Buffers, Titrations, and
Aqueous Equilibria
Common Ion Effect
The shift in equilibrium that occurs because
of the addition of an ion already involved in
the equilibrium reaction. (Le Chatelier’s
principle)
AgCl(s)  Ag+(aq) + Cl(aq)
adding


NaCl( aq ) shifts equilibrium position
This affects the concentrations of other ions,
notably H+
Calculations with ICE
What is the [H+] and % ionization of 1.0 M HF mixed
with 1.0 M NaF. (1.0 M HF alone, [H+]=2.7 x 10-2, %ion=2.7%)
HF(aq) <====> H+ (aq) + F- (aq)
I 1.0
0
1.0
C -x
+x
+x
E 1.0-x
x
1.0 + x
Ka = 7.2 x 10-4 = x (1.0+x) / (1.0 - x) = x/1
x=[H+]= 7.2 x 10-4
% ion = 7.2 x 10-4 /1.0 = .072%
A Buffered Solution
. . . resists change in its pH when either H+ or
OH are added.
1.0 L of 0.50 M H3CCOOH
+ 0.50 M H3CCOONa
pH = 4.74
Adding 0.010 mol solid NaOH raises the pH
of the solution to 4.76, a very minor change.
Key Points on Buffered
Solutions
1. They are weak acids or bases containing a
common ion.
2. After addition of strong acid or base, deal
with stoichiometry first, then equilibrium.
Demonstration of Buffer Action
A buffered solution of 1.0 L of 0.5 M HC2H3O2 Ka=1.8 x
10-5 and 0.5 M NaC2H3O2 has 0.01 mol of solid NaOH
added. What is the new pH?
HC2H3O2 (aq)<====> C2H3O2-(aq) + H+ (aq)
I
0.5
0.5
0
C
-x
+x
+x
0.5 +x
x
E 0.5 -x
Ka=1.8 x 10-5 = 0.5(x) / 0.5 x= 1.8 x 10-5 pH= 4.74
When OH- is added, it takes away an equal amount of H+ ions,
and will affect also the acid concentration by the same amount.
It will also add to the common ion.
HC2H3O2 (aq)<====> C2H3O2-(aq) + H+ (aq)
I 0.5
0.5
S -0.01
+0.01
1.8 x 10-5
- 1.8 x 10-5 due to adding base
I 0.49
0.51
0
C -x
+x
+x
E 0.49-x
0.51 +x
x
Ka= 1.8 x 10-5 = x (.51)/ .49
x= 1.73 x 10-5 pH= 4.76
Practically no change in pH after base is added.
Henderson-Hasselbalch
Equation
-
Useful for calculating pH when the
[A]/[HA] ratios are known.
pH  pKa  log( A  / HA ) 
pKa  log( base / acid )
Base Buffers With H/H
Equation
pOH = pKb + log ( [HB+] / [B] ) =
pKb + log ([acid]/ [base])
Base buffers can be calculated in similar fashion
to acid buffers. pH = 14 - pOH
Buffered Solution Characteristics
-
Buffers contain relatively large amounts of
weak acid and corresponding base.
-
Added H+ reacts to completion with the weak
base.
-
Added OH reacts to completion with the
weak acid.
-
The pH is determined by the ratio of the
concentrations of the weak acid and weak
base.
H/H Calculations
From the previous example, [HC2H3O2] = .49 M after the
addition of base, and [C2H3O2-] = .51 M. Using the H/H
equation,
pH= pKa + log (.51)/(.49)
= 4.74 + 0.02 = 4.76 same as using ICE, but easier
A weak base buffer is made with 0.25 M NH3 and 0.40 M
NH4Cl. What is the pH of this buffer? Kb = 1.8 x 10-5
NH3 (aq) + H2O (l) <===> NH4+ (aq) + OH- (aq)
pOH = pKb +log ([NH4+] / [NH3]) = 4.74 +log (.40/.25)= 4.94
pH = 14- 4.94 = 9.06
H/H Calculations
The weak base buffer with pH of 9.06 from the previous
example has 0.10 mol of HCl added to it. What is the new
pH?
NH3 (aq) + H2O (l) <===> NH4+ (aq) + OH- (aq)
I
.25
S
- .10
I
.15
.40
+ .10
1.15 x 10-5
- 1.15 x 10-5 due to acid
.50
Using H/H, pOH = 4.74 + log (.5)/(.15) = 5.26
pH= 14 - 5.26 = 8.74
0
Buffering Capacity
. . . represents the amount of H+
or OH the buffer can absorb
without a significant change
in pH.
Titration (pH) Curve
A plot of pH of the solution being analyzed as
a function of the amount of titrant added.
Equivalence (stoichiometric) point: Enough
titrant has been added to react exactly with the
solution being analyzed.
15_327
pH
13.0
Equivalence
point
7.0
1.0
0
50.0
100.0
Vol NaOH added (mL)
Strong Acid/Base Titration
In titrations, it is easier to calculate millimoles(mmol), which
would be Molarity x mL, in stoichiometry.
What would be the pH of 50.0 mL of a 0.2 M solution of HCl
after 20.0 mL of 0.1 M NaOH have been added?
HCl---50.0 x 0.2 =10.0 mmol NaOH--- 20.0 x 0.1= 2.0 mmol
H+
+
OH- ----> H2O
I 10mmol
2 mmol
S -2 mmol
-2mmol
End 8mmol
0 mmol
volumes must be added
50 + 20 = 70
[H+] = 8 mmol/ 70 mL = 0.11 M
pH= 0.95
Weak Acid - Strong Base
Titration
Step 1 - A stoichiometry problem - reaction is
assumed to run to completion - then
determine remaining species.
