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1
CHE 1302
Tro 2nd Ed.
Chapter 16: Aqueous Ionic Equilibrium
Dr. Bruce E. Hodson
The Common Ion Effect
Ionization of the weak acid is suppressed: pH INCREASES
1. Consider a solution of the WEAK acid acetic acid, CH3COOH….
3. LeChâtelier’s principle: What happens to [H3O+] ?
NaC2H3O2 → Na+ + CH3COO–
2. … then add acetate ion as a second solute
CH3COOH(aq) at equilibrium: a few H3O+ ions and a
few CH3COO– ions
Add CH3COO– , reestablish
equilibrium: more acetate ions, fewer
H3O+ ions
The Common Ion Effect
Common ion effect : the suppression of the ionization of a weak electrolyte by the presence of a common ion from a strong electrolyte
The Common Ion Effect
Ionization of the weak base is suppressed
1. Consider a solution of the WEAK base ammonia, NH3 ….
NH4Cl → NH4+ + Cl–
2. … then add ammonium ion (a COMMON ION) as a second solute
3. LeChâtelier’s principle:
What happens to [OH–] ?
EXAMPLES
5
16.2 Buffers: Solutions that Resist pH Change
6
Pure water has no ‘buffer capacity’
16.2 Buffers: Solutions that Resist pH Change
In contrast, for a buffer solution the pH barely changes
7
A buffer solution is one whose pH changes very little when small amounts of strong acid or strong base are added, or when diluted
Weak acid and a salt of its conjugate base
Weak base and a salt of its conjugate acid
16.2 Buffers: Solutions that Resist pH Change
A weak acid and a salt of it’s conjugate base…
16.2 Buffers: Solutions that Resist pH Change
Note that the COMMON ION EFFECT ….. is in effect …
16.2 Buffers: Solutions that Resist pH Change
Buffer pH higher than
pure acid and lower than pure salt
10
A buffer maintains approximately constant pH because of the acid and base components which are present at all times .….
Added OH– reacts with acid present HC2H3O2 → H2O + C2H3O2–
Added H3O+ reacts with base present C2H3O2– → H2O + HC2H3O2
16.2 Buffer Solutions – Why do they work ?
weak acidHC2H3O2
conjugate base C2H3O2
–
Na+ spectator ions
16.2 Buffers: Solutions that Resist pH Change
A weak base and a salt of it’s conjugate acid …
12
Added OH– reacts with acid present NH4+ → H2O + NH3
Added H3O+ reacts with base present NH3 → H2O + NH4
+
weak baseNH3
conjugate acid NH4
+
Cl– spectator ions
16.2 Buffer Solutions – Why do they work ?
A buffer maintains approximately constant pH because of the acid and base components which are present at all times .….
13
So…. Buffers contain components which can neutralize (small amounts of) added strong acids and bases
Buffers only work because the components do not completely neutralized each another… e.g HCl + NaOH doesn’t work ….
H3O+ + OH– → 2 H2O
pH of a Buffer: Equilibrium Approach EXAMPLE
16.2 Buffers: Solutions that Resist pH Change
14
The Henderson-Hasselbalch equation gives the pH of a ‘weak acid + salt of conjugate base’ buffer directly:
[conjugate base]pH = pKa + log –––––––––––––– [weak acid]
16.2 Buffers: Solutions that Resist pH Change
15
Generally, a buffer is most effective if [weak acid] = [conjugate base] …..
