WOLPA/AP CHEMISTRY/CDO Chapter 19 PRECIPITATION REACTIONS

WOLPA/AP CHEMISTRY/CDO

Chapter 19PRECIPITATION

REACTIONS


Solubility of Ionic Solids

• Depends on the balance of two forces:

– Attraction between H2O molecules and ions of solid.

– Force of attraction between oppositely charged ions within solid.


Solubility Rules

• Use in predicting results of precipitation reactions. MEMORIZE THE SOLUBILITY RULES!!!!!– Determine ions present and possible

products.– Use solubility rules to determine if any are

insoluble.


Example

• Ba(NO3)2(aq) + Na2CO3(aq)


Solubility Rules• 1. Salts containing Group I elements are soluble (Li+, Na+,

K+, Cs+, Rb+). Exceptions to this rule are rare. Salts containing the ammonium ion (NH4

+) are also soluble. •

2. Salts containing nitrate ion (NO3-) are generally soluble.

•3. Salts containing Cl-, Br-, I- are generally soluble. Important exceptions to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+. Thus, AgCl, PbBr2, and Hg2Cl2 are all insoluble.

•


Solubility Rules Continued• 4. Most silver salts are insoluble. AgNO3 and Ag(C2H3O2) are

common soluble salts of silver; virtually anything else is insoluble. •

5. Most sulfate salts are soluble. Important exceptions to this rule include BaSO4, PbSO4, Ag2SO4 and SrSO4.

• 6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are not soluble.


Solubility Rules Continued• 7. Most sulfides of transition metals are highly insoluble. Thus,

CdS, FeS, ZnS, Ag2S are all insoluble. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.

•8. Carbonates are frequently insoluble. Group II carbonates (Ca, Sr, and Ba) are insoluble. Some other insoluble carbonates include FeCO3 and PbCO3.

•9. Chromates are frequently insoluble. Examples: PbCrO4, BaCrO4

•10. Phosphates are frequently insoluble. Examples: Ca3(PO4)2, Ag2PO4

• 11. Fluorides are frequently insoluble. Examples: BaF2, MgF2 PbF2.


Example Continued

• Ba(NO3)2(aq) + Na2CO3(aq)


Stoichiometry

• Mole Relations– Coefficients in the net ionic equation can be

used in the usual way to relate the moles of reactants and products.

– Moles of ions can be deduced from solute concentrations.


Example

• What is the molar concentration of Ba2+ and F- in a solution containing 0.0075 M BaF2


Precipitation Titrations

• Used to determine the concentration of species in solution or in a solid mixture.

• Indicator shows, usually by color change, when the species being analyzed for has been consumed


General Principles

• Involves formation of a precipitate• Must determine the volume of a

standardized titrant needed to justprecipitate all of the ion.

• Need an indicator or electrode todetermine when the precipitation iscomplete


Solubility Equilibria

• Solubility Product Constant, Ksp– Precipitation reactions like all reactions, reach

a position of equilibrium.

– Expression for Ksp

MaXb <----------> aM+b + bX-a

Ksp = [M+b]a [X-a]b


Solubility Product Principle

• In any water solution in equilibrium with a slightly soluble ionic compound, the product of the concentrations of its ions, each raised to a power equal to its coefficient in the solubility equation is a constant. This constant, Ksp, has a fixed value at a given temperature, independent of the concentrations of the individual ions.


Two Ion Compound

• AgCl


Three Ion Compound

• PbCl2


Four Ion Compound

• Al(OH)3


Calculation of Ksp

• Calculated from measured solubility– AgCl Ksp = (s) (s) = s2

– PbCl2 Ksp = (s) (2s)2 = 4s3

– Al(OH)3 Ksp = (s) (3s)3 = 27s4

•


Example

• At 20 oC, a saturated aqueous solution of silver acetate, AgC2H3O2, contains 1.0 g dissolved in 100.0 mL of solution. Calculate the Ksp for AgC2H3O2.


