Transcript Chapter Sixteen - DePaul University
Chapter Sixteen
More Equilibria in Aqueous Solutions: Slightly Soluble Salts and Complex Ions
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2
The Solubility Product Constant, K
sp
• 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: BaSO 4 (s) Ba 2+ (aq) + SO 4 2– (aq) •
Solubility product constant, K
sp
: the equilibrium constant expression for the dissolving of a slightly soluble solid.
K
sp = [ Ba 2+ ][ SO 4 2– ]
Prentice Hall © 2005 Chapter Sixteen
General Chemistry
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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, FePO 4 , and
(b)
chromium(III) hydroxide, Cr(OH) 3 .
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K
sp
and Molar Solubility
• •
K
sp 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
K
sp , but molar solubility and
K
sp are
not
the same thing.
• In fact, “smaller
K
sp ” doesn’t always mean “lower molar solubility.” • Solubility depends on both
K
sp 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 Ag 2 CO 3
K
sp for Ag 2 CO 3 per liter of solution. Calculate at 20 °C. The balanced equation is Ag 2 CO 3 (s) 2 Ag + (aq) + CO 3 2– (aq)
K
sp = ?
Example 16.3
From the
K
sp value for silver sulfate, calculate its molar solubility at 25 °C.
Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2– (aq)
K
sp = 1.4 x 10 –5 at 25 °C 6 Prentice Hall © 2005 Chapter Sixteen
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Example 16.4
A Conceptual Example
Without doing detailed calculations,
Table 16.1, establish the order of but using data from
increasing
solubility of these silver halides in water: AgCl, AgBr, AgI.
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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 Ag 2 SO 4 .
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Na 2 SO 4 (aq) Saturated Ag 2 SO 4 (aq) Ag 2 SO 4 precipitates
General Chemistry
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Common Ion Effect Illustrated
When Na 2 SO 4 (aq) is added to the saturated solution of Ag 2 SO 4 …
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… [Ag + ] attains a new, lower equilibrium concentration as Ag + reacts with SO 4 2– to produce Ag 2 SO 4 .
Chapter Sixteen
Example 16.5
Calculate the molar solubility of Ag 2 SO 4 Na 2 SO 4 (aq).
in 1.00 M 11 Prentice Hall © 2005
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12
Solubility and Activities
• Ions that are
not
common to the precipitate can also affect solubility.
– CaF 2 water.
is
more
soluble in 0.010 M Na 2 SO 4 than it is in • Increased solubility occurs because of interionic attractions.
• Each Ca 2+ and F – is surrounded by ions of opposite charge, which impede the reaction of Ca 2+ with F – .
• The
effective
and F – concentrations, or
activities
, of Ca 2+ are lower than their actual concentrations.
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Will Precipitation Occur? Is It Complete?
• •
Q
ip is the
ion product reaction quotient
initial conditions of the reaction.
and is based on
Q
ip and
Q
c : new look, same great taste!
Q
ip can then be compared to
K
sp .
• Precipitation
should
occur if
Q
ip • Precipitation
cannot
occur if
Q
ip >
K
sp .
<
K
sp .
• A solution is
just saturated
if
Q
ip =
K
sp .
• In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered.
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Example 16.6
If 1.00 mg of Na 2 CrO 4 is added to 225 mL of 0.00015 M AgNO 3 , will a precipitate form? Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2– (aq)
K
sp = 1.1 x 10 –12 14 Prentice Hall © 2005
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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(NO 3 ) 2 . What is the solid that first appears? Explain why it then disappears.
Example 16.8
If 0.100 L of 0.0015 M MgCl 2 and 0.200 L of 0.025 M NaF are mixed, should a precipitate of MgF 2 form?
MgF 2 (s) Mg 2+ (aq) + 2 F – (aq)
K
sp = 3.7 x 10 –8 Prentice Hall © 2005 Chapter Sixteen
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• • • 1.
2.
3.
To Determine Whether Precipitation Is Complete
A slightly soluble solid does not precipitate
totally
solution … from … 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: A very small value of
K
sp .
A high initial concentration of the target ion.
A concentration of common ion that greatly exceeds that of the target ion.
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Example 16.9
To a solution with [Ca 2+ ] = 0.0050 M, we add sufficient solid ammonium oxalate, (NH 4 ) 2 C 2 O 4 (s), to make the initial [C 2 O 4 2– ] = 0.0051 M. Will precipitation of Ca 2+ CaC 2 O 4 (s) be complete? as CaC 2 O 4 (s) Ca 2+ (aq) + C 2 O 4 2– (aq)
K
sp = 2.7 x 10 –9 17 Prentice Hall © 2005
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AgNO 3 added to a mixture containing Cl – and I –
Selective Precipitation
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Example 16.10
An aqueous solution that is 2.00 M in AgNO 3 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 solution?
