Chapter 17b
Download
1 / 78

Chapter 17b - PowerPoint PPT Presentation


  • 192 Views
  • Uploaded on

Chapter 17b. Ionic Equilibria III: The Solubility Product Principle. Chapter Goals. Solubility Product Constants Determination of Solubility Product Constants Uses of Solubility Product Constants Fractional Precipitation Simultaneous Equilibria Involving Slightly Soluble Compounds

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 ' Chapter 17b' - neola


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
Chapter 17b

Chapter 17b

Ionic Equilibria III:

The Solubility Product Principle


Chapter goals
Chapter Goals

  • Solubility Product Constants

  • Determination of Solubility Product Constants

  • Uses of Solubility Product Constants

  • Fractional Precipitation

  • Simultaneous Equilibria Involving Slightly Soluble Compounds

  • Dissolving Precipitates


Solubility product constants
Solubility Product Constants

  • Silver chloride, AgCl,is rather insoluble in water.

  • Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.


Solubility product constants1
Solubility Product Constants

  • Silver chloride, AgCl,is rather insoluble in water.

  • Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.


Solubility product constants2
Solubility Product Constants

  • The equilibrium constant expression for this dissolution is called a solubility product constant.

    • Ksp = solubility product constant


Solubility product constants3
Solubility Product Constants

  • The solubility product constant, Ksp, for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound.

  • Consider the dissolution of silver sulfide in water.


Solubility product constants4
Solubility Product Constants

  • The solubility product expression for Ag2S is:


Solubility product constants5
Solubility Product Constants

  • The dissolution of solid calcium phosphate in water is represented as:

  • The solubility product constant expression is:

    You do it!


Solubility product constants6
Solubility Product Constants

  • In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as:


Solubility product constants7
Solubility Product Constants

  • The same rules apply for compounds that have more than two kinds of ions.

  • One example of a compound that has more than two kinds of ions is calcium ammonium phosphate.


Determination of solubility product constants
Determination of Solubility Product Constants

  • Example 20-1: One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl.

  • The molar solubility can be easily calculated from the data:


Determination of solubility product constants1
Determination of Solubility Product Constants

  • The equation for the dissociation of silver chloride, the appropriate molar concentrations, and the solubility product expression are:


Determination of solubility product constants2
Determination of Solubility Product Constants

  • Substitution of the molar concentrations into the solubility product expression gives:


Determination of solubility product constants3
Determination of Solubility Product Constants

  • Example 20-2: One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF2 at 25oC. Calculate the molar solubility of, and Ksp for, CaF2.

  • Calculate the molar solubility of CaF2.


Determination of solubility product constants4
Determination of Solubility Product Constants

  • From the molar solubility, we can find the ion concentrations in saturated CaF2. Then use those values to calculate the Ksp.

    • Note: You are most likely to leave out the factor of 2 for the concentration of the fluoride ion!


Uses of solubility product constants
Uses of Solubility Product Constants

  • The solubility product constant can be used to calculate the solubility of a compound at 25oC.

  • Example 20-3: Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC. For barium sulfate, Ksp= 1.1 x 10-10.


Uses of solubility product constants1
Uses of Solubility Product Constants


Uses of solubility product constants2
Uses of Solubility Product Constants

  • Make the algebraic substitution of x’s into solubility product expression and solve for x, giving the ion concentrations.


Uses of solubility product constants3
Uses of Solubility Product Constants

  • Finally, to calculate the mass of BaSO4 in 1.00 L of saturated solution, use the definition of molarity.


Uses of solubility product constants4
Uses of Solubility Product Constants

  • Example 20-4: The solubility product constant for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar solubility of magnesium hydroxide and the pH of a saturated magnesium hydroxide solution at 25oC.

    You do it!


Uses of solubility product constants5
Uses of Solubility Product Constants

  • Be careful, do not forget the stoichiometric coefficient of 2!


Uses of solubility product constants6
Uses of Solubility Product Constants

  • Substitute the algebraic expressions into the solubility product expression.


Uses of solubility product constants7
Uses of Solubility Product Constants

  • Solve for the pOH and pH.


The common ion effect in solubility calculations
The Common Ion Effect in Solubility Calculations

  • Example 20-5: Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC. Compare this to the solubility of BaSO4 in pure water. (Example 20-3). (What is the common ion? How was a common ion problem solved in Chapter 19?)


The common ion effect in solubility calculations1
The Common Ion Effect in Solubility Calculations

  • Write equations to represent the equilibria.


The common ion effect in solubility calculations2
The Common Ion Effect in Solubility Calculations

  • Substitute the algebraic representations of the concentrations into the Ksp expression and solve for x.


