Transcript Masterton and Hurley Chapter 16

William L Masterton
Cecile N. Hurley
Edward J. Neth
cengage.com/chemistry/masterton
Chapter 15
Complex Ion and Precipitation Equilibria
Edward J. Neth • University of Connecticut
Outline
1.
2.
3.
4.
Complex ion equilibria; formation constant (Kf)
Solubility; solubility product constant (Ksp)
Precipitate formation
Dissolving precipitates
Complex Ions
• Recall (chapter 13) that metal ions are Lewis acids
• Compounds with atoms containing lone pairs are
Lewis bases
• Metal ions can combine with molecules or ions to
produce complex ions
Copper ion and ammonia
• Cu2+ (aq) + 4NH3 (aq) ⇌ Cu(NH3)42+ (aq)
• The equilibrium constant for the formation of the
complex ion is called the formation constant, Kf
• For this complex ion, Kf = 2 X 1012, so the reaction goes
essentially to completion
2
[Cu(NH3 )4 ]
Kf 
2
4
[Cu ][NH3 ]
Table 15.1
Example 15.1
Example 15.1, (Cont’d)
Example 15.1, (Cont’d)
Example 15.1, (Cont’d)
Two coordination complexes of Co3+
• On the left: [Co(NH3)63+]
• On the right: [Co(NH3)5Cl2+]
Revisiting Solubility and Precipitation
• In Chapter 4 we learned that there are compounds
that do not dissolve in water
• These were called insoluble
• A reaction that produces an insoluble precipitate
was assumed to go to completion
• In reality, even insoluble compounds dissolve to
some extent, usually small
• An equilibrium is set up between the precipitate
and its ions
• Precipitates can be dissolved by forming complex
ions
Two Types of Equilibria
• AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
• Solid exists in equilibrium with the ions formed
when a small amount of solid dissolves
• AgCl (s) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq) + Cl- (aq)
• Formation of a stable complex ion can cause an
otherwise insoluble compound to dissolve
• There are multiple equilibria at work in this
example, in similar fashion to the equilibria
underlying the function of a buffer (Chapter 14)
Solubility Product Constant, Ksp
• Consider mixing two solutions:
• Sr(NO3)2 (aq)
• K2CrO4 (aq)
• The following net ionic equation describes the
reaction:
• Sr2+ (aq) + CrO42- (aq)  SrCrO4 (s)
Figure 15.3 – A precipitate of SrCrO4
Ksp Expression
• SrCrO4 (s) ⇌ Sr2+ (aq) + CrO42- (aq)
• The solid establishes an equilibrium with its ions once it forms
• We can write an equilibrium expression, leaving out the term for the
solid (recall that its concentration does not change as long as some
is present)
2
2
K sp  [Sr ][CrO4 ]
• Ksp is called the solubility product constant
Interpreting the Solubility Expression and Ksp
• Ksp has a fixed value at a given temperature
• For strontium chromate, Ksp = 3.6 X 10-5 at 25 ºC
• The product of the two concentrations at equilibrium
must have this value regardless of the direction from
which equilibrium is approached
Example 15.2
Ksp and the Equilibrium Concentration of Ions
• Ksp SrCrO4 = [Sr2+][CrO42-] = 3.6 X 10-5
• This means that if we know one ion concentration, the other
one can easily be calculated
• If [Sr2+] = 1.0 X 10-4 M, then
[CrO4
2
3.6 X 10 5
]
 0.36M
4
1.0 X 10
• If [CrO42-] = 2.0 X 10-3, then
5
3
.
6
X
10
2
[Sr 2 ] 

1
.
8
X
10
M
3
2.0 X 10
Example 15.3
Example 15.3, (Cont’d)
Example 15.3, (Cont’d)
Table 15.2
Ksp and Water Solubility
• One way to establish a solubility equilibrium
• Stir a slightly soluble solid with water
• An equilibrium is established between the solid and its ions
• BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq)
• If we set the concentration of the ions equal to a variable, s:
2
[Ba 2 ]  [SO4 ]  s
2
K sp BaSO4  [Ba 2 ][SO4 ]  s 2
s  K sp  (1.1X 10
1
10 2
)  1.0 X 10 5 M
Example 15.4
Example 15.4, (Cont’d)
Example 15.4, (Cont’d)
Calculating Ksp Given Solubility
• Instead of calculating solubility from Ksp, it is possible
to calculate Ksp from the solubility
• Recall that solubility may be given in many
different sets of units
• Convert the solubility to moles per liter for use in
the Ksp expression
Example 15.5
Ksp and the Common Ion Effect
• BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq)
