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CH160 General Chemistry II
Lecture Presentation
Solubility Equilibria
Chapter 17
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Chapter 17
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Why Study Solubility Equilibria?
Many natural processes involve precipitation or dissolution
of salts. A few examples:
Dissolving of underground limestone deposits (CaCO3)
forms caves
Note: Limestone is water “insoluble” (How can this be?)
Precipitation of limestone (CaCO3) forms stalactites and
stalagmites in underground caverns
Precipitation of insoluble Ca3(PO4)2 and/or CaC2O4 in the
kidneys forms kidney stones
Dissolving of tooth enamel, Ca5(PO4)3OH, leads to tooth
decay (ouch!)
Precipitation of sodium urate, Na2C5H2N4O2, in joints
results in gouty arthritis.
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Why Study Solubility Equilibria?
Many chemical and industrial processes involve
precipitation or dissolution of salts. A few examples:
Production/synthesis of many inorganic compounds involves their
precipitation reactions from aqueous solution
Separation of metals from their ores often involves dissolution
Qualitative analysis, i.e. identification of chemical species in
solution, involves characteristic precipitation and dissolution
reactions of salts
Water treatment/purification often involves precipitation of metals
as insoluble inorganic salts
Toxic Pb2+, Hg2+, Cd2+ removed as their insoluble sulfide (S2-) salts
PO43- removed as insoluble calcium salts
Precipitation of gelatinous insoluble Al(OH)3 removes suspended
matter in water
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Why Study Solubility Equilibria?
To understand precipitation/dissolution processes in
nature, and how to exploit precipitation/dissolution
processes for useful purposes, we need to look at the
quantitative aspects of solubility and solubility
equilibria.
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Solubility of Ionic Compounds
Solubility Rules
general rules for predicting the solubility of ionic
compounds
strictly qualitative
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Solubility of Ionic Compounds
Solubility Rule Examples
All alkali metal compounds are soluble
Most hydroxide compounds are insoluble. The
exceptions are the alkali metals, Ba2+, and Ca2+
Most compounds containing chloride are soluble. The
exceptions are those with Ag+, Pb2+, and Hg22+
All chromates are insoluble, except those of the alkali
metals and the NH4+ ion
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Solubility of Ionic Compounds
large excess added
+
NaOH
Fe(OH)3 Cr(OH)3
Fe3+
Precipitation of both Cr3+
and Fe3+ occurs
Cr3+
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Solubility of Ionic Compounds
small excess added
slowly
+
NaOH
Cr3+
Fe(OH)3
Fe3+
less soluble salt
precipitates only
Cr3+
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Solubility of Ionic Compounds
Solubility Rules
general rules for predicting the solubility of ionic
compounds
strictly qualitative
Do not tell “how” soluble
Not quantitative
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Solubility Equilibrium
saturated
solution
solid
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My+
xMy+
yAx-
My+
Ax-
AxMxAy
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Solubility of Ionic Compounds
Solubility Equilibrium
MxAy(s) <=> xMy+(aq) + yAx-(aq)
The equilibrium constant for this reaction is the
solubility product, Ksp:
Ksp = [My+]x[Ax-]y
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Solubility Product, Ksp
Ksp is related to molar solubility
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Solubility Product, Ksp
Ksp is related to molar solubility
qualitative comparisons
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Solubility Product, Ksp
Ksp used to compare relative solubilities
smaller Ksp = less soluble
larger Ksp= more soluble
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Solubility Product, Ksp
Ksp is related to molar solubility
qualitative comparisons
quantitative calculations
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Calculations with Ksp
Basic steps for solving solubility equilibrium
problems
Write the balanced chemical equation for the
solubility equilibrium and the expression for Ksp
Derive the mathematical relationship between Ksp
and molar solubility (x)
Make an ICE table
Substitute equilibrium concentrations of ions into Ksp
expression
Using Ksp, solve for x or visa versa, depending on
what is wanted and the information provided
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Example 1
(1 on Example Problems Handout)
Calculate the Ksp for MgF2 if the molar
solubility of this salt is 2.7 x 10-3 M.
