slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Solubility Equilibria PowerPoint Presentation
Download Presentation
Solubility Equilibria

Loading in 2 Seconds...

play fullscreen
1 / 69

Solubility Equilibria - PowerPoint PPT Presentation


  • 186 Views
  • Uploaded on

Solubility Equilibria. Why Study Solubility Equilibria?. Many natural processes involve precipitation or dissolution of salts. A few examples: Dissolving of underground limestone deposits (CaCO 3 ) forms caves Note: Limestone is water “insoluble” (How can this be?)

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Solubility Equilibria' - upton-camacho


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
why study 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 equilibria1
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 equilibria2
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 equilibria3
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 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 compounds1

Fe(OH)3

Cr(OH)3

Solubility of Ionic Compounds

large excess added

+

NaOH

Fe3+

Precipitation of both Cr3+ and Fe3+ occurs

Cr3+

solubility of ionic compounds2
Solubility of Ionic Compounds

small excess added slowly

+

NaOH

Cr3+

Fe(OH)3

Fe3+

less soluble salt precipitates only

Cr3+

solubility of ionic compounds3
Solubility of Ionic Compounds
  • Solubility Rules
    • general rules for predicting the solubility of ionic compounds
    • strictly qualitative
      • Do not tell “how” soluble
      • Not quantitative
slide11

Solubility Equilibrium

My+

yAx-

saturated solution

xMy+

My+

Ax-

Ax-

solid

MxAy

solubility of ionic compounds4
Solubility of Ionic Compounds

Solubility Equilibrium

MxAy(s) <=> xMy+(aq) + yAx-(aq)

The equilibrium constant for this reaction is the solubilityproduct, Ksp:

Ksp = [My+]x[Ax-]y

solubility product k sp
Solubility Product, Ksp
  • Ksp is related to molar solubility
solubility product k sp1
Solubility Product, Ksp
  • Ksp is related to molar solubility
    • qualitative comparisons
solubility product k sp2
Solubility Product, Ksp
  • Ksp used to compare relative solubilities
    • smaller Ksp = less soluble
    • larger Ksp= more soluble
solubility product k sp3
Solubility Product, Ksp
  • Ksp is related to molar solubility
    • qualitative comparisons
    • quantitative calculations
calculations with k sp
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
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
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
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
  • Precipitation reaction
    • exchange reaction
      • one product is insoluble
  • Example

Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)

precipitation1
Precipitation
  • Precipitation reaction
    • exchange reaction
      • one product is insoluble
  • Example

Overall:CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)

Na+ and Ca2+ “exchange” anions

precipitation2
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)

precipitation3
Precipitation
  • Compare precipitation to solubility equilibrium

Ca2+(aq) + CO32-(aq) <=> CaCO3(s) prec.

vs

CaCO3(s) <=> Ca2+(aq) + CO32-(aq) sol. Equil.

saturated solution

precipitation4
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.

precipitation5
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
Predicting Precipitation
  • To determine if solution is supersaturated:
    • Compare ion product (Q or IP) to Ksp
      • ForMxAy(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 precipitation1
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
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 compounds5
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
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
Factors that Affect Solubility
  • Common Ion Effect
  • pH
  • Complex-Ion Formation
common ion effect and solubility
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 solubility1
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 Cl-

Cl- is common ion

example 6
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 solubility2
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

common ion effect and solubility3
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 solubility4
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-

AgCl(s) <=> Ag+ + Cl-

common ion effect and solubility5
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-

AgCl(s) <=> Ag+ + Cl-

Common-Ion Effect

ph and solubility
pH and Solubility
  • How can pH influence solubility?
    • Solubility of “insoluble” salts will be affected by pH changes if the anion of the salt isat least moderately basic
      • Solubility increases as pH decreases
      • Solubility decreases as pH increases
ph and solubility1
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 solubility2
pH and Solubility
  • Example:
    • Fe(OH)2

Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)

ph and solubility3
pH and Solubility
  • Example:
    • Fe(OH)2-Add acid

Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)

ph and solubility4
pH and Solubility
  • Example:
    • Fe(OH)2-Add acid

Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)

2H3O+(aq) + 2OH-(aq)  4H2O

ph and solubility5
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 solubility6
pH and Solubility
  • Example:
    • Fe(OH)2-Add acid

Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)

2H3O+(aq) + 2OH-(aq)  4H2O

More Fe(OH)2 dissolves in response

Solubility increases

Decrease = stress

Stress relief = increase [OH-]

ph and solubility7
pH and Solubility
  • Example:
    • Fe(OH)2

Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)

2H3O+(aq) + 2OH-(aq)  4H2O(l)

Fe(OH)2(s) + 2H3O+(aq) <=> Fe2+(aq) + 4H2O(l)

overall

ph and solubility8
pH and Solubility
  • Example:
    • Fe(OH)2

Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)

2H3O+(aq) + 2OH-(aq)  4H2O(l)

Fe(OH)2(s) + 2H3O+(aq) <=> Fe2+(aq) + 4H2O(l)

overall

decrease pH

solubility increases

increase pH

solubility decreases

ph solubility and tooth decay
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 decay1
pH, Solubility, and Tooth Decay

Enamel (hydroxyapatite) = Ca10(PO4)6(OH)2

(insoluble ionic compound)

strong base

weak base

Ca10(PO4)6(OH)2  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)

ph solubility and tooth decay2
pH, Solubility, and Tooth Decay

metabolism

+ food organic acids

(H3O+)

bacteria in mouth

ph solubility and tooth decay3
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 decay4
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)2dissolves in response

Solubility increases

Leads to tooth decay

Decrease = stress

Decrease = stress

ph solubility and tooth decay5
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

ph solubility and tooth decay6
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!

ph solubility and tooth decay7
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
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 solubility1
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 solubility2
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 solubility3
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)

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

Complex ion

Lewis base

Lewis acid

complex ion formation and solubility4
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 solubility5
Complex-Ion Formation and Solubility
  • Example
    • AgCl

AgCl(s)  Ag+(aq) + Cl-(aq)

complex ion formation and solubility6
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 solubility7
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 solubility8
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

complex ion formation and solubility9
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

summary factors that influence solubility
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