Equilibria ppt

1
Aspects of Aqueous Equilibria
2
Aspects of Aqueous Equilibria The Common Ion
Effect
Recall that salts like sodium acetate are strong
electrolytes
NaC2H3O2(aq) ? Na(aq) C2H3O2-(aq)
and that the C2H3O2- ion is a conjugate base of
a weak acid
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
H3O C2H3O2-
Ka
HC2H3O2
3
Aspects of Aqueous Equilibria The Common Ion
Effect
Now, lets think about the problem from the
perspective of
LeChateliers Principle
What would happen if the concentration of the
acetate ion were increased?
H3O C2H3O2-
Ka
HC2H3O2
Q gt K and the reaction favors reactant
Addition of C2H3O2- shifts equilibrium, reducing
H
4
Aspects of Aqueous Equilibria The Common Ion
Effect
H3O C2H3O2-
Ka
HC2H3O2
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
Since the equilibrium has shifted to favor the
reactant, it would appear as if the dissociation
of the weak acid(weak electrolyte) had decreased.
5
Aspects of Aqueous Equilibria The Common Ion
Effect
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
So where might the additional C2H3O2-(aq) come
from? Remember we are not adding H. So its not
like we can add more acetic acid.
How about from the sodium acetate?
6
Aspects of Aqueous Equilibria The Common Ion
Effect
NaC2H3O2(aq) ? Na(aq) C2H3O2-(aq)
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
In general, the dissociation of a weak
electrolyte (acetic acid) is decreased by adding
to the solution a strong electrolyte that has an
ion in common with the weak electrolyte
The shift in equilibrium which occurs is called
the COMMON ION EFFECT
7
Aspects of Aqueous Equilibria The Common Ion
Effect
Lets explore the COMMON ION EFFECT in a little
more detail
Suppose that we add 8.20 g or 0.100 mol sodium
acetate, NaC2H3O2, to 1 L of a 0.100 M solution
of acetic acetic acid, HC2H3O2. What is the pH of
the resultant solution?
NaC2H3O2(aq) ? Na(aq) C2H3O2-(aq)
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
8
Aspects of Aqueous Equilibria The Common Ion
Effect
Calculate the pH of a solution containing 0.06 M
formic acid (HCH2O, Ka 1.8 x 10-4) and 0.03 M
potassium formate, KCH2O.
Now you try it!
9
Aspects of Aqueous Equilibria The Common Ion
Effect
Calculate the fluoride ion concentration and pH
of a solution containing 0.10 mol of HCl and 0.20
mol HF in 1.0 L
HCl H2O(aq) ? H3O(aq) Cl-(aq)
HF(aq) H2O ? H3O(aq)
F-(aq)
Note who the strong electrolyte is this time!
10
Aspects of Aqueous Equilibria The Common Ion
Effect
Now, lets think about the problem from the
perspective of
LeChateliers Principle
But this time lets deal with a weak base and a
salt containing its conjugate acid.
NH3(aq) H2O ? NH4(aq) OH-
NH4 OH-
Kb
NH3
Q gt K and the reaction favors reactant
Addition of NH4 shifts equilibrium, reducing OH-
11
Aspects of Aqueous Equilibria The Common Ion
Effect
Calculate the pH of a solution produced by mixing
0.10 mol NH4Cl with 0.40 L of 0.10 M NH3(aq), pKb
4.74?
NH4Cl(aq) ? NH4(aq) Cl- (aq)
NH3(aq) H2O ? NH4(aq) OH-
12
Aspects of Aqueous Equilibria Common
Ions Generated by Acid-Base Reactions
The common ion that affects a weak-acid or
weak-base equilibrium may be present because it
is added as a salt, or the common ion can be
generated by reacting an acid and base directly
(no salt would be necessary.which is kind of
convenient if you think about it)
Lets take a look at a weak acid-strong base
combination first
Suppose we react 0.20 mol of acetic acid with
0.10 mol of sodium hydroxide
HC2H3O2(aq) OH- ? H2O C2H3O2
0
0.20 mol
0.10 mol
-0.10 mol
-0.10 mol
-0.10 mol
0.10 mol
0.10 mol
0
13
Aspects of Aqueous Equilibria Common
Ions Generated by Acid-Base Reactions
Suppose we react 0.20 mol of acetic acid with
0.10 mol of sodium hydroxide
HC2H3O2(aq) OH- ? H2O C2H3O2
0
0.20 mol
0.10 mol
-0.10 mol
-0.10 mol
0.10 mol
0.10 mol
0.10 mol
0
Lets suppose that all this is occurring in 1.0 L
of solution
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
0
0.10 M
0.10 M
14
Aspects of Aqueous Equilibria Common
Ions Generated by Acid-Base Reactions
Sample problem Calculate the pH of a solution
produced by mixing 0.60 L of 0.10 M NH4Cl with
0.40 L of 0.10 M NaOH
NH4Cl(aq) ? NH4(aq) Cl- (aq)
NH4 OH- ? NH3
H2O
0.04 mol
0
0.06 mol
-0.04 mol
0.04 mol
-0.04 mol
0.04 mol
0.02 mol
0
NH4 H2O ? H3O NH3
0.04 M
0.02 M
0
Dont forget to convert to MOLARITIES
15
Aspects of Aqueous Equilibria Common
Ions Generated by Acid-Base Reactions
Calculate the pH of a solution formed by mixing
0.50 L of 0.015 M NaOH with 0.50 L of 0.30 M
benzoic acid (HC7H5O2, Ka 6.5 x 10-5)
Now you try it!
16
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
A buffered solution is a solution that resists
change in pH upon addition of small amounts of
acid or base.
HERES HOW IT WORKS !
Suppose we have a salt
MX ? M (aq) X- (aq)
HX H2O ? H3O X-
And weve added the salt to a weak
acid containing the same conjugate base as the
salt, HX
H X-
Ka
And the equilibrium expression for this reaction
is
HX
Note that the concentration of the H
is dependent upon the Ka and the ratio between
the HX and X- (the conjugate acid-base pair)
HX
Ka
H
X-
CONTINUED ON NEXT SLIDE
17
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
Two important characteristics of a buffer are
buffering capacity and pH. Buffering capacity is
the amount of acid or base the buffer can
neutralize before the pH begins to change to an
appreciable degree.