The common ion effect in solubility calculations3
The Common Ion Effect in Solubility Calculations

  • The molar solubility of BaSO4 in 0.010 M Na2SO4 solution is 1.1 x 10-8M.

  • The molar solubility of BaSO4 in pure water is 1.0 x 10-5M.

    • BaSO4 is 900 times more soluble in pure water than in 0.010 M sodium sulfate!

    • Adding sodium sulfate to a solution is a fantastic method to remove Ba2+ ions from solution!

  • If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.


The reaction quotient in precipitation reactions
The Reaction Quotient in Precipitation Reactions

  • The reaction quotient, Q, and the Ksp of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form.

  • Example 20-6: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form?


The reaction quotient in precipitation reactions1
The Reaction Quotient in Precipitation Reactions

  • Write out the solubility expressions.


The reaction quotient in precipitation reactions2
The Reaction Quotient in Precipitation Reactions

  • Calculate the Qsp for PbSO4.

    • Assume that the solution volumes are additive.

    • Concentrations of the important ions are:


The reaction quotient in precipitation reactions3
The Reaction Quotient in Precipitation Reactions

  • Finally, calculate Qsp for PbSO4 and compare it to the Ksp.


The reaction quotient in precipitation reactions4
The Reaction Quotient in Precipitation Reactions

  • Example 20-7: Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO3)2. We can do this by precipitating mercury (II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8M? For HgS, Ksp=3.0 x 10-53.

    You do it!



The reaction quotient in precipitation reactions6
The Reaction Quotient in Precipitation Reactions

  • Example 20-8: Refer to example 20-7. What volume of the solution (1.0 x 10-8M Hg2+ ) contains 1.0 g of mercury?


Fractional precipitation
Fractional Precipitation

  • The method of precipitating some ions from solution while leaving others in solution is called fractional precipitation.

    • If a solution contains Cu+, Ag+, and Au+, each ion can be precipitated as chlorides.


Fractional precipitation1
Fractional Precipitation

  • Example 20-9: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu+, Ag+, and Au+ ions, which compound precipitates first? Calculate the concentration of Cl- required to initiate precipitation of each of these metal chlorides.



Fractional precipitation3
Fractional Precipitation

  • Repeat the calculation for silver chloride.


Fractional precipitation4
Fractional Precipitation

  • Finally, for copper (I) chloride to precipitate.


Fractional precipitation5
Fractional Precipitation

  • These three calculations give the [Cl-] required to precipitate AuCl ([Cl-] >2.0 x 10-11 M), to precipitate AgCl ([Cl-] >1.8 x 10-8 M), and to precipitate CuCl ([Cl-] >1.9 x 10-5 M).

  • It is also possible to calculate the amount of Au+ precipitated before the Ag+ begins to precipitate, as well as the amounts of Au+ and Ag+ precipitated before the Cu+ begins to precipitate.


Fractional precipitation6
Fractional Precipitation

  • Example 20-10: Calculate the percentage of Au+ ions that precipitate before AgCl begins to precipitate.

    • Use the [Cl-] from Example 20-9 to determine the [Au+] remaining in solution just before AgCl begins to precipitate.


Fractional precipitation7

The percent of Au+ ions unprecipitated just before AgCl precipitates is:

Therefore, 99.9% of the Au+ ions precipitates before AgCl begins to precipitate.

Fractional Precipitation


Fractional precipitation8
Fractional Precipitation

  • A similar calculation for the concentration of Ag+ ions unprecipitated before CuCl begins to precipitate is:


Fractional precipitation9

The percent of Ag+ ions unprecipitated just before AgCl precipitates is:

Thus, 99.905% of the Ag+ ions precipitates before CuCl begins to precipitate.

Fractional Precipitation


Simultaneous equilibria involving slightly soluble compounds
Simultaneous Equilibria Involving Slightly Soluble Compounds

  • Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.


Simultaneous equilibria involving slightly soluble compounds1
Simultaneous Equilibria Involving Slightly Soluble Compounds

  • Example 20-12: If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution?

    • For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5.

  • Calculate Qsp for Mg(OH)2 and compare it to Ksp.

    • Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M.

    • Aqueous ammonia is a weak base that we can calculate [OH-].



Simultaneous equilibria involving slightly soluble compounds3
Simultaneous Equilibria Involving Slightly Soluble Compounds

  • Once the concentrations of both the magnesium and hydroxide ions are determined, the Qsp can be calculated and compared to the Ksp.


Simultaneous equilibria involving slightly soluble compounds4
Simultaneous Equilibria Involving Slightly Soluble Compounds

  • Example 20-13: How many moles of solid ammonium chloride, NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO3)2 ? (Note the similarity between this problem and Example 20-12.)

    You do it!

  • Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+.