[conjugate base]pH = pKa + log –––––––––––––– [weak acid]
pH = pKa
Under these conditions, the pH of the buffer is approx. equal to the pKa of the weak acid
16.2 Buffer Solutions – The BUFFER CONDITION
.. and within a range of 2 pH units, meaning 0.1 through 10 for the ratio base to acid
16
Capacity of a buffer is the amount of strong acid, or strong base, it can consume before it no longer acts as a buffer
….the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize
16.2 Buffer Solutions - CAPACITY
More Buffer EXAMPLES
17
16.4 Titration and pH Curves
4.8 Acid-Base Titrations (Stoichiometry in Solutions)
18
A titration is a method for determining the concentration of a solute in a sample of known volume ………
4.8 Titrations (Stoichiometry in Solutions)
Titrations can be applied to precipitation, acid-base, or redox reactions
Known volume (L) of unknown concentration
(mol/L) reactant B (analyte)
Add a known volume (L) of a known
concentration (mol/L) reactant A (titrant)
1. Add enough A to react completely with B
2. Calculate moles of A added
3. Use a balanced equation to find moles
of B present
4. Moles B / volume B = molarity of B
19
4.8 Acid – Base Titration
42.00 mL of
0.150 M NaOH(aq)
added
A TITRANT of known concentration is added until the ANALYTE solution becomes
neutral…..
An INDICATOR is added which changes color at its END
POINT
At the EQUIVALENCE POINT, the acid and base have been
brought together in exact stoichiometric proportions
HCl(aq) ? conc. (50.00 mL)
Neutral Solution
H+
20
A pH indicator is a weak acid that is one color when protonated (HIn), and another color when deprotonated (In–) (the conjugate base)
HIn(aq) + acid HIn(aq) + base
16.4 Acid-Base Indicators
HIn + H2O In– + H3O+
Adding H3O+ to HIn(aq) (pink) favors HIn (RED)Adding OH– to HIn(aq) (pink) favors In– (PURPLE)
RED PURPLE
21
16.4 Acid-Base Indicators
‘H’Phenolphthalein + H2O
Phenolphthalein– + H3O+
Adding OH– favors Phenolphthalein–
Adding H3O+ favors ‘H’Phenolphthalein
22
Different indicators have different values of Ka, so they exhibit color changes at different values of pH …
16.4 Acid-Base Indicators
23
Bromthymol blue changes color at the
equivalence point (approx. 20.0 mL NaOH(aq) added)
So … Ideally we want the indicator’s color change (end point) to be centered around the equivalence point
16.4 Titrations: Strong Acid with Strong Base (MONOPROTIC)
Titration Curve
HCl + NaOH → NaCl + H2O
24
pH values can be calculated at any point by assessing moles of acid and base present and using a balanced equation for the neutralization
16.4 Titrations: Strong Acid with Strong Base (MONOPROTIC)
pH << 7
pH >> 7
HCl + NaOH → NaCl + H2O
25
16.4 Titrations: Strong Base with Strong Acid (MONOPROTIC)
26
Equivalence-point pH is NOT 7.00
16.4 Titrations: Weak Acid with Strong Base (MONOPROTIC)
HC2H3O2 + NaOH → NaC2H3O2 + H2O
pH low
pH >> 7
pH = pKa ± 1: buffer solution (weak acid
and its conjugate base): H-H equation
27
16.4 Titrations: Weak BASE with Strong Acid (MONOPROTIC)
Titration curve for the titration of 0.100 M NH3 with 0.100 M HCl
EXAMPLES
28
16.4 Titrations: Weak Acid with Strong Base (POLYPROTIC)
Titration of 50.0 mL of 0.10 M H3PO3 with 0.10 M NaOH
H3PO3 + NaOH → H2PO3– + H2O
H2PO3– + NaOH → HPO3
2– + H2O
(Multiple Equivalence
points)
29
16.5 Solubility Equilibria and the Solubility Product Constant
30
Solid begins to dissolve
As solid dissolves, some dissolved solute begins to crystallize
Eventually, the rates of dissolving and of crystallization are equal; no more solute appears to dissolve.
Longer standing does not change the amount
of dissolved solute
Equilibria Review
For a saturated solution ……..