(ans.: 7.9 x 10-8)
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Example 2
(2 on Example Problems Handout)
Calculate the Ksp for Ca3(PO4)2 (FW = 310.2)
if the solubility of this salt is 8.1 x 10-4 g/L.
(ans.: 1.3 x 10-26)
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Example 3
(4 on Example Problems Handout)
The Ksp for CaF2 (FW = 78 g/mol) is 4.0 x 10-11.
What is the molar solubility of CaF2 in water?
What is the solubility of CaF2 in water in g/L?
(ans.: 2.2 x 10-4 M, 0.017 g/L)
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Precipitation
Precipitation reaction
exchange reaction
one product is insoluble
Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)
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Precipitation
Precipitation reaction
exchange reaction
one product is insoluble
Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)
Na+ and Ca2+ “exchange” anions
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Precipitation
Precipitation reaction
exchange reaction
one product is insoluble
Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)
Net Ionic: Ca2+(aq) + CO32-(aq) <=> CaCO3(s)
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Precipitation
Compare precipitation to solubility equilibrium
Ca2+(aq) + CO32-(aq) <=> CaCO3(s)
prec.
vs
CaCO3(s) <=> Ca2+(aq) + CO32-(aq)
sol. Equil.
saturated solution
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Precipitation
Compare precipitation to solubility equilibrium:
Ca2+(aq) + CO32-(aq) <=> CaCO3(s)
vs
CaCO3(s) <=> Ca2+(aq) + CO32-(aq)
saturated solution
Precipitation occurs until solubility equilibrium is
established.
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Precipitation
Ca2+(aq) + CO32-(aq) <=> CaCO3(s)
vs
CaCO3(s) <=> Ca2+(aq) + CO32-(aq)
saturated solution
Key to forming ionic precipitates: Mix ions so
concentrations exceed those in saturated solution
(supersaturated solution)
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Predicting Precipitation
To determine if solution is supersaturated:
Compare ion product (Q or IP) to Ksp
For MxAy(s) <=> xMy+(aq) + yAx-(aq)
= [My+]x[Ax-]y
 Q calculated for initial conditions
Q
Q > Ksp  supersaturated solution, precipitation
occurs, solubility equilibrium established (Q =
Ksp)
= Ksp  saturated solution, no precipitation
 Q < Ksp  unsaturated solution, no precipitation
Q
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Predicting Precipitation
Basic Steps for Predicting Precipitation
Consult solubility rules (if necessary) to determine
what ionic compound might precipitate
Write the solubility equilibrium for this substance
Pay close attention to the stoichiometry
Calculate the moles of each ion involved before
mixing
moles = M x L or moles = mass/FW
Calculate the concentration of each ion involved
after mixing assuming no reaction
Calculate Q and compare to Ksp
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Example 4
(7 and 8 on Example Problems Handout)
Will a precipitate form if (a) 500.0 mL of 0.0030
M lead nitrate, Pb(NO3)2, and 800.0 mL of
0.0040 M sodium fluoride, NaF, are mixed, and
(b) 500.0 mL of 0.0030 M Pb(NO3)2 and 800.0
mL of 0.040 M NaF are mixed?
(ans.: (a) No, Q = 7.5 x 10-9; (b) Yes, Q = 7.5 x 10-7)
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Solubility of Ionic Compounds
Solubility Rules
All alkali metal compounds are soluble
The nitrates of all metals are soluble in water.
Most compounds containing chloride are soluble. The
exceptions are those with Ag+, Pb2+, and Hg22+
Most compounds containing fluoride are soluble. The
exceptions are those with Mg2+, Ca2+, Sr2+, Ba2+, and
Pb2+
Ex. 4: Possible precipitate = PbF2 (Ksp = 4.1 x 10-8)
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Example 5
(10 on Example Problem Handout)
A student carefully adds solid silver nitrate,
AgNO3, to a 0.0030 M solution of sodium
sulfate, Na2SO4. What [Ag+] in the solution is
needed to just initiate precipitation of silver
sulfate, Ag2SO4 (Ksp = 1.4 x 10-5)?
(ans.: 0.068 M)
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Factors that Affect Solubility
Common Ion Effect
pH
Complex-Ion Formation
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Factors that Affect Solubility
Common Ion Effect
pH
Complex-Ion Formation
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These sure sound
familiar. Where
have I seen them
before?
