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Goes with chapter 19: Silberberg’s Principles of General Chemistry
AP Chemistry
Mrs. Laura Peck, 2013
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Objectives/Study Guide
 Write balanced equations for the dissolution of a salt and its
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corresponding solubility product expression.
Predict the relative solubilities of salts which dissolve to give the same
number of ions from their Ksp values
Calculate the Ksp value from the solubility of a salt and also calculate the
solubility of the salt in units of mol/L or g/L from the given Ksp value
Predict the effect of a common ion on the solubility of a salt and perform
calculations.
Perform calculations to predict if a precipitate will form when two
solutions are mixed
Do problems involving selective precipitation.
Perform calculations involving complex ions and solubility
Use qualitative analysis to separate a mixture of ions
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AP tip:
 Solubility problems would appear in the first free-
response question, since they would be considered an
equilibrium problem. Awareness of the types of
problems outlined in this Topic and the methods used
to solve them will lead you to succeed in this topic on
the AP exam.
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Solubility Product
 You should have already memorized the solubility rules and know which salts
are soluble.
 For slightly soluble or insoluble salts, an equilibrium exists between the solid
and its aqueous ions.
 Ex: PbCl2(s)   Pb2+(aq) + 2Cl-(aq)
 At first, when the salt is added to the water, there are no ions present.
 As the solid dissolves, the concentration of the ions increases.
 A simultaneous competing process is the reverse of the dissolution, that is, the reforming of
the solid called crystallization.
 At some point, the maximum amount of dissolution is achieved, which is called the
saturation point.
 However, remember that on a molecular level, a dynamic equilibrium exists between
dissolved solute and undissolved solid (Rdissolution=Rcrystallization)
 The solution is saturated when no more solid dissolves and equilibrium is reached.
 Keq = Ksp = [Pb2+][Cl-]2
 The constant, Ksp, is the solubility product constant.
 For salts producing the same number of ions, the Ksp value can be used to
measure the extent to which the solid dissolves.
 The larger the Ksp value, the more soluble the salt.
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Example #1
 Given the following salts and their Ksp values, which
salt is the most soluble? Which salt is the least
soluble?
Formula
Ksp
NiCO3
1.4x10-7
MnS
2.3x10-13
CaSO4
6.1x10-5
The most soluble salt is the salt with the largest Ksp value, CaSO4
The least soluble salt is the salt with the lowest value of Ksp, MnS. You
are able to compare the Ksp values to determine the relative solubilities
of the salts because they all produce the same number of ions.
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Calculations involving solubility
 The solubility of a salt that will dissolve in 1 L of water.
 The solubility of a salt can be given in units of mol/L or
g/L.
 The solubility of a salt can be used to determine the
Ksp value for the salt.
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Example #2: Calculating Ksp from solubility.
 The solubility of Pb3(PO4)2 is 6.2x10-12M. Calculate
the Ksp value for the solid.
Step 1: write the reaction for the dissolution of the solid.
Step 2: Underneath the reaction,
make an ICE chart.
I.
0
0
Or you can use stoich to solve this problem:
In the ICE chart, x represents x mol/L
C.
+3x
+3x
Of Pb3(PO4)2(s) dissolving to reach
E.