Determination of Solubility

• In pure water

– Ksp = s2 s = (Ksp)1/2

– Ksp = 4s3 s = (Ksp/4)1/3

– Ksp = 27s4 s = (Ksp/27)1/4


Example

• Estimate the solubility of lead (II) bromide in (a) moles per liter and (b) grams per liter of pure water. Ksp = 6.3 x 10-6


Uses of Ksp

• Calculation of concentration of one ion, knowing that of the other


Example

• You have a solution that has a lead (II) concentration of 0.0012 M. What is the maximum concentration of chloride ions that would be present? Ksp = 1.7 x 10-5


Uses of Ksp

• Determination of whether a precipitate will form

• Compare original concentration product, P, to Ksp– if P < Ksp, no precipitate will form– if P > Ksp, precipitate forms until P becomes

equal to Ksp


Example

• You have 100.0 m of a solution that has a lead (II) concentration of 0.0012 M. Does PbCl2 precipitate when 1.20 g of solid NaCl is added?


Determination of Solubility

• In solution containing a common ion

– Solubility is much less than in pure water


Example

• Calculate the solubility of silver carbonate, Ag2CO3, in moles per liter, in pure water. Compare this with the molar solublity of Ag2CO3 in 225 mL of water to which 0.15 g of Na2CO3 has been added.


Simultaneous Equilibria

• Two or more reactions occur at the same time is a solution, all of them being described as equilibrium processes.

• The equilibrium for the overall reaction is the product of the equilibrium constants for the summed reactions.

That is Knet = K1 x K2


Solubility and pH

• Any salt containing an anion that is the conjugate base of a weak acid dissolves in water to a greater extent than that given by Ksp because the ions undergo a hydrolysis reaction.


Example

PbS• PbS(s) <----------> Pb2+(aq) + S2-(aq)

Ksp = 8.4 x 10-28

• S2-(aq) + H2O(l) <----------> HS-(aq) + OH-(aq) Kb = 1 x 105

• Overall• PbS(s) + H2O(l) <---> Pb2+(aq) + HS-(aq) +

OH-(aq) Knet = 8.4 x 10-23


• In general, the solubility of a salt containing the conjugate base of a weak acid is increased by addition of a stronger acid to the solution. In contrast, the salts are not soluble in strong acid if the anion is the conjugate base of a strong acid.


• CaCO3 <--------> Ca2+(aq)+ CO32-(aq)

K = Ksp = 3.8 x 10-9

• CO32-(aq) + H2O(l) <--------> HCO3

-(aq) + OH-(aq) K = Kb = 2.1 x 10-4

• OH-(aq) + H3O+(aq) <--------> 2H2O(l)K = 1/Kw = 1 x 1014


• NET

• CaCO3(s) + H3O+(aq) <------> Ca2+(aq) + HCO3

-(aq) + H2O(l)

• Knet = (Ksp) (Kb)/(Kw) = 79.8


Solubility and Complex Ions

• Examples of complex ions: AgCl2-, Ag(S2O3)23-, Ag(CN)2

-

• The solubility of certain “insoluble “ compounds can be increased when a complex ion is formed. Complex ions usually refer to cations in which surrounding water molecules have been replaced by some other electron pair donor. The equilibrium constant will equal the solubility constant times the formation constant for the complex ions. (Note - Chapter 23 in your textbook covers complex ions - their formation and nomenclature)


Solubility and Complex ions

• AgCl(s) + 2NH3 Ag(NH3)2+ + Cl-

• K = (Ksp) (Kf)

• = (1.8 x 10-10) (1.6 x 107) = 0.00288 = 0.0029


Example

• Solid gold (I) chloride AuCl, is dissolved when excess cyanide ions, CN-, are added to give a wter souble complex ion.

• AuCl(s) + 2CN- (aq) Au(CN)-(aq) + Cl-

• Show that this equation is the sum of twp other equations and calculate the Knet.


Solubility, Ion Separations, and Qualitative Analysis

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WOLPA/AP CHEMISTRY/CDO Chapter 19 PRECIPITATION REACTIONS