Transcript جلسه هفتم

Solubility Equilibria
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
– Precipitation of sodium urate, Na2C5H2N4O2, in
joints results in gouty arthritis.
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
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.
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.
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
Solubility of Ionic Compounds
large excess added
+
NaOH
Fe(OH)3 Cr(OH)3
Fe3+
Cr3+
Precipitation of both Cr3+
and Fe3+ occurs
Solubility of Ionic Compounds
small excess added
slowly
+
NaOH
Cr3+
Fe(OH)3
Fe3+
Cr3+
less soluble salt
precipitates only
Solubility of Ionic Compounds
• Solubility Rules
– general rules for predicting the solubility of ionic
compounds
– strictly qualitative
• Do not tell “how” soluble
• Not quantitative
Solubility Equilibrium
saturated
solution
solid
My+
xMy+
yAx-
My+
Ax-
AxMxAy
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
Solubility Product, Ksp
• Ksp is related to molar solubility
Solubility Product, Ksp
• Ksp is related to molar solubility
–qualitative comparisons
Solubility Product, Ksp
• Ksp used to compare relative
solubilities
–smaller Ksp = less soluble
–larger Ksp= more soluble
Solubility Product, Ksp
• Ksp is related to molar solubility
–qualitative comparisons
–quantitative calculations
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
Example 1
• 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)
Example 2
• 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)
Example 3
• 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)
Precipitation
• Precipitation reaction
– exchange reaction
• one product is insoluble
• Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)
Precipitation
• Precipitation reaction
–exchange reaction
• one product is insoluble
• Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) +
2NaCl(aq)
Na+ and Ca2+ “exchange” anions
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)
Precipitation
• Compare precipitation to solubility equilibrium
Ca2+(aq) + CO32-(aq) <=> CaCO3(s) prec.
vs
CaCO3(s) <=> Ca2+(aq) + CO32-(aq) sol. Equil.
saturated solution
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.
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)
Predicting Precipitation
• To determine if solution is supersaturated:
– Compare ion product (Q or IP) to Ksp
• For MxAy(s) <=> xMy+(aq) + yAx-(aq)
– Q = [My+]x[Ax-]y
– Q calculated for initial conditions
• Q > Ksp  supersaturated solution, precipitation
occurs, solubility equilibrium established (Q = Ksp)
– Q = Ksp  saturated solution, no precipitation
– Q < Ksp  unsaturated solution, no precipitation
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
Example 4
• 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)
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)
Example 5
• 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)
Factors that Affect Solubility
• Common Ion Effect
• pH
• Complex-Ion Formation
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)
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
Example 6
• 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)
Common Ion Effect and Solubility
• How does adding excess NaCl affect
the solubility equilibrium of AgCl?
1.3 x 10-5 M
+ 0.020 M NaCl
Molar solubility
Molar solubility
AgCl in H2O
AgCl in 0.020
M NaCl
8.5 x 10-9 M
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-
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-
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
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
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
pH and Solubility
• Example:
–Fe(OH)2
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
pH and Solubility
• Example:
–Fe(OH)2-Add acid
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
pH and Solubility
• Example:
–Fe(OH)2-Add acid
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O
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
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
pH and Solubility
• Example:
–Fe(OH)2
overall
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O(l)
Fe(OH)2(s) + 2H3O+(aq) <=> Fe2+(aq) + 4H2O(l)
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
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)
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)
pH, Solubility, and Tooth Decay
+ food
metabolism
organic acids
(H3O+)
bacteria in mouth
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)
pH, Solubility, and Tooth Decay
Solubility increases
Leads to tooth decay
OH-(aq) + H3O+(aq)  2H2O(l)
PO43-(aq) + H3O+(aq)  HPO43-(aq) + H2O(l)
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)
Tooth Decay
pH, Solubility, and Tooth Decay
• Why fluoridation?
–F- replaces OH- in enamel
Ca10(PO4)6(F)2(s)  10Ca2+(aq) + 6PO43-(aq)
(aq)
+
2F
fluorapatite
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 OHmore resistant to acid
attack
Factors together fight tooth decay!
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
Complex Ion Formation and Solubility
• Metals act as Lewis acids
• Example
Fe3+(aq) + 6H2O(l)  Fe(H2O)63+(aq)
Lewis acid
Lewis base
Complex Ion Formation and Solubility
• Metals act as Lewis acids
– 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
Complex Ion Formation and Solubility
• Metals act as Lewis acids
– 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.
Complex Ion Formation and Solubility
• Metals act as Lewis acids
– 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
Lewis base
Complex ion
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.
Complex-Ion Formation and Solubility
• Example
–AgCl
AgCl(s)  Ag+(aq) + Cl-(aq)
Complex-Ion Formation and Solubility
• Example
–AgCl Add NH3
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
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?
Complex-Ion Formation and Solubility
• Example
– AgCl-Add NH3
AgCl(s)  Ag+(aq) + Cl-(aq)
More AgCl dissolves in response
Solubility increases
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
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
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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