31
Many ionic compounds do not dissolve well in aqueous solution – these are termed ‘sparingly soluble’ rather than ‘insoluble’
A Dynamic Equilibrium is established between the solid solute and the ions of that solute in solution
16.5 The Solubility Product Constant, Ksp
32
The Solubility Product, Ksp , is the equilibrium constant for the dissolution of a sparingly soluble ionic compound
One mole of solid is the only reactant and the concentration of the solid is not included in the Ksp expression
FeS(s) Fe2+(aq) + S2–
(aq)
Ksp = [Fe2+] [S2–]
16.5 The Solubility Product Constant, Ksp
33
Ksp values for a wide variety of sparingly soluble ionic compounds are known for a given temperature …… (25 °C in the table below)
16.5 The Solubility Product Constant, Ksp
34
Molar solubility is simply the solubility of a compound in moles per liter – it can be computed directly from Ksp and vice versa
AgCl(s) Ag+(aq) + Cl –
(aq) Ksp = 1.77 × 10–10
1 mol AgCl ≡ 1 mol Ag+ ≡ 1 mol Cl –
Ksp = [Ag+] [Cl–]
Molar Solubility = S = [Ag+] = [Cl–]
Ksp = [Ag+] [Cl–] = S × S = S2
S = Ksp = 1.77 × 10–10 = 1.33 × 10–5 M
16.5 Relationship between Ksp and Molar Solubility
EXAMPLES
3535
16.5 The Common Ion Effect and Solubility Equilibria
36
If we add one of its ions to a saturated solution of a sparingly soluble compound, we predict that precipitate will form :
This follows Le Chatelier’s principle (more product shifts equilibrium to the left)
PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO4
2-(aq)
K2CrO4 added
18.3 The Common Ion Effect and Solubility Equilibria
PbCrO4(s) Pb2+(aq) + CrO42-(aq)
37
When K2CrO4 is added to the
saturated solution of
PbCrO4 (aq) …
… [Pb2+] attains a new, lower equilibrium
concentration as Pb2+ reacts with
CrO42–
to produce PbCrO4 (s)
The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution
Pb2+
CrO42–
EXAMPLES
18.3 The Common Ion Effect and Solubility Equilibria
38
16.5 Solubility and pH
39
One way to dissolve a sparingly soluble compound is often to add a strong acid ……….
If the anion of the compound is the conjugate base of a weak acid, the H3O+ added will react with it lower its concentration
Added H3O+ reacts with, and removes F–; LeChâtelier’s principle says more F– forms
CaF2(s) Ca2+(aq) + 2 F–
(aq)
H3O+(aq) + F–
(aq) HF(aq) + H2O(l)
16.5 Solubility and pH
40
Qsp = [Ca2+] [F–]2 < Ksp
equilibrium shifts to the rightSome CaF2 (s) dissolves (increase [F–] and [Ca2+]), so that Qsp becomes equal to Ksp. Continues until all acid is consumed or all solid dissolves.
16.5 Solubility and pH
CaF2(s) Ca2+(aq) + 2 F–
(aq)
Ksp = [Ca2+] [F–]2 = 5.3 × 10–9
What about Ksp and Qsp ? Adding H3O+ reduces the [F–], hence Qsp < Ksp
Remove product
41
AgCl(s) Ag+ (aq) + Cl–
(aq)
This method will not work when the anion of the sparingly soluble compound is the conjugate base of a strong acid ……..
…….. because the anion is therefore a very weak base
H3O+(aq) + Cl–
(aq)
BaSO4(s) Ba2+ (aq) + SO4
2– (aq)
H3O+(aq) + SO4
2– (aq)
NO REACTION
NO REACTION
16.5 Solubility and pH
EXAMPLES
42
Limitations of the Ksp Concept
43
Ksp can only be used for sparingly soluble compounds if we wish to use molarities in the equilibrium constant expression …….
…… in more concentrated solutions, stoichiometric and effective concentrations (activities) are not equal
Limitations of the Ksp Concept
Ksp = [A+] [B–]
… if [AB] = 1.0 × 10–5 M … is [A+] = [B–] = 1.0 × 10–5 M ???