33
Common Ion Effect and Solubility
Consider the solubility equilibrium of AgCl.
AgCl(s) <=> Ag+(aq) + Cl-(aq)
How does adding excess NaCl affect the
solubility equilibrium?
NaCl(s)  Na+(aq) + Cl-(aq)
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Common Ion Effect and Solubility
Consider the solubility equilibrium of AgCl.
AgCl(s) <=> Ag+(aq) + Cl-(aq)
How does adding excess NaCl affect the
solubility equilibrium?
NaCl(s)  Na+(aq) + Cl-(aq)
2 sources of ClCl- is common ion
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Example 6
(11 on Example Problem Handout)
What is the molar solubility of AgCl (Ksp = 1.8 x
10-10) in a 0.020 M NaCl solution? What is the
molar solubility of AgCl in pure water?
(ans.: 8.5 x 10-9, 1.3 x 10-5)
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Common Ion Effect and Solubility
How does adding excess NaCl affect the
solubility equilibrium of AgCl?
1.3 x 10-5 M
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+ 0.020 M NaCl
Chapter 17
Molar solubility
Molar solubility
AgCl in H2O
AgCl in 0.020
M NaCl
8.5 x 10-9 M
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Common Ion Effect and Solubility
Why does the molar solubility of AgCl decrease
after adding NaCl?
Understood in terms of LeChatelier’s principle:
NaCl(s) --> Na+ + Cl-
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Common Ion Effect and Solubility
Why does the molar solubility of AgCl decrease
after adding NaCl?
Understood in terms of LeChatelier’s principle:
NaCl(s) --> Na+ + ClAgCl(s) <=> Ag+ + Cl-
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Common Ion Effect and Solubility
Why does the molar solubility of AgCl decrease
after adding NaCl?
Understood in terms of LeChatelier’s principle:
NaCl(s) --> Na+ + ClAgCl(s) <=> Ag+ + ClCommon-Ion Effect
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pH and Solubility
How can pH influence solubility?
Solubility of “insoluble” salts will be affected by pH
changes if the anion of the salt is at least moderately
basic
Solubility increases as pH decreases
Solubility decreases as pH increases
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pH and Solubility
Salts contain either basic or neutral anions:
basic anions
Strong bases: OH-, O2Weak bases (conjugate bases of weak molecular acids):
F-, S2-, CH3COO-, CO32-, PO43-, C2O42-, CrO42-, etc.
Solubility affected by pH changes
neutral anions (conjugate bases of strong
monoprotic acids)
Cl-, Br-, I-, NO3-, ClO4Solubility not affected by pH changes
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pH and Solubility
Example:
Fe(OH)2
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
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pH and Solubility
Example:
Fe(OH)2-Add acid
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
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pH and Solubility
Example:
Fe(OH)2-Add acid
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O
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pH and Solubility
Example:
Fe(OH)2-Add acid
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O
Which way does this reaction shift the solubility equilibrium? Why?
Understood in terms of LeChatlier’s principle
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pH and Solubility
Example:
Fe(OH)2-Add acid
More Fe(OH)2 dissolves in response
Solubility increases
Decrease = stress
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
Stress relief = increase [OH-]
2H3O+(aq) + 2OH-(aq)  4H2O
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pH and Solubility
Example:
Fe(OH)2
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O(l)
overall Fe(OH)2(s) + 2H3O+(aq) <=> Fe2+(aq) + 4H2O(l)
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pH and Solubility
Example:
Fe(OH)2
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O(l)
overall Fe(OH)2(s) + 2H3O+(aq) <=> Fe2+(aq) + 4H2O(l)
decrease pH
solubility increases
increase pH
solubility decreases
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pH, Solubility, and Tooth Decay
Enamel (hydroxyapatite) = Ca10(PO4)6(OH)2
(insoluble ionic compound)
Ca10(PO4)6(OH)2  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
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pH, Solubility, and Tooth Decay
Enamel (hydroxyapatite) = Ca10(PO4)6(OH)2
(insoluble ionic compound)
weak base
strong base
Ca10(PO4)6(OH)2  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
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pH, Solubility, and Tooth Decay
+ food
metabolism
(Yummy)
organic acids
(H3O+)
bacteria in mouth
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pH, Solubility, and Tooth Decay
Ca10(PO4)6(OH)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
OH-(aq) + H3O+(aq)  2H2O(l)
PO43-(aq) + H3O+(aq)  HPO43-(aq) + H2O(l)
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pH, Solubility, and Tooth Decay
Solubility increases
Leads to tooth decay
Decrease = stress
More Ca10(PO4)6(OH)2 dissolves in response
Decrease = stress
Ca10(PO4)6(OH)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
OH-(aq) + H3O+(aq)  2H2O(l)
PO43-(aq) + H3O+(aq)  HPO43-(aq) + H2O(l)
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Tooth Decay
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pH, Solubility, and Tooth Decay
Why fluoridation?