3x
3x
-11Mequals
6.2x10-12 x molPb3(PO4)2 x 3mol Pb2+ =K1.9x10
Pb2+ 6.2x10-12 M
eq which
Step 3: Write the Keq expression
for the
reaction
1L
1mol
Pb3(PO4)2
And plug in the values from the E line of the ICE For every 1mol/L of Pb (PO ) , which
3
4 2
-12
3-11
32+
x mol3-]Pb
2 108x
mol 5Pb4 Dissolves,
= 1.2x10 M
3 mols/L
PO4 of Pb and 2 mol/L
2 =3(PO
2x 2=
Keq = Ksp =6.2x10
[Pb2+]3[PO
(3x)43)(2x)
4
1L
1mol Pb3(PO4)2Of PO43- form
The value of x is the solubility of Pb3(PO4)2, which
-12M
2+
Equals 6.2x10
Plug
these values into the Ksp expression3xand
is the
solve
mol/L
for Kof
sp:Pb produced when
The solid Pb3(PO4)2 dissolves
-12
5
-55
Ksp = 108(6.2x10
) 43-=]2 9.9x10
Ksp = [Pb2+]3[PO
= (1.9x10-11)3(1.2x10-11)2 = 9.9x10-55
2x is the mol/L of PO43- produced when
The solid, Pb3(PO4)2 dissolves
Pb3(PO4)2(s)   3Pb2+(aq) + 2PO43-(aq)
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Calculating solubility from Ksp
 If you are given a group of salts which do not all have
the same cation to anion ratio and asked which is more
soluble, you must perform a calculation to determine
the solubility of each salt.
 If the salts each provide a number of ions in solution,
so you cannot directly compare the Ksp values to
predict which is more soluble.
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Example #3: Given the two salts in the table
below, which is more soluble? Show calculation to
Step 1: Write the
Salt
Ksp
support your answer.
reaction for the
FeC2O4(s)   Fe2+(aq) + C2O42-(aq)
I.
C.
E.
0
+x
X
FeC2O4
2.1x10-7
Cu(IO4)2
1.4x10-7
0
+x
x
Ksp = [Fe2+][C2O42-]  2.1x10-7 = x2  x = 4.6x10-4 mol Fe2+/L
4.6x10-4 mol Fe2+ x 1mol FeC2O4/1L = 4.6x10-4 mol FeC2O4/L
Cu(IO4)2(s)   Cu2+(aq) + IO4-(aq)
I.
C.
E.
0
+x
X
0
+2x
2x
dissolution of the
solid Salt #1
Step 2:
Underneath the
reaction, make
an ICE chart
Step 3: Write the
equilibrium
expression for
Ksp, plug in The
equilibrium line,
and solve for x
Ksp = [Cu2+][IO4-]2  1.4x10-7 = x(2x)2  4x3 = 1.4x10-7  x = 3.3x10-3 mol/L Step 4: Repeat
process for salt
3.3x10-3mol Cu2+ x 1mol Cu(IO4)2/1L = 3.3x10-3mol Cu(IO4)2/L
#2
Turns out that Cu(IO4)2/L is greater than FeC2O4/L so its more soluble!
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Common Ion Effect
 When a salt is dissolved in water containing a common
ion, its solubility is decreased.
 Consider the solubility equilibrium of silver sulfate:
 Ag2SO4(s)   2Ag+(aq) + SO42-(aq)
 When silver sulfate is dissolved in 0.100M AgNO3, the
Ag+ ion from silver nitrate causes the equilibrium to shift
to the left, decreasing the solubility of silver sulfate.
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Example #4
 Calculate the molar solubility of Ag2SO4 in 0.10M
AgNO3. Ksp for Ag2SO4(s) is 1.2x10-5
Ag2SO4(s)   2Ag+(aq) + SO42-(aq)
I…
C…
E…
0.10
+2x
0.10+2x
0
+x
x
Ksp = [Ag+]2[SO42-] = (0.10 + 2x)2(x)
First, fill out the ICE chart. Be sure
To include the initial [ ] of the ions
From the soluble salt, AgNO3.
Second, plug in the equilibrium line
Into the Ksp expression.
Assume that 0.10 x 2x is about 0.10, since Ksp is small you can assume that the
change (2x) from the initial concentration (0.10) is negligible.
