CH160 General Chemistry II Lecture Presentation Solubility Equilibria

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CH160 General Chemistry IILecture
PresentationSolubility Equilibria

• 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/dissol
  ution 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
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
My
yAx-
saturated solution
xMy
My
Ax-
Ax-
solid
MxAy
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Solubility of Ionic Compounds

• Solubility Equilibrium
• MxAy(s) ltgt xMy(aq) yAx-(aq)
• The equilibrium constant for this reaction is the
  solubility product, Ksp
• Ksp MyxAx-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) --gt CaCO3(s)
  2NaCl(aq)

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Precipitation

• Precipitation reaction
• exchange reaction
• one product is insoluble
• Example
• Overall CaCl2(aq) Na2CO3(aq) --gt 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) --gt CaCO3(s)
  2NaCl(aq)
• Net Ionic Ca2(aq) CO32-(aq) ltgt CaCO3(s)

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Precipitation

• Compare precipitation to solubility equilibrium
• Ca2(aq) CO32-(aq) ltgt CaCO3(s) prec.
• vs
• CaCO3(s) ltgt Ca2(aq) CO32-(aq) sol. Equil.

saturated solution
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Precipitation

• Compare precipitation to solubility equilibrium
• Ca2(aq) CO32-(aq) ltgt CaCO3(s)
• vs
• CaCO3(s) ltgt Ca2(aq) CO32-(aq)

saturated solution
Precipitation occurs until solubility equilibrium
is established.
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Precipitation

• Ca2(aq) CO32-(aq) ltgt CaCO3(s)
• vs
• CaCO3(s) ltgt 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) ltgt xMy(aq) yAx-(aq)
• Q MyxAx-y
• Q calculated for initial conditions
• Q gt Ksp ? supersaturated solution, precipitation
  occurs, solubility equilibrium established (Q
  Ksp)
• Q Ksp ? saturated solution, no precipitation
• Q lt Ksp ? unsaturated solution, no precipitation

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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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Problem Solving Strategy

• Precipitation does not occur until Q exceeds Ksp.
  (Q gt Ksp)
• We need to add enough Ag to make the solution
  supersaturated
• Use the saturated solution (Q Ksp) as a
  reference point
• Calculate the Ag needed to give a saturated
  solution.
• Add more Ag than this to give a precipitate

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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

These sure sound familiar. Where have I seen
them before?
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Common Ion Effect and Solubility

• Consider the solubility equilibrium of AgCl.
• AgCl(s) ltgt 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) ltgt Ag(aq) Cl-(aq)
• How does adding excess NaCl affect the solubility
  equilibrium?
• NaCl(s) ? Na(aq) Cl-(aq)

2 sources of Cl- Cl- 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?

AgCl in H2O
1.3 x 10-5 M
0.020 M NaCl
Molar solubility
AgCl in 0.020 M NaCl
Molar solubility
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 LeChateliers principle
• NaCl(s) --gt 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 LeChateliers principle
• NaCl(s) --gt Na Cl-
• AgCl(s) ltgt 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 LeChateliers principle
• NaCl(s) --gt Na Cl-
• AgCl(s) ltgt Ag Cl-

Increase stress
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Common Ion Effect and Solubility

• Why does the molar solubility of AgCl decrease
  after adding NaCl?
• Understood in terms of LeChateliers principle
• NaCl(s) --gt Na Cl-
• AgCl(s) ltgt Ag Cl-

Increase stress
Stress relief remove some 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 LeChateliers principle
• NaCl(s) --gt Na Cl-
• AgCl(s) ltgt Ag Cl-

reacts w/some Cl-
Reverse reaction removes some excess
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Common Ion Effect and Solubility

• Why does the molar solubility of AgCl decrease
  after adding NaCl?
• Understood in terms of LeChateliers principle
• NaCl(s) --gt Na Cl-
• AgCl(s) ltgt Ag Cl-

Shifts towards reactants Equilibrium
reestablished More AgCl present less dissolved
lower solubility
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Common Ion Effect and Solubility

• Why does the molar solubility of AgCl decrease
  after adding NaCl?
• Understood in terms of LeChateliers principle
• NaCl(s) --gt Na Cl-
• AgCl(s) ltgt Ag Cl-

Common-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-, O2-
• Weak 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-, ClO4-
• Solubility not affected by pH changes

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pH and Solubility

• Example
• Fe(OH)2
• Fe(OH)2(s) ltgt Fe2(aq) 2OH-(aq)

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pH and Solubility

• Example
• Fe(OH)2-Add acid
• Fe(OH)2(s) ltgt Fe2(aq) 2OH-(aq)

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pH and Solubility

• Example
• Fe(OH)2-Add acid
• Fe(OH)2(s) ltgt 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) ltgt Fe2(aq) 2OH-(aq)
• 2H3O(aq) 2OH-(aq) ? 4H2O

Which way does this reaction shift the solubility
equilibrium? Why? Understood in terms of
LeChatliers principle
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pH and Solubility

• Example
• Fe(OH)2-Add acid
• Fe(OH)2(s) ltgt Fe2(aq) 2OH-(aq)
• 2H3O(aq) 2OH-(aq) ? 4H2O

More Fe(OH)2 dissolves in response Solubility
increases
Decrease stress
Stress relief increase OH-
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pH and Solubility

• Example
• Fe(OH)2
• Fe(OH)2(s) ltgt Fe2(aq) 2OH-(aq)
• 2H3O(aq) 2OH-(aq) ? 4H2O(l)
• Fe(OH)2(s) 2H3O(aq) ltgt Fe2(aq) 4H2O(l)

overall
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pH and Solubility

• Example
• Fe(OH)2
• Fe(OH)2(s) ltgt Fe2(aq) 2OH-(aq)
• 2H3O(aq) 2OH-(aq) ? 4H2O(l)
• Fe(OH)2(s) 2H3O(aq) ltgt Fe2(aq) 4H2O(l)

overall
decrease pH
solubility increases
increase pH
solubility decreases
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pH, Solubility, and Tooth Decay
Enamel (hydroxyapatite) Ca10(PO4)6(OH)2 (insolub
le 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 (insolub
le ionic compound)
strong base
weak base
Ca10(PO4)6(OH)2 ? 10Ca2(aq) 6PO43-(aq)
2OH-(aq)
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pH, Solubility, and Tooth Decay
metabolism
food organic acids
(Yummy)
(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
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)
More Ca10(PO4)6(OH)2 dissolves in
response Solubility increases Leads to tooth decay
Decrease stress
Decrease stress
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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)

Less soluble (has lower Ksp) than Ca10(PO4)6(OH)2
weaker base than OH- more resistant to acid attack
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
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 gtgt 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)
• Ni2(aq) 6NH3(aq) ? Ni(NH3)62(aq)
• Ag(aq) 2S2O32-(aq) ? Ag(S2O3)23-(aq)

Lewis acid
Lewis base
Complex ion
Lewis acid
Lewis base
Complex ion
Lewis acid
Lewis base
Complex ion
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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
• AgCl(s) ? Ag(aq) Cl-(aq)
• Ag(aq) 2NH3(aq) ? Ag(NH3)2(aq)

More AgCl dissolves in response Solubility
increases
Decrease stress
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Complex-Ion Formation and Solubility

• Example
• AgCl
• AgCl(s) ? Ag(aq) Cl-(aq)
• Ag(aq) 2NH3(aq) ? Ag(NH3)2(aq)
• AgCl(s) 2NH3(aq) ? Ag(NH3)2(aq) Cl-(aq)

overall
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