Transcript IonicEq_final

Selective Precipitation
 a solution containing several different cations can often be
separated by addition of a reagent that will form an insoluble salt
with one of the ions, but not the others
 a successful reagent can precipitate with more than one of the
cations, as long as their Ksp values are significantly different
1
What is the minimum [OH−] necessary to just begin to
precipitate Mg2+ (with [0.059]) from seawater assuming
Ksp=2.06x10-13)?
precipitating may just occur when Q = Ksp
Q = [Mg2+ ][OH- ]2
Q = K sp
[( 0.059 )][OH - ]2 = 2.06 ´ 10 -13
( 2.06 ´ 10 ) = 1.9 ´ 10
-13
[OH - ] =
( 0.059 )
-6
M
2
What is the [Mg2+] when Ca2+ (with [0.011]) just begins
to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Q = [Ca 2+ ][OH- ]2
Q = K sp
[( 0.011)][OH - ]2 = 4.68 ´ 10 -6
( 4.68 ´ 10 ) = 2.06 ´ 10
-6
[OH - ] =
( 0.011)
-2
M
3
What is the [Mg2+] when Ca2+ (with [0.011]) just begins
to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M
Q = [Mg2+ ][OH- ]2
when Ca2+ just begins to precipitate
out, the [Mg2+] has dropped from
0.059 M to 4.8 x 10-10 M
when Q = K sp
[Mg 2+ ][(2.06 ´10-2 )]2 = 2.06 ´10-13
2.06 ´10 )
(
]=
= 4.8 ´10
(2.06 ´10 )
-13
[Mg 2+
-2 2
-10
M
4
Qualitative Analysis
 an analytical scheme that utilizes selective precipitation to identify the
ions present in a solution is called a qualitative analysis scheme
 wet chemistry
 a sample containing several ions is subjected to the addition of several
precipitating agents
 addition of each reagent causes one of the ions present to precipitate
out
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Qualitative Analysis
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Complex Ion Formation
 transition metals tend to be good Lewis acids
 they often bond to one or more H2O molecules to form a hydrated ion
 H2O is the Lewis base, donating electron pairs to form coordinate
covalent bonds
Ag+(aq) + 2 H2O(l)  [Ag(H2O)2+](aq)
 ions that form by combining a cation with several anions or neutral
molecules are called complex ions
 e.g., Ag(H2O)2+
 the attached ions or molecules are called ligands
 e.g., H2O
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Complex Ion Equilibria
 if a ligand is added to a solution that forms a stronger bond than the current
ligand, it will replace the current ligand
Ag(H2O)2+(aq) + 2 NH3(aq)
Ag(NH3)2+(aq) + 2 H2O(l)
 generally H2O is not included, since its complex ion is always present in
aqueous solution
Ag+(aq) + 2 NH3(aq)
Ag(NH3)2+(aq)
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Formation Constant
 the reaction between an ion and ligands to form a complex ion is
called a complex ion formation reaction
Ag+(aq) + 2 NH3(aq)
Ag(NH3)2+(aq)
 the equilibrium constant for the formation reaction is called the
formation constant, Kf
+
[Ag(NH 3 ) 2 ]
Kf =
[Ag + ][NH 3 ]2
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Formation Constants
16
200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of
0.20 M NH3. What is the [Cu2+] at equilibrium?
Write the formation
reaction and Kf
expression.
Look up Kf value
Determine the
concentration of
ions in the diluted
solutions
Cu2+(aq) + 4 NH3(aq)
Cu(NH3)42+(aq)
[Cu(NH3 )4 2+ ]
13
Kf =
=
1.7´10
[Cu 2+ ][NH3 ]4
1.5 ´10 -3 mol
0.200 L ´
2+
1L
[Cu ] =
= 6.7 ´10-4 M
(0.200 L + 0.250 L )
2.0 ´10 -1 mol
0.250 L ´
1L
[NH 3 ] =
= 1.1´10-1 M
(0.200 L + 0.250 L )
17
200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of
0.20 M NH3. What is the [Cu2+] at equilibrium?
Cu2+(aq) + 4 NH3(aq)
Cu(NH3)42+(aq)
[Cu 2+ ][NH 3 ]4
13
Kf =
=
1.7
´10
2+
[Cu(NH 3 ) 4 ]
Create an ICE
table. Since Kf is
large, assume all
the Cu2+ is
converted into
complex ion, then
the system returns
to equilibrium
Initial
Change
Equilibrium
[Cu2+]
[NH3]
[Cu(NH3)22+]
6.7E-4
0.11
0
-≈6.7E-4
-4(6.7E-4)
+ 6.7E-4
x
0.11
6.7E-4
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200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of
0.20 M NH3. What is the [Cu2+] at equilibrium?
Substitute in and
solve for x
Cu2+(aq) + 4 NH3(aq)
Cu(NH3)22+(aq)
2+
[Cu(NH 3 ) 4 ]
13
Kf =
=
1.7
´10
[Cu 2+ ][NH 3 ]4
confirm the “x
is small”
approximation
-4
6.7´10
(
)
13
1.7´10[Cu
=2+]
4 3]
[NH
[Cu(NH3)22+]
0.11
0
( x)( 0.11)
Initial
x=
Change
( 6.7´10-4)
6.7E-4
(1.7´10 )( 0.11)
Equilibrium
-≈6.7E-4
13
x
4
-13
=
2.7´10
-4(6.7E-4)
0.11
+ 6.7E-4
6.7E-4
since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid
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The Effect of Complex Ion Formation on
Solubility
 the solubility of an ionic compound that contains a metal cation that
forms a complex ion increases in the presence of aqueous ligands
AgCl(s)
Ag+(aq) + Cl−(aq)
Ag+(aq) + 2 NH3(aq)
Ag(NH3)2+(aq)
Ksp = 1.77 x 10-10
Kf = 1.7 x 107
 adding NH3 to a solution in equilibrium with AgCl(s) increases the
solubility of Ag+
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Solubility of Amphoteric
Metal Hydroxides
 many metal hydroxides are insoluble eg Fe(OH)3, Al(OH)3, Co(OH)2
 all metal hydroxides become more soluble in acidic solution
 shifting the equilibrium to the right by removing OH−
Fe(OH)3(s)
[Fe(OH)2+](aq) + OH-(aq)
H3O+(aq)+OH-(aq)  2H2O(l)
 Amphoteric metal hydroxides also become more soluble in basic
solution
 acting as a Lewis base forming a complex ion
 some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+,
Pb2+, and Sb2+
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Al3+
 Al3+ is hydrated in water to form an acidic solution
Al(H2O)63+(aq) + H2O(l)
Al(H2O)5(OH)2+(aq) + H3O+(aq)
 addition of OH− drives the equilibrium to the right and continues to remove
H from the molecules
Al(H2O)5(OH)2+(aq) + OH−(aq)
Al(H2O)4(OH)2+(aq) + OH−(aq)
Al(H2O)4(OH)2+(aq) + H2O (l)
Al(H2O)3(OH)3(s) + H2O (l)
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