28
CuS
6 x 10-37
AgCl
1.8 x 10-10
PbCl2
1.7 x 10-5
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-?
Slide 78 / 91
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.
Slide 79 / 91
Separation of Ions 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.
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.
Slide 80 / 91
Separation of Ions Problems
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.
Slide 81 / 91
Slide 82 / 91
Separation of Ions Problems
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
Slide 83 / 91
Separation of Ions Problems
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.
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
Slide 84 / 91
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 85 / 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.
Slide 85 (Answer) / 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
B
for BaCrO4 is 2.1 x 10 -10 and
the Ksp for PbCrO4 is
22+
BaCrO
Ba
+
CrO
#
4(s)
4 (aq)
(aq)
2.8 x 10-13.
Ksp =[Ba2+][CrO42-] = 2.1 x 10-10
A BaCrO4
B PbCrO4
B PbCrO4
[CrO42-] > 2.10 x 10-10 /2.0 x 10-5.
C They will precipitate at the same time.
C They will precipitate
at the same
time.
precipitate. Completing
the same
calculation
D It's impossible to determine
with the information provided.
D It's impossible
] >determine
2.8 x 10-13 /1.8 x 10-4. PbCrO4 will
[CrO42-to
precipitate when
[CrO42-] > 1.56 x 10 -9. It
with the information
provided.
Answer
A BaCrO4
In order for this salt to precipitate Q>Ksp
therefore [CrO42-] > Ksp/[Ba2+]
for PbCrO4, we find that
2takes much
less
CrO
[This
object
is a4pullto precipitate the
PbCrO4 sotab]
it will precipitate first.
Slide 86 / 91
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?
When the [CrO42-] ≥ 1.05 x 10-5 BaCrO4 will
Slide 86 (Answer) / 91
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 A(Ksp = 1.8 x 10 -10) or PbCI2
(Ksp = 1.7 xAgCl
10-5(s))#precipitate
Ag+(aq) + Cl-(aq) first?
Ksp =[Ag+][Cl-] = 1.8 x 10-10
In order for this salt to precipitate Q>Ksp
therefore [Cl-] > Ksp/[Ag+]
A AgCl
B PbCl2
B PbCl2
C They will precipitate at the same time.
C They
D It is impossible to determine with
the information provided.
-2. It takes
precipitate when
[Cl-] > 8.5 x 10with
D It is impossible
to determine
much less of Cl- to precipitate the AgCl so it
[This
object is a pull
the information
provided.
will precipitate
tab] first.
Answer
A AgCl
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)
When the [Cl-] > 9.0 x 10-7 AgCl will
precipitate. Completing the same calculation
will
precipitate
at the same time.
for PbCl
2, we find that
[Cl-] > 1.7 x 10-5 /2.0 x 10-4. PbCl2 will
Slide 87 (Answer) / 91
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)
Answer
Slide 87 / 91
[Cl-] > 1.8 x 10-10 /2.0 x 10-4.
When the [Cl-] > 9.0 x 10-7, AgCl
will begin to precipitate.
[This object is a pull
tab]
Slide 88 / 91
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)
Answer
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)
Slide 88 (Answer) / 91
PbCl2 begins to precipitate when
the [Cl-] > 8.5 x 10-2. We can solve
for [Ag+] at this concentration of Cl-.
Ksp (AgCl) = 1.8 x 10-10 = [Ag+][Cl-]
[Ag+] = 1.8 x 10 -10 / 8.5 x 10-2
[Ag+] = 2.1 X 10-9
[This object is a pull
tab]
Slide 89 / 91
39 Will Co(OH)2 precipitate from Bsolution if the pH of a
0.002M solutionCo(OH)
of Co(NO
+ 2OH- (aq) to 8.4? Ksp for
3)22+is
(aq) adjusted
2 (s) # Co
Co(OH)2 is 2.5 xCo(OH)
10 -142. will precipitate if Q>Ksp.
A Yes
A Yes
B No
B No
Answer
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 89 (Answer) / 91
pH = 8.4, pOH=5.6, [OH-] = 2.5 x 10-6
Q = [Co2+][OH-]2 = (0.002)(2.5 x 10-6)2
Q = 1.26 x 10-14
Q<Ksp , therefore a precipitate will not
form.
[This object is a pull
tab]
Slide 90 / 91
Students type their answers here
40 Will a precipitate
form
if you
mix 25.0 mL of 0.250 M
+ CrO
CaCrO
# Ca
calcium chloride,
and
50.0
of you
0.155
M lithium
In order
to solve
this mL
problem
have to
determine the concentrations of Ca and
-9
chromate? The
K
sp for calcium chromate is 4.5 x 10 .
CrO in solution after mixing. Then you will
2+
(aq)
4(s)
24 (aq)
2+
4
Answer
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.
Slide 90 (Answer) / 91
2-
need to calculate Q and then compare it to Ksp.
[Ca2+]: M2= M1V1/V2 = (0.250M)(0.025 L)/0.075L
M2=0.0833
2Students type their
answers
here
[CrO
4 ]: M2= M1V1/V2 = (0.155M)(0.050 L)/
0.075L
M2= 0.1033
Q = (0.0833)(0.1033) = 8.61 x 10-3
Q>Ksp therefore
CaCrO
will precipitate.
[This object
is a4 pull
tab]
Slide 91 / 91