• How would you expect the solubility of barium
sulfate in water to compare to its solubility in
0.10 M Na2SO4?
• Solubility must be less than it is in pure water
• Recall LeChâtelier’s Principle
• The presence of the common ion, SO42-, will drive the
equilibrium to the left
• Common ions reduce solubility
Visualizing the Common Ion Effect
Figure 15.2
Example 15.6
Example 15.6, (Cont’d)
Ksp and Precipitate Formation
• Ksp values can be used to predict whether a
precipitate will form when two solutions are mixed
• Recall the use of Q, the reaction quotient, from
Chapter 12
• We can calculate Q at any time and compare it to
Ksp
• The relative magnitude of Q vs. Ksp will indicate
whether or not a precipitate will form
Figure 15.3: Precipitation of strontium chromate
Q and Ksp
• If Q > Ksp, a precipitate will form, decreasing the ion
concentrations until equilibrium is established
• If Q < Ksp, the solution is unsaturated; no precipitate
will form
• If Q = Ksp, the solution is saturated just to the point of
precipitation
Figure 15.4
Example 15.7
Example 15.7, (Cont’d)
Example 15.7, (Cont’d)
Example 15.7, (Cont’d)
Precipitation Visualized
Selective Precipitation
• Consider a solution of two cations
• One way to separate the cations is to add an
anion that precipitates only one of them
• This approach is called selective precipitation
• Related approach
• Consider a solution of magnesium and barium
ions
Selective Precipitation, (Cont'd)
• Ksp BaCO3 = 2.6 X 10-9
• Ksp MgCO3 = 6.8 X 10-6
• Carbonate ion is added
• Since BaCO3 is less soluble than MgCO3, BaCO3
precipitates first, leaving magnesium ion in
solution
• Differences in solubility can be used to separate
cations
Figure 15.5 – Selective Precipitation
Example 15.8
Example 15.8, (Cont'd)
Example 15.8, (Cont’d)
Dissolving Precipitates
• Bringing water-insoluble compounds into solution
• Adding a strong acid to react with basic anions
• Adding an agent that forms a complex ion to react
with a metal cation
Strong Acid
• Zn(OH)2 (s) + 2H+ (aq)  Zn2+ (aq) + 2H2O
• This reaction takes place as two equilibria:
• Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq)
• 2H+ (aq) + 2OH- (aq) ⇌ 2H2O
• Because the equilibrium constant for the
neutralization is so large, the reaction goes
essentially to completion
• Note that for the second equilibrium, K = (1/Kw)2 =
1 X 1028
Example 15.9
Example 15.9, (Cont'd)
Example 15.9, (Cont'd)
Insoluble Compounds that Dissolve in Strong Acid
• Virtually all carbonates
• The product of the reaction is H2CO3, a weak acid
that decomposes to carbon dioxide
• H2CO3 (aq)  H2O + CO2 (g)
• Many sulfides
• The product of the reaction is H2S, a gas that is
also a weak acid
• H2S (aq) ⇌ H+ (aq) + HS- (aq)
Visualizing Selective Dissolving of Precipitates
Example 15.10
Complex Formation
• Ammonia and NaOH can dissolve compounds
whose metal cations form complexes with NH3 and
OH• As with the addition of a strong acid, multiple
equilibria are at work:
• Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq)
Ksp
• Zn2+ (aq) + 4NH3 (aq) ⇌ Zn(NH3)42+ (aq)
Kf
• Net: Zn(OH)2 (s) + 4NH3 (aq)  Zn(NH3)42+ (aq) + 2OH- (aq)
• Knet = KspKf = 4 X 10-17 X 3.6 X 108 = 1 X 10-8
Table 15.3
Visualizing Dissolving by Complex Formation
Example 15.11
Example 15.11, (Cont'd)
Example 15.11, (Cont’d)
Example 15.12
Key Concepts
1. Relate Kf to the ratio of concentration of complex
ion to metal ion
2. Write the Ksp expression for any insoluble ionic solid
2. Use the value of Ksp to
A. Calculate the concentration of one ion, knowing
the other
B. Determine whether a precipitate will form
C. Calculate the water solubility of a compound
D. Calculate the solubility of a compound in a
solution of a common ion
E. Determine which ion will precipitate first
Key Concepts
3. Calculate K for
A. Dissolving a metal hydroxide in a strong acid
B. Dissolving a precipitate in a complexing agent
4. Write balanced, net ionic equations to explain why a
precipitate dissolves in
A. Strong acid
B. Ammonia or hydroxide solution