• The pH of the buffer depends upon the Ka

• This capacity depends on the amount of acid and
  base from which the
• buffer is made

• The greater the amounts of the conjugate
  acid-base pair, the more
• resistant the ratio of their concentrations, and
  hence the pH, to change

HX
Ka
H
Henderson-Hasselbalch Equation
X-
HX
pH
pKa
-log Ka
X-
-logH
log

X-
HX
pH
pKa
-
log
18
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
NaC2H3O2(aq) ? Na(aq) C2H3O2-(aq)
0.1 M
0.1 M
0.1 M
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
0
0.1
0.1 M
x
x
-x
x
0.1 - x
0.1 x
x(0.1 x )
1.8 x 10-5
pH 4.74
0.1 - x
x 1.8 x 10-5
Henderson-Hasselbalch Equation
Note that these are initial concentrations
.1
log
pH
4.74

.1
19
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
A liter of solution containing 0.100 mol of
HC2H3O2 and 0.100 mol NaC2H3O2 forms a buffered
solution of pH 4.74. Calculate the pH of this
solution (a) after 0.020 mol NaOH is added, (b)
after adding 0.020 mol HCl is added.
NaC2H3O2(aq) ? Na(aq) C2H3O2-(aq)
0.1 M
0.1 M
0.1 M
HC2H3O2(aq) OH- ? H2O C2H3O2-(aq)
0.1 M
0.1 M
0.02 M
Note that the OH- reacts with the HX
-0.02 M
-0.02 M
Step 1
0.02 M
0.12 M
0.00 M
0.08 M
Henderson-Hasselbalch Equation
Note that these are initial concentrations
.12
Step 2
pH
4.74
log

.08
pH 4.92
20
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
A liter of solution containing 0.100 mol of
HC2H3O2 and 0.100 mol NaC2H3O2 forms a buffered
solution of pH 4.74. Calculate the pH of this
solution (a) after 0.020 mol NaOH is added, (b)
after adding 0.020 mol HCl is added.
C2H3O2-(aq) H ? HC2H3O2
0.10 M
0.10 M
0.02 M
Step 1
0.02 M
-0.02 M
-0.02 M
0.08M
0.00 M
0.12 M
Henderson-Hasselbalch Equation
Note that these are initial concentrations
.08
pH
4.74

log
Step 2
.12
pH 4.56
21
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
Now consider, for a moment, what would have
happened if I had added 0.020 mol of NaOH or
0.02 mol HCl to .1 M HC2H3O2.
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
0
0
0.1 M
x
x
-x
x
0.1 - x
x
x2
1.8 x 10-5
pH 2.9
0.1 - x
x 0.0013
22
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
HC2H3O2(aq) OH- ? H2O C2H3O2-(aq)
0.1 M
0.00
0.02 M
-0.02 M
-0.02 M
0.02 M
0.02 M
0.00 M
0.08 M
Henderson-Hasselbalch Equation
.02
pH
4.74
log

