Chapter 15 solubility equilibrium
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Chapter 15: Solubility Equilibrium. Sections 15.6-15.8. Solubility of various compounds plays important roles. Sugar and table salt’s ability to dissolve in water help us to flavor food. Calcium sulfate is less soluble in hot water than cold water, which allows us to coat tubes in boilers.

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Chapter 15: Solubility Equilibrium

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Chapter 15 solubility equilibrium

Chapter 15: Solubility Equilibrium

Sections 15.6-15.8


15 6 solubility equilibria

Solubility of various compounds plays important roles.

Sugar and table salt’s ability to dissolve in water help us to flavor food.

Calcium sulfate is less soluble in hot water than cold water, which allows us to coat tubes in boilers.

The low solubility of barium sulfate allows a safe way to X-ray the gastrointestinal tract.

15.6 Solubility Equilibria


Aqueous equilibria

Aqueous Equilibria

  • Assume that when an ionic solid dissolves, that it dissociates completely.

    CaF2(s) → Ca2+ (aq) + 2F- (aq)

  • There is a possibility for the ions to collide and re-form the solid phase.

    Ca2+ (aq) + 2F- (aq)→ CaF2(s)

  • Finally, equilibrium is reached, and no more solid will dissolve (saturated).

    CaF2(s) ↔ Ca2+ (aq) + 2F- (aq)


Solubility product constant

Solubility Product Constant

  • Remember that solids are left out of equilibrium expressions

    Ksp=[Ca2+][F-]2

  • Kspis the solubility product constant (sometimes referred to as just the “solubility product”).

    *Notice that the amount of solid present nor the size of particles effect the equilibrium.


Solubility vs solubility product

Solubility is an equilibrium position.

Solubility product is an equilibrium constant and has ONE value for a particular solid at a particular temperature.

Solubility vs. Solubility Product


Calculating k sp from solubility

Calculating Ksp from Solubility

Calculate Ksp for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x 10-15mol/L at 25°C.

Ksp= [Bi3+]2[S2-]3

=(2.0 x 10-15)2 (3.0 x 10-15)3 = 1.1 x 10-73

HW 15.75-15.79 odd on page 743


Calculating solubility from k sp

Calculating Solubility from Ksp

The Ksp value from Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility.

Ksp=[Cu2+][IO3-]2=[x][2x]2=(x)(4x2)=4x3

1.4 x 10-7=4x3

X=3.3 x 10-3mol/L= Cu(IO3)2 solubility

HW 15.81 on page 743


Relative solubilities

Relative Solubilities

  • Use Ksp values to predict relativesolubilities of a group of salts.

    • Salts producing the same number of ions

      • Ex. AgI (s), CuI (s), CaSO4 (s)

      • Solubility is directly proportional to Kspbecause each compound produces 2 ions.

        • Largest Ksp= most soluble

        • Smallest Ksp= least soluble

  • Salts producing differing numbers of ions.

    • Ex. CuS (s), Ag2S (s), Bi2S3 (s)

    • Solubility NOT directly proportional to Ksp and you have to calculate the solubilities.


Practice problems

Practice Problems

  • HW 15.83,15.87 on page 743


Common ion effect

Common Ion Effect

Consider, the solubility of solid Ag2CrO4(Ksp=9.0x10-12) in a 0.100 M aqueous solution of AgNO3.

  • What is the common ion?

  • What is the [Ag+]0?

  • What is the [CrO4-2]0?

  • What is the Ksp expression?


Chapter 15 solubility equilibrium

Ksp = [Ag+]2[CrO4]2-

9.0 x 10-12 = (0.100+2x)2 (x)

*Since Ksp is small, x is considered negligible compared to 0.100 M; therefore 0.100+2x ≈ 0.100

9.0 x 10-12 = (0.100)2 (x)

X=9.0 x 10-10 M= solubility of Ag2CrO4 (s)

[Ag+]=0.100 M and [CrO4-]=9.0 x 10-10 M


Chapter 15 solubility equilibrium

Solubility of Ag2CrO4 in pure water = 1.3 x 10-4 M

Solubility of Ag2CrO4 in 0.100 M AgNO3=9.0x10-10 M

*Note that solubility of a solid is lowered due to the common ion effect.

Example 15.85 on page 743

HW 15.91 and 15.93 on page 743.


Ph and solubility

pH and Solubility

  • If the anion (X-) is an effective base (or HX is a weak acid), the salt MX will increase solubility in an acidic solution.

    • Effective bases include OH-, S2-, CO3-, C2O42-, and CrO42-.

    • Salts with these anions are more soluble in acidic solutions that pure water.

      HW 15.95 (in reference to 15.81) on page 743


Precipitation

Precipitation

  • Precipitation: the formation of a solid from solution

  • Ion Product (Q)=defined the same as Ksp except initial concentrations are used

    Ex. Ca(NO3)2 (aq) is mixed with NaF (aq). What is the ion product for CaF2?

    Q=[Ca2+]0[F-]02


Chapter 15 solubility equilibrium

To predict if precipitation will occur:

  • If Q>Ksp, precipitation occurs until concentrations are reduced to satisfy Ksp.

  • If Q<Ksp, no precipitation occurs.