Ksp = 1.2x10-5 = (0.10)2(x)  x = 1.2x10-3 mol SO42-/L
1.2x10-3mol/L SO42- x (1mol Ag2SO4)/(1mol SO42-) = 1.2x10-3mol/L Ag2SO4
The solubility of silver sulfate in 0.100M silver nitrate, 1.2x 10-3 mol/L is less
Than the solubility of silver sulfate in pure water, 1.4x10-2 mol/L
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pH and Solubility
 Chromium (III) hydroxide dissolves according to the
equilibrium:
 Cr(OH)3(s)   Cr3+(aq) + 3OH-(aq)
 An increase in pH, caused by the addition of OH- ions, will shift the
equilibrium to the left, decreasing the solubility of Cr(OH)3
 A decrease in pH, caused by the addition of H+ ions, will shift the
equilibrium to the right, increasing the solubility of Cr(OH)3.
 The H+ ions remove the OH- ions from the solution.
 A salt with the general formula, MX, will show increased
solubility in acidic solution if the anion, X-, is an effective base
(if HX is a weak acid).
 Common anions that make effective bases include S2-, OH-, and
CO3212
Example #5
 Calculate the solubility of Fe(OH)3
in a solution with a pH equal to 5.0.
Ksp = 4.0x10-38
pOH = 14 – pH = 14.0-5.0 = 9.0
[OH-] = 1.0x10-9 M
Fe(OH)3(s)   Fe3+(aq) + 3OH-(aq)
I..
C..
E..
0
X
X
1.0x10-9
3x
1.0x10-9
Ksp = 4.0x10-38 = [Fe3+][OH-]3 = x(1.0x10-9)3
First steps are to write out the reaction
Then fill out an ICE chart. Calculate
The initial hydroxide concentration
From the pH.
Second, plug E values into Ksp equation
Because pH is 5.0, the equilibrium
[OH-] must be equal to 1.0x10-9
X = 4.0x10-11 M
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Precipitate
formation
 A precipitate may or may not form when two solutions are mixed,
depending on the concentration of the ions involved in the formation
of the solid.
 Ion Product:
 The ion product, Q, is written in the same way as the Ksp expression.
 F0r Lead(II) chloride, Q = [Pb2+][Cl-]2
 Calculation of the value, Q, involves the use of the initial
concentrations of the solutions mixed, [Pb2+]0 and [Cl-]0, instead of the
equilibrium concentrations.
 A comparison of the value of Q to Ksp determines if a precipitate is
formed.
 Q>Ksp – precipitation occurs
 Q<Ksp – no precipitation occurs
 Q=Ksp – the solution is saturated.
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 Will a precipitate for when 100.0 mL
Example #6
of 4.0x10-4M Mg(NO3)2 is added to
100.0 mL of 2.0x10-4M NaOH?
Mg(OH)2 is the possible precipitate.
NaNO3 is always soluble.
[Mg2+]0 = (0.100L x (4.0x10-4mol/L))/0.200L
= 2.0x10-4M
[OH-]0 = (0.100L x(2.0x10-4mol/L))/0.200L
= 1.0x10-4M
Step 1: Determine the identity of
the precipitate formed.
Step 2: Determine the
concentration of the ions after they
are Mixed and before any reaction
occurs.
Determine the moles of
concentration of each solute
Q = [Mg2+][OH-]2 = (2.0x10-4M)(1.0x10-4)2
present. Be sure to divide by the
= 2.0x10-12
total volume of the two solutions
Since Q<Ksp – no precipitate will form mixed.
Step 3: Write the ion product
expression, calculate its value, and
Compare it to Ksp, which equals
8.8x10-12
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Selective Precipitation
 A reagent is added to a mixture of metal ions, thus forming a
precipitate.
 One metal ion will precipitate before the other, allowing the mixture
to be separated.
 Example #7: a solution contains 0.25M Ni(NO3)2 and 0.25M
Cu(NO3)2. A solution of Na2CO3 is slowly added to this solution.
 A) Will NiCO3 (Ksp = 1.4x10-7) or CuCO3 (Ksp = 2.5x10-10) precipitate
first?
CuCO3 will precipitate first because its Ksp is smaller
 B) Calculate the concentration of CO32- necessary to begin the
precipitation of each salt.