Step 2 - An equilibrium problem - determine
position of weak acid equilibrium and
calculate pH.
15_329
12.0
Equivalence
point
pH
9.0
3.0
25
50
Vol NaOH added (mL)
Weak Acid/Strong Base
First do a stoichiometry, then equilibrium using H/H equation
What is the pH of 50.0 mL of a 0.100M HCN solution after
8.00 mL of 0.100 M NaOH has been added? Ka= 6.2 x 10-10
HCN + OH- -----> H2O + CNI
5 mmol
0.8 mmol
0 mmol
Volume
50 + 8 = 58
S - 0.8 mmol -0.8 mmol
+0.8 mmol
I 4.2 mmol /58mL
0.8 mmol /58 mL
0.072M
0.0138 M
pH = pKa + log (0.0138/0.072) = 8.49
15_330
pH
Weak acid
Strong acid
Vol NaOH
15_331
12.0
Ka = 10– 10
10.0
Ka = 10– 8
8.0
pH
Ka = 10– 6
6.0
Ka = 10– 4
4.0
Ka = 10– 2
2.0
Strong acid
0
10
20
30
40
50
60
Vol 0.10 M NaOH added (mL)
15_328
14.0
pH
Equivalence
point
7.0
50.0 mL
Vol 1.0 M HCl added
15_332
12
pH
10
Equivalence
point
8
6
4
2
0
10
20
30 40 50 60 70
Vol 0.10 M HCl (mL)
Acid-Base Indicator
. . . marks the end point of a titration by
changing color.
The equivalence point is not necessarily the
same as the end point.
15_333
OH
HO
C
C
–
O
O
OH
C
O–
C
O
(Colorless acid form, HIn)
O–
O
(Pink base form, In – )
pH
15_334
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Crystal Violet
Cresol Red
Thymol Blue
Erythrosin B
2,4-Dinitrophenol
Bromphenol Blue
Methyl Orange
Bromcresol Green
Methyl Red
Eriochrome* Black T
Bromcresol Purple
Alizarin
Bromthymol Blue
Phenol Red
m - Nitrophenol
o-Cresolphthalein
Phenolphthalein
Thymolphthalein
Alizarin Yellow R
* Trademark
CIBA GEIGY CORP.
The pH ranges shown are approximate. Specific transition ranges depend on the indicator solvent chosen.
15_335AB
14
14
12
12
10
10
Phenolphthalein
Equivalence
point
8
pH
Equivalence
point
6
Methyl red
pH
8
Phenolphthalein
6
4
4
2
2
0
0
0
20 40 60 80 100 120
Vol 0.10 M NaOH added (mL)
Methyl red
0
20 40 60 80 100 120
Vol 0.10 M NaOH added (mL)
Homework !!
p. 761 33, 35, 45 (for 35 only)
Solubility Product
For solids dissolving to form aqueous solutions.
Bi2S3(s)  2Bi3+(aq) + 3S2(aq)
Ksp = solubility product constant
and
Ksp = [Bi3+]2[S2]3
Solubility Product
“Solubility” = s = concentration of Bi2S3 that
dissolves, which equals 1/2[Bi3+] and 1/3[S2].
Bi2S3 (s) <==> 2 Bi3+ (aq)+ 3 S2(aq)
2s
3s
Note: Ksp is constant (at a given temperature)
s is variable (especially with a common
ion present)
Relation of Ksp to s
Ionic compounds will dissociate according to
set equations, which will give set Ksp to s
relation. Use ICE to find it.
A2B <==> 2A + B
2s
s
Ksp = (2s)2(s)=4s3
AB3 <==> A + 3B
s
3s
Ksp= (s)(3s)3=27s4
Ksp from s
If 4.8 x 10-5 mol of CaC2O4 dissolve in 1.0 L of solution, what is its
Ksp?
[CaC2O4]= mol / L = 4.8 x 10-5 / 1.0 L = 4.8 x 10-5 M
CaC2O4 (s) <==> Ca+2 (aq) + C2O4-2 (aq)
s
s
Ksp= (s)(s)= s2 = (4.8 x 10-5 )2 = 2.3 x 10-9
If the molar solubility (s) of BiI3 is 1.32 x 10-5, find its Ksp.
BiI3 (s) <==> Bi+3 + 3Is
3s
Ksp= (s)(3s)3= 27 s4
Ksp = 27(1.32 x 10-5)4= 8.20 x 10-19
s from Ksp
If Ksp for Ag2CO3 is 8.1 x 10-12, find its molar solubility.
Ag2CO3 (s) <==> 2 Ag+ (aq) + CO3-2 (aq)
2s
s
Ksp = (2s)2(s) = 4 s3 = 8.1 x 10-12 s= 1.27 x 10-4
If Ksp for Al(OH)3 is 2 x 10-32, find its molar solubility.
Al(OH)3 (s) <==> Al+3 (aq) + 3 OH- (aq)
s
3s
Ksp = (s)(3s)3 = 27s4 = 2 x 10-32 s = 5.22 x 10-9
Equilibria Involving Complex
Ions
Complex Ion: A charged species consisting of a
metal ion surrounded by ligands (Lewis bases).
Coordination Number: Number of ligands attached
to a metal ion. (Most common are 6 and 4.)
Formation (Stability) Constants: The equilibrium
constants characterizing the stepwise addition of
ligands to metal ions.
Complex Ion Examples
Coordination number=2
Ag(NH3)2+
Coordination number=4
Al(OH)4-1
Cu(NH3)4+2
Coordination number=6
Fe(CN)6-4
Fe(SCN)6-3
Homework and XCR
p. 762ff 50, 51, 59, 60
XCR 72, 76, 86
Super XCR 90