.08
pH 4.13
23
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
C2H3O2-(aq) H ? HC2H3O2
0.00
0.10 M
0.02 M
Complete dissociation
pH -log 0.02
pH 1.7
24
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
25
Aspects of Aqueous Equilibria BUFFERED SOLUTIONS
Sample exercise Consider a litter of buffered
solution that is 0.110 M in formic acid (HC2H3O
)and 0.100 M in sodium formate (NaC2H3O ).
Calculate the pH of the buffer (a) before any
acid or base are added, (b) after the addition of
0.015 mol HNO3, (3) after the addition of 0.015
mol KOH
26
Aspects of Aqueous Equilibria Titration Curves
Stoichiometrically equivalent quantities of acid
and base have reacted
HCl(aq) NaOH(aq) ? H2O NaCl
27
Titration of a weak acid and a strong base
results in pH curves that look similar to those
of a strong acid-strong base curve except that
the curve (a) begins at a higher pH, (b) the pH
rises more rapidly in the early part of the
titration, but more slowly near the equivalence
point, and (3) the equivalence point pH is not 7.0
28
Aspects of Aqueous Equilibria Titration Curves
of Both Strong and Weak Acids
29
Aspects of Aqueous Equilibria Calculating pHs
from Titrations
Calculate the pH in the titration of acetic acid
by NaOH after 30.0 ml of 0.100 M NaOH has been
added to 50 mL of 0.100 acetic acid
HC2H3O2 (aq) OH- ? H2O(l) C2H3O2
0.005 mol
0
0.003 mol
0.003 mol
-0.003 mol
-0.003 mol
0.003 mol
-0.002 mol
0
.0370
log
pH
4.74

.0250
pH 4.91
30
Aspects of Aqueous Equilibria Determining the
Ka From the Titration Curve
pKa pH 4.74
31
Aspects of Aqueous Equilibria Titrations of
Polyprotic Acids
Na2CO3 ? 2Na(aq) CO32-
H(aq) CO32- ? HCO3- (aq)
H(aq) HCO- ? H2CO3 (aq)
32
Aspects of Aqueous Equilibria
Solubility Equilibria and Solubility Product
Consider the following reaction
CaF2(s) ? Ca2(aq) 2F-(aq)
At equilibrium, the law of mass action expression
is
Solubility describes the equilibrium
concentrations
Ksp Ca2F-2
Where Ksp is called the solubility product
constant. The term solubility, on the other hand,
is a measure of the degree to which a substance
dissolves in pure water. For example, the
solubility of NaCl in water increases as the
temperature of the water increases. Solubility
also varies according the the concentration of a
common ion.
33
Aspects of Aqueous Equilibria
Solubility Equilibria and Solubility Product
Two sample Problems
Copper (I) bromide has a measured solubility of
2.0 x 10-4 M at 25C. Calculate the Ksp value.
The Ksp value for copper (II) iodate, Cu(IO3)2 is
1.4 x 10-7 at 25C. Calculate the solubility at
25C.
34
Aspects of Aqueous Equilibria
Solubility Equilibria and Solubility Product
Suppose you are comparing three salts the produce
the same number of ions
AgI(s) ? Ag1(aq) I-(aq) Ksp 1.5
x 10-16
CuI(s) ? Cu1(aq) I-(aq) Ksp 5.0
x 10-12
CaSO4(s) ? Ca2(aq) SO42-(aq) Ksp 6.1 x 10-5
Lets talk a little about the size of the Ksp
value
Under these conditions, CaSO4 is more soluble
than CuI, which is more soluble than AgI.
The larger the Ksp the more soluble the compound
35
Aspects of Aqueous Equilibria
Solubility Equilibria and Solubility Product
CuS(s) ? Cu2(aq) S-(aq) Ksp 8.5
x 10-45
Ag2S(s) ? 2Ag2(aq) S2-(aq) Ksp 1.6
x 10-49
Bi2S3(s) ? 2Bi3(aq) 3S2-(aq) Ksp 1.1
x 10-73
The rules change, however, if different number
so ions are produced
8.5 x 10-45 x2 x 9.2 x 10-23
1.6 x 10-49 (2x)2(x) 4x3 x 3.4 x 10-17
1.1 x 10-73 (2x)2(3x)3 36x5 x 1.3 x 10-15
Note that the order of solubility is different
than that of the solubility product constants.
36
Aspects of Aqueous Equilibria
Solubility Equilibria and Solubility Product
Suppose you are asked to determine the solubility
of solid silver chromate in a solution 0.100 M
solution of AgNO3
We have two things going on
AgNO3 ? Ag(aq) NO3-(aq) 0.10M 0.10M
0.10M
Ag2CrO4(s) ? 2Ag(aq) CrO42-(aq)
Ksp 9.0 x 10-12
Note how things change when we take the common
ion effect into account
9.0 x 10-12 (2x 0.1)2(x)
x 9.0 x 10-10
? Ag 0.10 M, CrO42- 9.0 x 10-10
37
Aspects of Aqueous Equilibria
Solubility Equilibria and Solubility Product
A solution is prepared by adding 750.0 ml of 4.00
x 10-3 M Ce(NO3)3 to 300 ml. of 2.0 x 10-2 M
KIO. Will Ce(IO3)3 (Ksp 1.9 x 10-10)
precipitate from this reaction?
Ce(IO3)3(s) ? Ce3(aq) 3IO-(aq)
Ksp 1.9 x 10-10
4.00 x 10-3 mol
Ce3
0.750 L
2.86 x 10-3 M
1.050 L
2.00 x 10-2 mol
IO3-
0.300 L
5.71 x 10-3 M
Then there is the Q vs K stuff
1.050 L
Q 2.86 x 10-3 M 5.71 x 10-3 M 3 5.32 x
10-10
Since Q gt K, the reaction will favor the
production of precipitate