Ch. 16: Equilibrium in
Acid-Base Systems
16.3a: Acid-Base strength and
equilibrium law
Definitions

Arrhenius




A: produce H+ in aqueous solution
B: produces OH- in aqueous solution
very limited
Bronsted-Lowry



A: H+ donor
B: H+ acceptor
more general
Acid ionization constant


equilibrium expression where H+ is
removed to form conjugate base
so for: HA + H2O <--> H3O+ + A-




[ H 3O ][ A ] [ H ][ A ]
Ka 

[ HA]
[ HA]
Strength


determined by equilibrium position of
dissociation reaction
strong acid:




lies far to right, almost all HA is dissociated
large Ka values
creates weak conjugate base
weak acid:



lies far to left, almost all HA stays as HA
small Ka values
creates strong conjugate base
Water is a stronger base than the CB of a strong
acid but a weaker base than the CB of a weak acid
Water is a stronger acid than the CA of a strong
base but a weaker acid than the CA of a weak base
[H2O], pH and Kw

conc. of liquid water is omitted from the Ka
expression




we assume that this conc. will remain constant in
aqueous sol’n that are not highly concentrated
pH= -log[H+]
pOH = -log[OH-]
14.00= pH + pOH


K w  [ H 3O ][OH ]  1.0 10
14
Example 1

The [OH-] of a solution at 25oC is 1.0x10-5 M.
Determine the [H+], pH and pOH.






Kw = 1.0x10-14 = [OH-] x [H+]
[H+] = 1.0x10-9
pH= -log(1.0x10-9) = 9.00
pOH = -log(1.0x10-5) = 5.00
acidic or basic?
basic
Approximations



If K is very small, we can assume that the
change (x) is going to be negligible
“rule of thumb” is if initial conc. of the acid is
>1000 times its Ka value then cancel x
this makes the answer true to +/- 5% and why Ka
values are given to 2 sig. digs
2
2
( x)( 2 x) 0 ( x)( 2 x)
3
K

 4x
2
2
(1.0  2 x)
(1.0)
Calculating Weak Acids
1.
2.
3.
4.
5.
6.
Write major species
Decide on which can provide H+ ions
Make ICE table
Put equilibrium values in Ka
expression
Check validity of assumption (x must
be less than 5% of initial conc)
Find pH
Example 2

Calculate the pH of 1.00 M solution of
HF (Ka = 7.2 x 10-4)
 HF, H2O
+
 HF  H + F
Ka = 7.2x10-4
+
 H2O  H + OH
Kw = 1.0 x 10-14
+
 HF will provide much more H than
H2O – ignore H2O

HF
Example 2
H+
F-
+
I
1.00 M
0
0
C
-x
+x
+x
E
1.00 -x
x
x


2
[ F ][ H ] ( x)( x)
x
4
Ka 


 7.2  10
[ HF ]
1.00  x 1.00
x  7.2 10
2
4
 x  0.027 M
Example 2

Check assumption:
x
100  5%
[ HA]
0.027
100  2.7%  5%
1.00
pH
= -log(0.027) = 1.57
Example 3

Find pH of 0.100 M solution of HOCl (Ka = 3.5x10-8)
 HOCl, H2O
+
 HOCl will provide much more H than H2O, so we ignore
H 2O
HOCl

H+
+
OCl-
I
0.100 M
0
0
C
-x
+x
+x
E
0.100 -x
x
x
Example 3


2
[OCl ][ H ]
( x)( x)
x
8
Ka 


 3.5 10
[ HOCl ]
0.100  x 0.100
9
5
x  3.5 10  x  5.9 10 M
2

Check assumption:
5
5.9 10
100  0.059%  5%
0.100
pH
= -log(5.9x10-5) = 4.23
Homework


Textbook p743 #2a,c,e 5,7,9
LSM 16.3A and 16.3D
Ch. 16: Equilibrium in
Acid-Base Systems
16.3b: Base strength and
equilibrium law
Base Strength and Kb


follows same standard rules as for
calculating Ka for acids
Kb is used with weak bases that react
only partially with water (<50%)
Bases

Kb




base ionization constant
refers to reaction of base with water to make
conjugate acid and OHliquid water is again ignored like in Ka
B(aq) + H2O (l)  BH+ (aq) + OH- (aq)


[ BH ][OH ]
Kb 
[ B]
Example 4

Find the pH for 15.0 M solution of NH3
(Kb = 1.8x10-5)

NH3 will create more OH- than water so
self- ionization (H2O) can be ignored
NH3 + H2O <--> NH4+ + OHNH3
+ H2O  NH4+
+
OH-
I
15.0
0
0
C
-x
+x
+x
E
15.0-x
x
x
Example 4 con’t
2
[ x][ x]
x
5
Kb 
 1.8 10 
[15.0  x]
15.0
x  0.016
% ion.
0.016
100  0.11%  5%
15.0

x  [OH ]  0.016M
14
1.0 10
13
[H ] 
 6.2 10
0.016
13
pH   log(6.2 10 )  12.20

Example 5


Codeine (C18H21NO3) is a weak organic base. A
5.0x10-3 M solution of codeine has a pH of 9.95.
Calculate the Kb for this substance.




Sol’n
What is chemical reaction?
C18H21NO3 + H2O <--> HC18H21NO3+ + OHFind [OH-] using given pH


pOH= 14.00-9.95 = 4.05
[OH-] = 10-4.05 = 8.9x10-5
Example 5 con’t
C18H21NO3 + H O<--> HC18H21NO3+ +
2
5.0x10-3
-x
5.0x10-3 - x
I
C
E

0
+x
x
OH0
+x
x
x = [OH-] = 8.9x10-5
[ x][ x]
[8.9 10 5 ]2
Kb 

-3
-3
5
[5.0x10  x] [5.0x10  8.9 10 ]
K b  1.6 10 6
Ka - Kb relationship for conjugate pairs



Acid-Base strength tables do not give Kb values
we use the Ka - Kb relationship to solve this
problem
this can be used with any conjugate acid-base
pair
Kw = Ka x Kb
or
Kb = Kw / Ka
Kw = 1.0 x 10-14
Example 6

What is the Kb value for the weak base present when
sodium cyanide dissociates into an aqueous solution?
NaCN --> Na+ + CN- (complete dissociation)
the CN- is the weak base so we write the equilibrium equation
with water to determine its conjugate acid
CN- + H2O <--> HCN + OHthe conjugate acid is found to be HCN and its Ka is 6.2 x 10-10
Using Kb = Kw / Ka find Kb for CNKb = 1 x 10-14 / 6.2 x 10-10 = 1.61 x 10-5
Homework



Textbook p746 #11,12
Textbook p750 #2,6,9
LSM 16.3 A,B & D