Transcript Applications of aqueous equilibria

pH titration curves
 pH
measured as an acid is neutralized by
a base follows a characteristic curve that
enables the equivalence-point to be
determined with precision
Constructing curves
 Shapes
of curves will depend on the
strength of the acid and base
 Construct curve by calculating [H3O+] as
the base (acid) is added.
 The strong-strong case is easy: all species
are always completely ionized. At the
equivalence point, pH = 7
 Weak acid - strong base will use pKa for
the acid
Titration of weak acid with strong
base in four zones




Zone 1: Initial pH of acid
with no base added
Zone 2: Addition of OHup to equivalence point
Zone 3: At equivalence
point: all the HAc is
converted into AcZone 4: After equivalence
point
Constructing a pH curve for the
weak acid - strong base case
 Zone
1: Initial pH of acid with no base
added
HAc (aq)  H 2O(aq)  H 3O   Ac  (aq)...K a  1.8 x10 5
 Zone
2: Addition of OH- up to equivalence
point

Treat like a buffer solution
[ Base ]
pH  pK a  log
[ Acid ]

NOTE: Concentrations change as volume
increases with addition of base!
Explicit (contains adult themes)
calculation for zone 2
 Determination
point
of [H+] up to equivalence


[
HAc
]
V

[
OH
]o VOH 

o H
[H ]  Ka 


[OH ]o VOH


 [HAc]o
is initial concentration of HAc and
VH is the initial volume of HAc used
 [OH-] is initial concentration of the base
and VOH is the volume added at that point
in the titration
Weak acid – strong base continued

Zone 3: At equivalence point: all the HAc is
converted into Ac
Treat like solution of basic salt
Ac  (aq)  H 2O(aq)  OH   HAc(aq)
[ HAc][OH  ] K w
K


[ Ac ]
Ka


[H ] 
Ka Kw
[ Ac  ]
Where [Ac-] is the initial acetate concentration at the
neutralization point prior to dissociation
Weak acid – strong base continued
 Zone

4: After equivalence point
All OH- results from the excess base
Kw
[H ] 

[OH ]

pH curves assist in identifying
suitable indicator

The equivalence point with the weak acid is at
pH>7
 Above pH 7, both curves coincide (strong base
controls pH)
 The initial rise in pH is greater with the weak
acid (but at much lower [H3O+]
End-point detection gets harder
with weaker acids
 Initial
pH is higher
 Initial change in pH is greater
 Change in pH at equivalence point is lower
 Harder to detect equivalence point in weak
acid
Weak base - strong acid
Analogous to the weak acid – strong base case
 pH at equivalence point < 7
 Dictates use of different indicator
 pH after equivalence point controlled by strong
acid

Polyprotic acids – lots to note
 An
amino acid has two dissociations:
H2A+ + H2O = HA + H3O+
HA + H2O = A- + H3O+
Example for alanine
1.
2.
3.
Initially, pH determined by pKa1
Halfway to 1st equivalence pt
pH = pKa1
1st equivalence pt,
pH 
4.
5.
6.
pK a1  pK a 2
2
Halfway to 2nd equivalence pt,
pH = pKa2
2nd equivalence pt, pH
determined by pKb for the base
A- (Kb obtained from Ka2)
Beyond 2nd equivalence pt, pH
determined from OH- from
NaOH
Solubility products – equilibrium
between solute and solid
 An
electrolyte completely dissociated in
equilibrium with undissolved solid
n
y
M m X x  mM (aq )  xX (aq )
n m
K sp  [ M ] [ X
 The
y x
]
solid phase is ignored
 Writing Ksp expressions for salts
Values of Ksp at 298 K
Aluminium
hydroxide
Al(OH)3
1.9 x 10-33
Barium
carbonate
BaCO3
2.6 x 10-9
Calcium
carbonate
CaCO3
5.0 x 10-9
PbCl2
1.2 x 10-5
PbCrO4
2.8 x 10-13
Silver chloride
AgCl
1.8 x 10-10
Silver sulphate
Ag2SO4
1.2 x 10-5
Lead chloride
Lead(II)
chromate
Calculations
 Determine
Ksp from solution
concentrations (concentrations of
individual ions may not be equal to those
you would get by simple dissociation of
compound)
 Determine solubility from Ksp
MgF2 = Mg2+ + 2FKsp = x.(2x)2 x =[Mg2+]
Factors affecting solubility
ion effect – the addition of an ion
from another source
 Solution pH – where there is a weak acid
or base
 Complex ion formation
 Amphotericity
 Common
Common ion effect

In a solution of a salt AB, addition of a additional
B ions from another source will cause [A] to
decrease because of Ksp
Calculation of solubility under these
conditions
 Calculate
solubility of MgF2 in a solution of
0.1 M NaF(aq)?

At equilibrium, [Mg2+] = x, [F-] = 0.1 + 2x
K sp  x(0.10  2 x) 2  7.4 x10 11
x(0.10) 2  7.4 x10 11
x  7.4 x10 9 M
Solution pH

Basic anions are
protonated in acid





CaCO3 is insoluble in pure
water
In acid solution, H+
converts CO32- to HCO3More Ca2+ is drawn into
solution (Le Chatelier)
Salts of basic anions
increase solubility in acid
conditions
pH does not affect anions
that are not basic
Complex ion formation affects
solubility
 AgCl
is ordinarily highly insoluble (test for
chloride ions)
 Addition of aqueous ammonia redissolves
the precipitate by formation of the complex
ion Ag(NH3)2+
 Two
stage process of formation of highly
favoured complex
 Equilibrium lies to right
Ag  (aq)  NH 3 (aq)  Ag ( NH 3 )  (aq )


Ag ( NH 3 ) (aq)  NH 3 (aq)  Ag ( NH 3 ) 2 (aq)
Amphoteric substances

Dissolve in both acid and basic solutions
 Examples of complex ion formation
 Oxides and hydroxide of Al are examples
Al2O3 ( s)  6 H 3O   2 Al 3 (aq)  9 H 2O
Al2O3 ( s)  2OH  (aq)  3H 2O  2 Al (OH ) 4 (aq)
Prediction of precipitation and ion
products
 Predict
the formation of a precipitate when
solutions are mixed
 Ion product is not an equilibrium quantity
IP = [Ca2+][F-]2
 If IP is found to be larger than Ksp then
precipitation occurs
Selective precipitation

Mixtures of ions can be separated by combining with an
anion that has wide range of solubiity.


For example, sulphides of Zn, Pb, Cu and Hg will precipitate
leaving the sulphides of Mn, Fe, Co and Ni in solution
Ksp for former group are much larger than for latter group
MnS
3 x 1010
ZnS
3 x 10-2
FeS
6 x 102
PbS
3 x 10-7
CoS
3
CuS
6 x 10-16
NiS
8 x 10-1
HgS
2 x 10-32
Qualitative analysis

Apply a sequence of precipitation steps to divide a group
of many metal ions into smaller groups. These smaller
groups will be further analyzed to identify the members
therein