Download Chapter Sixteen

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Chapter Sixteen
More Equilibria in Aqueous
Solutions:
Slightly Soluble Salts and Complex Ions
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Chapter Sixteen
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The Solubility Product Constant, Ksp
• Many important ionic compounds are only slightly soluble
in water (we used to call them “insoluble” – Chapter 4).
• An equation can represent the equilibrium between the
compound and the ions present in a saturated aqueous
solution:
BaSO4(s)
Ba2+(aq) + SO42–(aq)
• Solubility product constant, Ksp: the equilibrium constant
expression for the dissolving of a slightly soluble solid.
Ksp = [ Ba2+ ][ SO42–]
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Chapter Sixteen
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Chapter Sixteen
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Example 16.1
Write a solubility product constant expression for
equilibrium in a saturated aqueous solution of the
slightly soluble salts (a) iron(III) phosphate, FePO4,
and (b) chromium(III) hydroxide, Cr(OH)3.
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Ksp and Molar Solubility
• Ksp is an equilibrium constant
• Molar solubility is the number of moles of compound that
will dissolve per liter of solution.
• Molar solubility is related to the value of Ksp, but molar
solubility and Ksp are not the same thing.
• In fact, “smaller Ksp” doesn’t always mean “lower molar
solubility.”
• Solubility depends on both Ksp and the form of the
equilibrium constant expression.
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Example 16.2
At 20 °C, a saturated aqueous solution of silver carbonate
contains 32 mg of Ag2CO3 per liter of solution. Calculate
Ksp for Ag2CO3 at 20 °C. The balanced equation is
Ag2CO3(s)
2 Ag+(aq) + CO32–(aq)
Ksp = ?
Example 16.3
From the Ksp value for silver sulfate, calculate its molar
solubility at 25 °C.
Ag2SO4(s)
2 Ag+(aq) + SO42–(aq)
Ksp = 1.4 x 10–5 at 25 °C
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Chapter Sixteen
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Example 16.4
A Conceptual Example
Without doing detailed calculations, but using data from
Table 16.1, establish the order of increasing solubility of
these silver halides in water: AgCl, AgBr, AgI.
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Chapter Sixteen
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The Common Ion Effect
in Solubility Equilibria
• The common ion effect affects solubility equilibria
as it does other aqueous equilibria.
• The solubility of a slightly soluble ionic
compound is lowered when a second solute that
furnishes a common ion is added to the solution.
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Common Ion Effect Illustrated
The added sulfate
ion reduces the
solubility of
Ag2SO4.
Na2SO4(aq)
Saturated
Ag2SO4(aq)
Ag2SO4
precipitates
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Common Ion Effect Illustrated
When Na2SO4(aq)
is added to the
saturated solution
of Ag2SO4 …
… [Ag+] attains a new,
lower equilibrium
concentration as Ag+
reacts with SO42– to
produce Ag2SO4.
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Example 16.5
Calculate the molar solubility of Ag2SO4 in 1.00 M
Na2SO4(aq).
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Solubility and Activities
• Ions that are not common to the precipitate can
also affect solubility.
– CaF2 is more soluble in 0.010 M Na2SO4 than it is in
water.
• Increased solubility occurs because of interionic
attractions.
• Each Ca2+ and F– is surrounded by ions of
opposite charge, which impede the reaction of
Ca2+ with F–.
• The effective concentrations, or activities, of Ca2+
and F– are lower than their actual concentrations.
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Chapter Sixteen
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Will Precipitation Occur? Is It
Complete?
• Qip is the ion product reaction quotient and is based on
initial conditions of the reaction.
Qip and Qc: new look,
same great taste!
• Qip can then be compared to Ksp.
• Precipitation should occur if Qip > Ksp.
• Precipitation cannot occur if Qip < Ksp.
• A solution is just saturated if Qip = Ksp.
• In applying the precipitation criteria, the effect of dilution
when solutions are mixed must be considered.
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Chapter Sixteen
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Example 16.6
If 1.00 mg of Na2CrO4 is added to 225 mL of 0.00015 M
AgNO3, will a precipitate form?
Ag2CrO4(s)
2 Ag+(aq) + CrO42–(aq) Ksp = 1.1 x 10–12
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Example 16.7
A Conceptual Example
Pictured here is the result of adding a few drops of
concentrated KI(aq) to a dilute solution of Pb(NO3)2.
What is the solid that first appears? Explain why it
then disappears.
Example 16.8
If 0.100 L of 0.0015 M MgCl2 and 0.200 L of 0.025 M NaF are
mixed, should a precipitate of MgF2 form?
MgF2(s)
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Mg2+(aq) + 2 F–(aq)
Ksp = 3.7 x 10–8
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Chapter Sixteen
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To Determine Whether
Precipitation Is Complete
•
•
•
A slightly soluble solid does not precipitate totally from
solution …
… but we generally consider precipitation to be
“complete” if about 99.9% of the target ion is precipitated
(0.1% or less left in solution).