Determining precipitation conditions

Determining Precipitation Conditions

A solution is prepared by adding 750.0 mL of 4.00 x 10-3 M Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 (Ksp=1.9 x 10-10)precipitate from this solution?

  • Calculate initial conditions.

  • Solve for Q.

    Q=[Ce3+]0[IO3-]03=(2.86x10-3)(5.71x10-3)3= 5.32 x 10-10

  • Compare Q to Ksp.

    Q>Ksp, so Ce(IO3)3 will precipitate.

    Example 15.97 on page 743 (Ksp values on page 718)


Precipitation and equilibrium concentrations

A solution is prepared by mixing 150.0 mL of 1.00x10-2 M Mg(NO3)2 and 250.0 mL of 1.00x10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp=6.4x10-9).

Precipitation and Equilibrium Concentrations


Chapter 15 solubility equilibrium

  • Find Initial Concentrations

    2. Find Q

    3. Compare Q vs. Ksp

    Q>Ksp, so solid MgF2 will form.

    * The next steps are used to determine equilibrium concentrations.


Chapter 15 solubility equilibrium

  • Run the reaction to completion. (BRA)

  • Calculate the concentration of excess reactant.

    [F-]excess=22.0 mmol / 400.0 mL = 5.50 x 10-2 M


Chapter 15 solubility equilibrium

  • Determine concentrations at equilibrium (ICE).

    Ksp= 6.4x10-9=[Mg2+][F-]2

    =(x)(5.50x10-2+2x)2≈ (x)(5.50x10-2)2

    X=2.1 x 10-6 M = [Mg2+]

    [F-]=5.50x10-2 M

    HW15.99 on page 743


Selective precipitation

Selective Precipitation

  • Using anions that form precipitates with only one or a few metal ions in a mixture in order to separate the metal ions.

  • Most insoluble sulfide salts can be precipitated in an acidic solutions.

  • Soluble sulfide salts can be precipitated by making the solution slightly basic.


Chapter 15 solubility equilibrium

A solution contains 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp=1.4x10-8) or CuI (Ksp=5.3x10-12) precipitate first? Specify the [I-] necessary to begin precipitation of each salt.

-For PbI2: Ksp=1.4x10-8=[Pb2+][I-]2

1.4x10-8=(2.0x10-3) [I-]2

[I-]=2.6 x 10-3 M is necessary to begin precipitation

-For CuI: Ksp=5.3x10-12=[Cu+][I-]

5.3x10-12=(1.0x10-4)[I-]

[I-]=5.3x10-8 M is necessary to begin precipitation

*CuI will precipitate first since [I-] required is less.

HW 15.101 on page 744


Qualitative analysis

Qualitative Analysis


Chapter 15 solubility equilibrium

Groups:

  • Insoluble Chlorides (Ag+, Pb2+, Hg22+)

  • Sulfides insoluble in acid solution (Hg2+, Cd2+, Bi3+, Cu2+, Sn4+)

  • Sulfides insoluble in basic solution (Co2+, Zn2+, Mn2+, Ni2+, Fe2+, Cr3+, Al3+)

  • Insoluble Carbonates (Group 2A)

  • Alkali Metal and ammonium ions (flame test)


Complex ion equilibria

Complex Ion Equilibria

  • Complex Ion: a charged species with a metal ion surrounded by ligands (Ag(NH3)+)

    • The ligand donates its lone electron pair to an empty orbital of the metal ion to form a covalent bond

  • Ligand: a Lewis base (H2O, NH3, Cl-, CN-)

  • Coordination number: the number of ligands attached to the metal ion


Chapter 15 solubility equilibrium

  • Formation constants or stability constants:

    • Metal ions add ligands one at a time in steps characterized by their own equilibrium constants

      Ag+ + NH3↔ Ag(NH3)+K1=2.1x103

      Ag(NH3)+ + NH3↔ Ag(NH3)2+K2=8.2x103

    • All species (Ag+, NH3, Ag(NH3)+, Ag(NH3)2+) exist at equilibrium.

      HW 15.107 on page 744


Chapter 15 solubility equilibrium

  • Usually, [ligand] is much larger than [metal ion] and approximations are used to simplify problems.

  • Assume both reactions go to completion

    Ag+ + NH3↔ Ag(NH3)+K1=2.1x103

    Ag(NH3)+ + NH3↔ Ag(NH3)2+K2=8.2x103

    Ag+ + 2 NH3 → Ag(NH3)2+K=K1 x K2

  • From here, use BRA to find concentrations and the equilibrium constant.

    • Assume the ligand amount consumed is neglible

      Examples 15.103a, 105, 109 on page 744

      HW 15.103 b, 107on page 744


Strategies for dissolving water insoluble ionic solids

Strategies for Dissolving Water-Insoluble Ionic Solids

  • If the anion of the solid is a good base, solubility is increased with the addition of an acid

  • If the anion is NOT a good base, solids can be dissolved in a solution containing a ligand to form stable complex ions with its cation.


Chapter 15 solubility equilibrium

Aqueous Ammonia is Added to Silver Chloride (white). Silver Chloride, Insoluble in Water, Dissolves to Form Ag(NH3)2+ (aq) and Cl-(aq)


Chapter 15 solubility equilibrium

Ex 15.111 on page 745

HW 15.113 and 15.115 on page 745


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