– or I – , is the first to precipitate from 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) Ag + (aq) + Cl – (aq)
K
sp = 1.8 x 10 –10 AgI(s) Ag + (aq) + I – (aq)
K
sp = 8.5 x 10 –17 Prentice Hall © 2005 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: CaF 2 (s) Ca 2+ (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) Ag + (aq) + Cl – (aq)
H + does not consume Cl – ; acid does not affect the equilibrium.
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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) Mg 2+ (aq) + 2 OH – (aq)
K
sp = 1.8 x 10 –11 21
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 NH 3
(b)
1.00 M NH 3 /1.00 M NH 4 +
(c)
1.00 M NH 4 Cl.
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Equilibria Involving Complex Ions
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Silver chloride becomes more soluble, not less soluble, in high concentrations of chloride ion.
Chapter Sixteen
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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, K
f
: Ag + (aq) + 2 Cl – (aq) [AgCl 2 ] – (aq)
K
f [AgCl 2 ] – = –––––––––– = 1.2 x [Ag + ][Cl – ] 2 10 8 Prentice Hall © 2005 Chapter Sixteen
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Complex Ion Formation
Concentrated NH 3 added to a solution of pale-blue Cu 2+ … … forms deep-blue Cu(NH 3 ) 4 2+ .
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Complex Ion Formation and Solubilities
But if the concentration of NH 3 is made high enough …
26
AgCl is insoluble in water.
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… the AgCl forms the soluble [Ag(NH 3 ) 2 ] + ion.
Chapter Sixteen
Example 16.13
Calculate the concentration of free silver ion, [Ag + ], in an aqueous solution prepared as 0.10 M AgNO 3 and 3.0 M NH 3 .
Ag + (aq) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq)
K
f = 1.6 x 10 7 27
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) Ag + (aq) + Br – (aq)
K
sp = 5.0 x 10 –13 Prentice Hall © 2005 Chapter Sixteen
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Example 16.15
What is the molar solubility of AgBr(s) in 3.0 M NH 3 ?
AgBr(s) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) + Br – (aq)
K
c = 8.0 x 10 –6
Example 16.16
A Conceptual Example
Figure 16.10 shows that a precipitate forms when HNO 3 (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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Complex Ions in Acid –Base Reactions
• Water molecules are commonly found as ligands in complex ions (H 2 O is a Lewis base).
[Na(H 2 O) 4 ] + [Al(H 2 O) 6 ] 3+ [Fe(H 2 O) 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.
30 [Fe(H 2 O) 6 ] 3+ + H 2 O [Fe(H 2 O) 5 OH] 2+ + H 3 O +
K
a = 1 x 10 –7 Prentice Hall © 2005
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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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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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Qualitative Analysis Outline
In acid, H 2 S produces very little S 2– , so only the most-insoluble sulfides precipitate.
33
In base, there is more S 2– , and the less-insoluble sulfides also precipitate.
Some hydroxides also precipitate here.
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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: Pb 2+ , Hg 2 2+ , 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 Pb
2+
• Precipitated PbCl 2 is slightly soluble in hot water.
• The precipitate is washed with hot water, then aqueous K 2 CrO 4 • If Pb 2+ is added to the washings.
is present, a precipitate of yellow lead chromate forms, which is less soluble than PbCl 2 .
• (If all of the precipitate dissolves in the hot water, what does that mean?) 35 Prentice Hall © 2005
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Cation Group 1
(cont’d)
Analyzing for Ag
+
and Hg
2 2+
• Next, any undissolved precipitate is treated with aqueous ammonia.
• If AgCl is present, it will dissolve, forming Ag(NH 3 ) 2 + dissolution may not be visually apparent).
(the • If Hg 2 2+ is present, the precipitate will turn dark gray/ black, due to a disproportionation reaction that forms Hg metal and HgNH 2 Cl.
• 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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Group 1 Cation Precipitates
PbCl 2 precipitates when HCl is added.
The presence of lead is confirmed by adding chromate ion; yellow PbCrO 4 precipitates.
37
Hg 2 Cl 2 reacts with NH 3 to form black Hg metal and HgNH 2 Cl.
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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.
• H 2 S 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, H 2 S is generally produced only in small quantities and directly in the solution where it is to be used.
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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 NH 3 /NH 4 + .
• 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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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 NH 3 is then dissolved in 0.500 L of water containing 225 ppm Ca 2 + as calcium chloride. Will a precipitate form in this water?
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