44
Ions that are not common to the precipitate can affect solubility (and hence Ksp values) :
CaF2 is more soluble in 0.010 M Na2SO4 than it is in waterdue to inter-ionic attractions
Ca2+
Na+
SO42–
F–
Na+
SO42–
Na+
Ca2+
F–
H2O Na2SO4 (aq)
Limitations of the Ksp Concept
45
16.6 Precipitation
16.6 Criteria for Precipitation and its completeness
46
To determine if precipitation will occur, we calculate the Ion Product Reaction Quotient, Qsp, and compare it to Ksp
…. the Qsp expression takes the same form of the Ksp expression, but concentrations are not equilibrium values
FeS(s) Fe2+(aq) + S2–
(aq)
Ksp = [Fe2+] [S2–]
precipitation should occur (reaction shifts left)Qsp > Ksp
Qsp < Ksp precipitation will not occur (reaction shifts right)
Qsp = Kspsolution is just saturated (equilibrium)
47
We generally consider precipitation to be “complete” if about 99.9% of the target ion is precipitated (0.1% or less left in solution)
In calculations, assume as much ppt. has formed as possible, calculate concentrations of remaining ions and then equilibrium
is reached from that point
16.6 Criteria for Precipitation and its completeness
EXAMPLES
48
16.6 Selective (Fractional) Precipitation
49
We can often completely precipitate one ion while leaving another ion in solution – Selective Precipitation
AgNO3 added to a mixture containing Cl– and I–
16.6 Selective (Fractional) Precipitation
EXAMPLES
… for this to work, there must be a fairly significant difference in the solubilities (Ksp values) of the substances being separated
50
16.8 Equilibria Involving Complex Ions
51
Complex ions have a more complex structure than the ions we have previously considered
Ligands are LEWIS BASES, the central metal atom/ion is a LEWIS ACID. Coordinate covalent bonds are formed.
Attached groups (LIGANDS) are ions or molecules with at
least one lone pair of electrons
3+
16.8 Equilibria Involving Complex Ions
Common ligands:Cl–, OH–, H2O, NH3
52
The formula of a complex ion is written in square [!!!!!] brackets, any charge is placed superscript right ……
A coordination compound is one which contains a complex ion, e.g. Na[AgCl2] or [Cu(NH3)4]Cl2
[AgCl2]–
–
[Cu(NH3)4]2+
2+
16.8 Equilibria Involving Complex Ions
53
The production of a complex ion can be described by an equilibrium reaction……..
The equilibrium constant is the FORMATION CONSTANT, Kf
Ag+(aq) + 2 NH3 (aq) [Ag(NH3)2]+
(aq)
Kf = [ [Ag(NH3)2]+ ]
Pb2+(aq) + 4 I–
(aq) [PbI4]2– (aq)
[Ag+] [NH3]2
Kf = [ [PbI4]2– ]
[Pb2+] [I–]4
16.8 Equilibria Involving Complex Ions
54
At a molecular level, the actual process (mechanism) is stepwise ……
M(H2O)42+ M(H2O)3(NH3)2+ M(NH3)4
2+
NH3
3NH3
The product of individual formation constants for each step gives the overall formation constant, Kf
16.8 Equilibria Involving Complex Ions
55
Sparingly soluble compounds can be dissolved by treatment with a complexing agent, which contains a ligand with great affinity for the metal ion
The above process is similar to dissolving sparingly soluble compound in strong acid
AgI (s) Ag+ (aq) + I–
(aq)
Ag+(aq) + 2 S2O3
2– (aq) [Ag(S2O3)2]3–
(aq)
KSP = 5.3 × 10–9
Kf = 1.7 × 1013
AgI(s) + 2 S2O32–
(aq) [Ag(S2O3)2]3– (aq) + I–
(aq)
Kc = Ksp × Kf = 1.4 × 10–3
16.8 Equilibria Involving Complex Ions
56
1. AgCl is insoluble in
water.…
2…but if the concentration of NH3 is made high
enough …
3 … the AgCl forms the soluble
[Ag(NH3)2]+ ion.