Three conditions generally favor completeness of
precipitation:
1. A very small value of Ksp.
2. A high initial concentration of the target ion.
3. A concentration of common ion that greatly exceeds that
of the target ion.
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Chapter Sixteen
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Example 16.9
To a solution with [Ca2+] = 0.0050 M, we add sufficient
solid ammonium oxalate, (NH4)2C2O4(s), to make the
initial [C2O42–] = 0.0051 M. Will precipitation of Ca2+ as
CaC2O4(s) be complete?
CaC2O4(s)
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Ca2+(aq) + C2O42–(aq)
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Ksp = 2.7 x 10–9
Chapter Sixteen
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Selective Precipitation
AgNO3 added to
a mixture
containing Cl–
and I–
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Example 16.10
An aqueous solution that is 2.00 M in AgNO3 is slowly
added from a buret to an aqueous solution that is
0.0100 M in Cl– and also 0.0100 M in I–.
a. Which ion, Cl– or I–, is the first to precipitate from
solution?
b. When the second ion begins to precipitate, what is the
remaining concentration of the first ion?
c. Is separation of the two ions by selective precipitation
feasible?
AgCl(s)
AgI(s)
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Ag+(aq) + Cl–(aq)
Ag+(aq) + I–(aq)
Ksp = 1.8 x 10–10
Ksp = 8.5 x 10–17
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Chapter Sixteen
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Effect of pH on Solubility
• If the anion of a precipitate is that of a weak acid, the
precipitate will dissolve somewhat when the pH is lowered:
CaF2(s)
Ca2+(aq) + 2 F–(aq)
Added H+ reacts with, and
removes, F–; LeChâtelier’s
principle says more F– forms.
• If, however, the anion of the precipitate is that of a strong acid,
lowering the pH will have no effect on the precipitate.
AgCl(s)
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Ag+(aq) + Cl–(aq)
H+ does not consume Cl– ;
acid does not affect the
equilibrium.
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Chapter Sixteen
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Example 16.11
What is the molar solubility of Mg(OH)2(s) in a buffer
solution having [OH–] = 1.0 x 10–5 M, that is, pH =
9.00?
Mg(OH)2(s)
Mg2+(aq) + 2 OH–(aq)
Ksp = 1.8 x 10–11
Example 16.12 A Conceptual Example
Without doing detailed calculations, determine in which
of the following solutions Mg(OH)2(s) is most soluble:
(a) 1.00 M NH3
(b) 1.00 M NH3 /1.00 M NH4+
(c) 1.00 M NH4Cl.
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Chapter Sixteen
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Equilibria Involving Complex Ions
Silver chloride becomes more
soluble, not less soluble, in high
concentrations of chloride ion.
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Complex Ion Formation
• A complex ion consists of a central metal atom or ion, with
other groups called ligands bonded to it.
• The metal ion acts as a Lewis acid (accepts electron pairs).
• Ligands act as Lewis bases (donate electron pairs).
• The equilibrium involving a complex ion, the metal ion,
and the ligands may be described through a formation
constant, Kf:
Ag+(aq) + 2 Cl–(aq)
[AgCl2]–(aq)
[AgCl2]–
Kf = –––––––––– = 1.2 x 108
[Ag+][Cl–]2
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Chapter Sixteen
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Complex Ion Formation
Concentrated NH3
added to a solution of
pale-blue Cu2+ …
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… forms deep-blue
Cu(NH3)42+.
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Complex Ion Formation
and Solubilities
But if the concentration
of NH3 is made high
enough …
… the AgCl forms
the soluble
[Ag(NH3)2]+ ion.
AgCl is insoluble
in water.
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Example 16.13
Calculate the concentration of free silver ion, [Ag+], in an
aqueous solution prepared as 0.10 M AgNO3 and 3.0 M NH3.
Ag+(aq) + 2 NH3(aq)
[Ag(NH3)2]+(aq)
Kf = 1.6 x 107
Example 16.14
If 1.00 g KBr is added to 1.00 L of the solution described in
Example 16.13, should any AgBr(s) precipitate from the
solution?
AgBr(s)
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Ag+(aq) + Br–(aq)
Ksp = 5.0 x 10–13
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Chapter Sixteen
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Example 16.15
What is the molar solubility of AgBr(s) in 3.0 M NH3?
AgBr(s) + 2 NH3(aq)
[Ag(NH3)2]+(aq) + Br–(aq)
Kc = 8.0 x 10–6
Example 16.16
A Conceptual Example
Figure 16.10 shows that a precipitate
forms when HNO3(aq) is added to
the solution in the beaker on the right
in Figure 16.9. Write the equation(s)
to show what happens.
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Chapter Sixteen
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Complex Ions in Acid–Base Reactions
• Water molecules are commonly found as ligands in
complex ions (H2O is a Lewis base).