Simultaneous equilibria involving slightly soluble compounds6
Simultaneous Equilibria Involving Slightly Soluble Compounds

  • Using the maximum [OH-] that can exist in solution, determine the number of moles of NH4Cl required to buffer 0.10 M aqueous ammonia so that the [OH-] does not exceed 3.9 x 10-5M.




Simultaneous equilibria involving slightly soluble compounds9
Simultaneous Equilibria Involving Slightly Soluble Compounds

  • Check these values by calculating Qsp for Mg(OH)2.


Simultaneous equilibria involving slightly soluble compounds10
Simultaneous Equilibria Involving Slightly Soluble Compounds

  • Use the ion product for water to calculate the [H+] and the pH of the solution.


Dissolving precipitates
Dissolving Precipitates

  • For an insoluble solid, if the ion concentrations (of either the cation or anion) are decreased, a solid precipitate can be dissolved.

    • The trick is to make Qsp < Ksp.

  • One method is to convert the ions into weak electrolytes.

    • Make these ions more water soluble.

  • If insoluble metal hydroxides are dissolved in strong acids, they form soluble salts and water.


Dissolving precipitates1
Dissolving Precipitates

  • For example, look at the dissolution of Mg(OH)2 in HCl.


Dissolving precipitates2
Dissolving Precipitates

  • A second method is to dissolve insoluble metal carbonates in strong acids.

    • The carbonates will form soluble salts, carbon dioxide, and water.


Dissolving precipitates3
Dissolving Precipitates

  • A third method is to convert an ion to another species by an oxidation-reduction reaction.

  • For example, the dissolution of insoluble metal sulfides in hot nitric acid causes the sulfide ions to be oxidized to elemental sulfur.


Dissolving precipitates4
Dissolving Precipitates

  • A fourth method is complex ion formation.

  • The cations in many slightly soluble compounds will form complex ions.

    • This is the method used to dissolve unreacted AgBr and AgCl on photographic film.

      • Photographic “hypo” is Na2S2O3.


Dissolving precipitates5
Dissolving Precipitates

  • Copper(II) hydroxide, which is light blue colored, dissolves in aqueous ammonia to form dark blue, [Cu(NH3)4]2+.


Complex ion equilibria
Complex Ion Equilibria

  • A metal ion coordinated to several neutral molecules or anions forms compounds called complex ions.

  • Familiar examples of complex ions include:


Complex ion equilibria1
Complex Ion Equilibria

  • The dissociation of complex ions can be represented similarly to equilibria.

  • For example:


Complex ion equilibria2
Complex Ion Equilibria

  • Complex ion equilibrium constants are called dissociation constants.

  • Example 20-14: Calculate the concentration of silver ions in a solution that is 0.010 M in [Ag(NH3)2]+.

    • Kd = 6.3 x 10-8

  • Write the dissociation reaction and equilibrium concentrations.


Complex ion equilibria3
Complex Ion Equilibria

  • Substitute the algebraic quantities into the dissociation expression.




Complex ion equilibria6
Complex Ion Equilibria

  • Example 20-15: How many moles of ammonia must be added to 2.00 L of water so that it will just dissolve 0.010 mole of silver chloride, AgCl?

  • The reaction of interest is:


Complex ion equilibria7
Complex Ion Equilibria

  • Two equilibria are involved when silver chloride dissolves in aqueous ammonia.


Complex ion equilibria8
Complex Ion Equilibria

  • The [Ag+] in the solution must satisfy both equilibrium constant expressions. Because the [Cl-] is known, the equilibrium concentration of Ag+ can be calculated from Ksp for AgCl.



Complex ion equilibria10
Complex Ion Equilibria

  • Substitute the maximum [Ag+] into the dissociation constant expression for [Ag(NH3)2]+ and solve for the equilibrium concentration of NH3.


Complex ion equilibria11
Complex Ion Equilibria

  • The amount just calculated is the equilibrium concentration of NH3 in the solution. But the total concentration of NH3 is the equilibrium amount plus the amount used in the complex formation.


Complex ion equilibria12
Complex Ion Equilibria

  • Finally, calculate the total number of moles of ammonia necessary.


Synthesis question
Synthesis Question

  • Most kidney stones are made of calcium oxalate, Ca(O2CCO2). Patients who have their first kidney stones are given an extremely simple solution to stop further stone formation. They are told to drink six to eight glasses of water a day. How does this stop kidney stone formation?



Group question
Group Question

  • The cavities that we get in our teeth are a result of the dissolving of the material our teeth are made of, calcium hydroxy apatite. How does using a fluoride based toothpaste decrease the occurrence of cavities?


End of chapter 20
End of Chapter 20

  • All of the solid structures in humans, teeth, bones, etc., are examples of the solubility product principle.


ad