Similarly for AgCl(s) and NH3 …….
16.8 Equilibria Involving Complex Ions
57
CaSO4 will not dissolve in strong acid, however adding EDTA …….
The complexation reaction competes successfully with the precipitation reaction and CaSO4 dissolves
CaSO4 (s) Ca2+ (aq) + SO4
2– (aq)
Ca2+ (aq) + EDTA 4–
(aq) [CaEDTA]2– (aq)
KSP = 9.1 × 10–6
Kf = 1.0 × 1011
16.8 Equilibria Involving Complex Ions
58
16.8 Equilibria Involving Complex Ions
EXAMPLES
59
A hydrated cation (a complex ion) can be a weak acid ….
High charge denisty of the metal ion tends to withdraw electron density from, and hence weaken, the O–H bonds e.g. Fe2+
(aq), Zn2+(aq), Al3+
(aq)
[Fe(H2O)6]3+ (aq) + H2O (l) [Fe(H2O)5OH]2+
(aq) + H3O+ (aq)
Ka = 9.0 × 10–4
16.8 Equilibria Involving Complex Ions
60
Certain metal hydroxides, insoluble in water, are amphoteric; they will react with both strong acids or strong bases
Al(OH)3(H2O)3 (s)[Al(H2O)6]3+ (aq) [Al(OH)4(H2O)2]–
(aq)
NaOH (aq)HCl (aq)
16.8 Equilibria Involving Complex Ions
61
The amphoteric behavior is a result of the balance between acidic and basic behavior of the Al–O–H unit:
+ OH–
(OH)2Al O H
(OH)2Al O+ H2O
–
In an amphoteric substance, M (Al above) is not strongly electronegative as it is for the oxoacids, nor is it highly electropositive as it is in strong bases
As an acid, O–H breaks
+ H3O+
(OH)2Al O H
(OH)2Al+ 2 H2O
As a base, M–O breaks
+
16.8 Equilibria Involving Complex Ions
62
16.7 Qualitative Cation Analysis
63
QIA is used to identify the cations and anions present in a sample
It has largely been replaced by modern instrumental methods
Add precipitating
ionC
entri
fuge
Add precipitating
ion
Cen
trifu
ge
16.7 Qualitative Cation Analysis
64
16.7 Qualitative Cation Analysis
65
Step 1 Add
NH3(aq)
Cen
trifu
ge
Cen
trifu
ge
Step 2 Add HCl
Step 3 Add NaOH
Cen
trifu
geStep 4
Add HCl, Na2HPO4
Step 5 Dissolve in
HCl and add KSCN
66
Chapter 16 Summaries
The Solubility Product Constant, Ksp, represents an equilibrium between a slightly soluble ionic compound and its ions in a saturated aqueous solution
The Common Ion Effect is responsible for the reduction in solubility of a slightly soluble ionic compound according to LeChatelier’s principle
Precipitation is assumed to be complete if no more than 0.1% of the target ion remains in solutionThe Solubilities of some slightly soluble compounds depends strongly on pH
Certain solutes become more soluble in the presence of species that canserve as ligands (Lewis Bases) in Complex Ions
The ability of water to donate protons when acting as a ligand accounts for theacidic character of some complex ions
Precipitation, acid–base, redox, complex-ion formation, and amphotericbehavior are all used for the Qualitative Analysis of common cations
67
Chapter 16 Summaries
A strong electrolyte that produces an ion common to the ionization equilibrium of a weak acid or a weak base suppresses the ionization of the weak electrolyte – this is called the COMMON ION EFFECT
Acid–base indicators are weak acids for which the acid and its conjugate base have different colors