[Na(H2O)4]+
[Al(H2O)6]3+
[Fe(H2O)6]3+
• The electron-withdrawing power of a small, highly
charged metal ion can weaken an O—H bond in one of
the ligand water molecules.
• The weakened O—H bond can then give up its proton to
another water molecule in the solution.
• The complex ion acts as an acid.
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Ionization of a Complex Ion
The highly-charged iron(III)
ion withdraws electron density
from the O—H bonds.
[Fe(H2O)6]3+ + H2O
[Fe(H2O)5OH]2+ + H3O+
Ka = 1 x 10–7
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Chapter Sixteen
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Amphoteric Species
• Certain metal hydroxides, insoluble in water, are
amphoteric; they will react with both strong acids and
strong bases.
• Al(OH)3, Zn(OH)2, and Cr(OH)3 are amphoteric.
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Chapter Sixteen
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Qualitative Inorganic Analysis
• Acid–base chemistry, precipitation reactions, oxidation–
reduction, and complex ion formation all apply to an area
of analytical chemistry called classical qualitative
inorganic analysis.
• “Qualitative” signifies that the interest is in determining
what is present.
– Quantitative analyses are those that determine how much of a
particular substance or species is present.
• Although classical qualitative analysis is not used as
widely today as are instrumental methods, it is still a good
vehicle for applying all the basic concepts of equilibria in
aqueous solutions.
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Chapter Sixteen
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Qualitative Analysis Outline
In acid, H2S produces
very little S2–, so only
the most-insoluble
sulfides precipitate.
In base, there is more S2–,
and the less-insoluble
sulfides also precipitate.
Some hydroxides also
precipitate here.
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Chapter Sixteen
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Cation Group 1
• If aqueous HCl is added to an unknown solution of cations,
and a precipitate forms, then the unknown contains one or
more of these cations: Pb2+, Hg22+, or Ag+.
• These are the only ions to form insoluble chlorides.
• Any precipitate is separated from the mixture and further
tests are performed to determine which of the three Group
1 cations are present.
• The supernatant liquid is also saved for further analysis (it
contains the rest of the cations).
• If there is no precipitate, then Group 1 ions must be absent
from the mixture.
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Cation Group 1 (cont’d)
Analyzing for Pb2+
• Precipitated PbCl2 is slightly soluble in hot water.
• The precipitate is washed with hot water, then aqueous
K2CrO4 is added to the washings.
• If Pb2+ is present, a precipitate of yellow lead chromate
forms, which is less soluble than PbCl2.
• (If all of the precipitate dissolves in the hot water, what
does that mean?)
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Cation Group 1 (cont’d)
Analyzing for Ag+ and Hg22+
• Next, any undissolved precipitate is treated with aqueous
ammonia.
• If AgCl is present, it will dissolve, forming Ag(NH3)2+ (the
dissolution may not be visually apparent).
• If Hg22+ is present, the precipitate will turn dark gray/
black, due to a disproportionation reaction that forms Hg
metal and HgNH2Cl.
• The supernatant liquid (which contains the Ag+, if present)
is then treated with aqueous nitric acid.
• If a precipitate reforms, then Ag+ was present in the
solution.
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Chapter Sixteen
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Group 1 Cation Precipitates
PbCl2 precipitates
when HCl is added.
The presence of lead is
confirmed by adding
chromate ion; yellow
PbCrO4 precipitates.
Hg2Cl2 reacts with NH3 to form
black Hg metal and HgNH2Cl.
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Chapter Sixteen
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Hydrogen Sulfide in the
Qualitative Analysis Scheme
• Once the Group 1 cations have been precipitated, hydrogen
sulfide is used as the next reagent in the qualitative
analysis scheme.
• H2S is a weak diprotic acid; there is very little ionization of
the HS– ion and it is the precipitating agent.
• Hydrogen sulfide has the familiar rotten egg odor that is
very noticeable around volcanic areas.
• Because of its toxicity, H2S is generally produced only in
small quantities and directly in the solution where it is to
be used.
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Chapter Sixteen
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Cation Groups 2, 3, 4, and 5
• The concentration of HS– is so low in a strongly acidic
solution, that only the most insoluble sulfides precipitate.
• These include the eight metal sulfides of Group 2.
• Five of the Group 3 cations form sulfides that are soluble
in acidic solution but insoluble in alkaline NH3/NH4+.
• The other three Group 3 cations form insoluble hydroxides
in the alkaline solution.
• The cations of Groups 4 and 5 are soluble.
• Group 4 ions are precipitated as carbonates.
• Group 5 does not precipitate; these must be determined by
flame test.
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Chapter Sixteen
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Cumulative Example
A solid mixture containing 1.00 g of ammonium
chloride and 2.00 g of barium hydroxide is heated to
expel ammonia. The liberated NH3 is then dissolved
in 0.500 L of water containing 225 ppm Ca2+ as
calcium chloride. Will a precipitate form in this water?
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Sixteen