For CuCO3 precip. begins: [CO32-]= Ksp/[Cu2+] = 2.5x10-10/0.25M = 1.0x10-9M
For NiCO3 precip. Begins: [CO32-]=Ksp/[Ni2+]=1.4x10-7/0.25M = 5.6x10-7M
 C) Determine the concentration of Cu2+ when NiCO3 begins to
precipitate.
[Cu2+] = Ksp/[CO32-] = 2.5x10-10/5.6x10-7M = 4.5x10-4M
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Complex Ion
Equilibria
 A complex ion consists of a metal ion surrounded by ligands which are
Lewis bases such as H2O, OH-, NH3, Cl-, and CN The coordination number is the number of ligands which attach to the
transition metal ion.
 Common coordination numbers include 4 for Cu2+ and Co2+ and 2 for
Ag+
 The ligands attach one at a time to the metal ion; each step has a
formation constant, Kf:
 Ag+ + S2O32-  Ag(S2O3)- Kf1 = 7.4x108
 Ag(S2O3)- + S2O32-  Ag(S2O3)23- Kf2 = 3.9x104
 Formation of a complex ion causes a precipitate to dissolve.
 The equilibrium, AgBr(s)  Ag+ + Br- is not affected by the addition
of H+.
 But the concentration of Ag+ can be lowered by the addition of excess
S2O32- forming the complex ion, Ag(S2O3)2317
Example #8
AgBr(s)  Ag+(aq) + Br-(aq)
Ag+ + S2O32-  Ag(S2O3)Ag(S2O3)- + S2O32-  Ag(S2O3)23-
 Calculate the mass of AgBr that can
dissolve in 1.00L of 0.500M Na2S2O3.
Ksp for AgBr = 5.0x10-13
Ksp = 5.0x10-13
Kf1 = 7.4x108
Kf2 = 3.9x104
AgBr(s) + 2S2O32-  Ag(S2O3)23- + BrI..
C..
E..
0.500
-2x
0.500-2x
0
+x
x
K’=14.4
0
+x
x
Step 1: Determine the overall
reaction by Adding up the
reactions for the solubility
Equilibrium and the stepwise
formation Of the complex ion.
Remember, to get the Equilibrium
constant for the overall reaction,
K, multiply together the K for each
step. (from Preceding slide)
K’ = 14.4 = [Ag(S2O3)23-][Br-]/[S2O32-]2
Step 2: Complete an ICE chart
with the overall reaction
= x2/(0.500-2x)2 take square root of both sides
3.79 = x/(0.500-2x)
X = 1.90-7.58x  x = 0.221MBr = 0.221M AgBr(s)
Step 3: Plug E line into Ksp
expression
1.00L x (0.221mol AgBr/1L) x (187.8g/1mol) = 41.5g AgBr = 42g AgBr
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Qualitative Analysis
 Qualitative analysis involves separating a mixture of
cations or anions based on their solubilities.
 Cations can be separated into five major groups based on
their solubilities.
 Group I: insoluble chlorides
 Group II: sulfides soluble in acidic solution
 Group II: sulfieds insoluble in basic solution
 Group IV: insoluble carbonates
 Group V: alkali metal and ammonium ions
 Each of these groups can be treated further to separate and
identify the individual ions.
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Example
#9
 Separate a mixture
Solution of Ag+,
Hg22+, Pb2+
containing the
Group I cations:
Ag+, Hg22+ and Pb2+
Add cold HCl(aq)
Precipitate of
AgCl(s), Hg2,
Cl2(s), PbCl3(s)
Heat
Precipitate of
AgCl(s), Hg2,
Cl2(s)
Solution of
Pb2+
Add CrO42-
Add NH3(aq)
Precipitate of
Hg(l) (black),
HgNH3Cl(s)
(white)
Precipitate
of
PbCrO4(s)
(yellow)
Solution of
Ag(NH3)2, Cl-
Add H+
Precipitate of
AgCl(s) (white)
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The End
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