Additional Aspects of Aqueous Equilibria
Aspects of Aqueous Equilibria:
The Common Ion Effect
Salts like sodium acetate are strong electrolytes
NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)
The C2H3O2- ion is a conjugate base of a weak acid
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
Ka =
[H3O+] [C2H3O2-]
[HC2H3O2]
The Common Ion Effect
Now, lets think about the problem from the
perspective of LeChatelier’s Principle
What would happen if the concentration of the
acetate ion were increased?
+] [C H O -]
[H
O
3
2 3 2
Ka =
[HC2H3O2]
Q > K and the reaction favors reactant
Addition of C2H3O2- shifts equilibrium, reducing H+
The Common Ion Effect K =
a
[H3O+] [C2H3O2-]
[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.
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 it’s
not like we can add more acetic acid.
How about from the
sodium acetate?
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
Let’s 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)
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!
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  H3O+(aq) + Cl-(aq)
(aq)
HF (aq) + H2O  H3O+ (aq) + F-(aq)
Now, lets think about the problem from the
perspective of LeChatelier’s Principle
But this time lets deal with a weak base
and a salt containing its conjugate acid.
NH3(aq) + H2O  NH4+(aq) + OH- (aq)
+] [OH-] Q > K and the
[NH
4
Kb =
reaction favors
[NH3]
reactant
Addition of NH4+ shifts equilibrium, reducing OH-
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-
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)
Suppose we react 0.20 mol of acetic acid (weak)
with 0.10 mol of sodium hydroxide strong)
HC2H3O2 (aq) + OH-(aq)  H2O + C2H3O2-(aq)
0.20 mol 0.10 mol
-0.10 mol -0.10 mol
0
0.10 mol
0
+0.10 mol
0.10 mol
Suppose we react 0.20 mol of acetic acid with
0.10 mol of sodium hydroxide
HC2H3O2(aq) + OH- (aq) H2O +
0.20 mol
-0.10 mol
0.10 mol
0.10 mol
-0.10 mol
0
C2H3O2-aq)
0
0.10 mol
0.10 mol
Let’s 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
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
0.06 mol 0.04 mol
-0.04 mol -0.04 mol 0.04 mol
0
0.02 mol
0.04 mol
Don’t
forget
to convert
to
MOLARI
TIES
NH4+ +
0.02 M
H2O  H3O+
0
+
NH3
0.04 M
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!
adding acid or base

calculate the pH of a solution that has .2
mol of NaOH added to a solution that is
.25 M HC2H3O2 and .32M NaC2H3O2
HC2H3O2(aq) + OH- (aq) H2O + C2H3O2-aq
.20
.32
..25
-.20
-.20
+.20
.05
0
.52





HC2H3O2(aq) 
.05
-X
H+
0
X
X (.52+X) =1.8 x 10-5
.05-X
+
C2H3O2-aq
.52
X
BUFFERED SOLUTIONS
A buffered solution is a solution that resists change in
pH upon addition of small amounts of acid or base.
Suppose we have a salt: MX  M+(aq) + X-(aq)
And we’ve added the salt to a
weak acid containing the
same conjugate base as the
HX +H2O  H3O+ + Xsalt, HX:
And the equilibrium
[H+ ] [ X-]
expression for this reaction
Ka =
[HX]
is
Note that the concentration of
the H+ is dependent upon the
Ka and the ratio between the
[HX]
HX and X (the conjugate
[H +] = Ka
[X-]
acid-base pair)
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
[HX]
hence the pH, to change
+
K
[H ] = a
[X-]
Henderson-Hasselbalch Equation
[HX]
+
-]
-log[H ] = -log Ka
[X
pH = pKa log +
[X-]
[HX]
NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)
0.1 M
0.1 M
0.1 M
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
0.1 M
0
-x
0.1 - x
x
x
0.1
x
0.1 + x
x(0.1
+
x
)
-5
pH = 4.74
-5
x
=
1.8
x
10
1.8 x 10 = 0.1 - x
Using the Henderson-Hasselbalch Equation
Note that these
[.1]
are initial
pH = 4.74 + log