F- replaces OH- in enamel
Ca10(PO4)6(F)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2F-(aq)
fluorapatite
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pH, Solubility, and Tooth Decay
Why fluoridation?
F- replaces OH- in enamel
Ca10(PO4)6(F)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2F-(aq)
weaker base than OHmore resistant to acid
attack
Less soluble (has
lower Ksp) than
Ca10(PO4)6(OH)2
Factors together fight tooth decay!
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pH, Solubility, and Tooth Decay
Why fluoridation?
F- replaces OH- in enamel
Ca10(PO4)6(F)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2F-(aq)
F- added to drinking water as NaF or Na2SiF6
1 ppm = 1 mg/L
F- added to toothpastes as SnF2, NaF, or Na2PO3F
0.1 - 0.15 % w/w
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Complex Ion Formation and Solubility
Metals act as Lewis acids (see Chapter 15)
Example
Fe3+(aq) + 6H2O(l)  Fe(H2O)63+(aq)
Lewis acid
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Lewis base
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Complex Ion Formation and Solubility
Metals act as Lewis acids (see Chapter 15)
Example
Fe3+(aq) + 6H2O(l)  Fe(H2O)63+(aq)
Complex ion
Complex ion/complex contains central metal ion bonded
to one or more molecules or anions called ligands
Lewis acid = metal
Lewis base = ligand
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Complex Ion Formation and Solubility
Metals act as Lewis acids (see Chapter 15)
Example
Fe3+(aq) + 6H2O(l)  Fe(H2O)63+(aq)
Complex ion
Complex ions are often water soluble
Ligands often bond strongly with metals
Kf >> 1: Equilibrium lies very far to right.
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Complex Ion Formation and Solubility
Metals act as Lewis acids (see Chapter 15)
Other Lewis bases react with metals also
Examples
Fe3+(aq) + 6CN-(aq)  Fe(CN)63-(aq)
Lewis acid
Lewis base
Complex ion
Ni2+(aq) + 6NH3(aq)  Ni(NH3)62+(aq)
Lewis acid
Lewis base
Complex ion
Ag+(aq) + 2S2O32-(aq)  Ag(S2O3)23-(aq)
Lewis acid
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Lewis base
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Complex ion
66
Complex-Ion Formation and Solubility
How does complex ion formation influence
solubility?
Solubility of “insoluble” salts increases with addition
of Lewis bases if the metal ion forms a complex with
the base.
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Complex-Ion Formation and Solubility
Example
AgCl
AgCl(s)  Ag+(aq) + Cl-(aq)
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Complex-Ion Formation and Solubility
Example
AgCl-Add NH3
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
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Complex-Ion Formation and Solubility
Example
AgCl-Add NH3
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
Which way does this reaction shift the solubility equilibrium? Why?
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Complex-Ion Formation and Solubility
Example
AgCl-Add NH3
More AgCl dissolves in response
Solubility increases
Decrease = stress
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
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Complex-Ion Formation and Solubility
Example
AgCl
overall
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
AgCl(s) + 2NH3(aq)  Ag(NH3)2+(aq) + Cl-(aq)
Addition of ligand
solubility increases
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Summary: Factors that Influence Solubility
Common Ion Effect
Decreases solubility
pH
pH decreases
Increases solubility
pH increases
Decreases solubility
Salt must have basic anion
Complex-Ion Formation
Increases solubility
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End of Presentation
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