[.1]
concentrations
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
Step 1
HC2H3O2(aq) + OH-  H2O + C2H3O2- (aq)
0.1 M
0.02 M
0.1 M
-0.02 M
0.08 M
-0.02 M
0.00 M
0.02 M
0.12 M
Step 2 Henderson-Hasselbalch Equation
Note that these are
[.12]
pH = 4.74 + log
initial concentrations
[.08]
pH = 4.92
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 0.02 M
0.10 M
Step 1
-0.02 M -0.02 M
0.02 M
0.12 M
0.08M
0.00 M
Henderson-Hasselbalch Equation
Step 2 pH = 4.74 + log [.08] Note that these are
pH = 4.56
[.12] initial concentrations
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
0.1 M
0
-x
x
x
0.1 - x
x
x
2
x
1.8 x 10-5 = 0.1 - x
x = 0.0013
pH = 2.9
HC2H3O2(aq) + OH- 
H2O
+
C2H3O2-(aq)
0.1 M
0.02 M
0.00
-0.02 M
-0.02 M
0.02 M
0.08 M
0.00 M
0.02 M
Henderson-Hasselbalch Equation
pH = 4.74 + log [.02]
[.08]
pH = 4.13
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-aq
0.10
0.02 M
0
H+ from HCl Completely dissociates therefore the
pH is calculated without regard for the weak acid
pH = -log [0.02]
pH = 1.7
add 2 ml
10 M HCl
to 1.8 x 10-5M
HCl
add 2 ml
10 M HCl
to .1M HC2H3O2
+ .1M NaC2H3O2
pH = 4.74
pH = 4.74
.02 M HCl
pH= 1.7
a drop of 3.04
to .12M HC2H3O2
+ .06M NaC2H3O2
pH = 4.56
a drop of .18
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,
(c) after the addition of 0.015 mol KOH
Titration Curves
End points
Stoichiometrically
equivalent quantities of
acid and base have
reacted
HCl(aq) + NaOH(aq)  H2O + NaCl
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
Calculating pH’s 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) + C2H3O20.005 mol 0.003 mol
0
0.003 mol
-0.003 mol -0.003 mol
-0.002 mol
0.003 mol
0
[.0370]
pH = 4.74 + log [.0250]
pH = 4.91
Determining the Ka From the Titration Curve
pKa = pH = 4.74
Solubility Equilibria Ksp
The equilibrium expression for the following
reaction is CaF2(s)  Ca2+(aq) + 2F-(aq)
[Ca2+] [F-]2
= Ksp
[CaF2]
Look at the table on page 759 or the
appendices A26, where you will find the
value of the ksp to be 4.1 x 10-11
Calculating Ksp from solubility
The solubility of Bi2S3 is 1.0 x 10-15M
what would be the Ksp?
Bi2S3(s) 
2Bi3+(aq) + 3S2-(aq)
1.0 x 10-15
2(1.0 x 10-15) 3(1.0x 10-15)
Ksp = [Bi3+]2 [S2-]3
Ksp = (2.0 x 10-15)2(3.0 x 10-15)3 = 1 x 10-73
Calculating solubility from Ksp
The Ksp of Cu(IO3)2 is 1.4 x 10-7 what
would be the solubility?
Cu(IO3)2(s)  Cu2+(aq) + 2IO31-(aq)
X
X
2X
Ksp= 1.4 x 10-7 = (X)(2X)2 = 4X3
X = 3.3 x 10-3 mol/L
Common ion effect
What would be the solubility of Ag2CrO4
in a solution that is .1M AgNO3,
the Ksp = 9.0 x 10-12
Ag2CrO4  2Ag1+ + CrO42X
2X + .1
X
9.0 x 10-12 = (2X+.1)2(X)
Drop the 2X as insignificant
X = 9.0 x 10-10 mol/L
pH and Solubility
Mg(OH)2  Mg2+ + 2OHIf the pH is raised (the OH- is raised)
then we have the common ion effect
and the solubility is decreased.
If the pH is lowered (the H+ is raised)
then OH- is reacted with H+ to make
water and the solubility is increased.
pH and solubility of salts
Ag3PO4  3Ag1+ + PO43If the pH is lowered (the H+ is raised) the
PO43- reacts with H+ to make HPO42- ,
which essentially removes PO43from the equation, shifting the reaction
to the right and the solubility is
increased.
Will a precipitate form?
If 750 mL of 4.00E-3M Ce(NO3)3 is mixed with 300
mL of 2.00E-2M KIO3, will a precipitate form?
Ce(NO3)3(aq) + KIO3 (aq) → Ce(IO3)3(s) + KNO3(aq)
(750)(4 x 10-3) = (1050) X; X= 2.86 x 10-3M for Ce3+
(300)(2 x 10-2) = (1050) Y; Y= 5.71 x 10-3 M for IO31Ce(IO3)3(s)  Ce3+ (aq) + 3IO31- (aq)
Ksp = 1.9 x 10-10
Q = (2.86 x 10-3)(5.71 x 10-3)3 = 5.32 x 10-10
Q is greater than K so a precipitate will form