WEAK ACIDS AND BASES
SALT OF WEAK ACID AND BASES
BUFFER
LECTURE 9
9 FEB 2011
Noorulnajwa Diyana Yaacob
noorulnajwa@unimap.edu.my
Weak Acid


A weak acid is an acid that does not completely
donate all of its hydrogens when dissolved in water.
These acids have higher pH compared to strong
acids, which release all of their hydrogens when
dissolved in water.
Weak Acid



The acidity constant for acetic acid at 25oC is 1.75 x 10-5
When acetic acid ionizes, it dissociates to equal portion of H+
and OAc- by such an amount will always be equal 1.75 x 10-5
If the original concentration of acetic acid is C and the
concentration of ionized acetic acid species (H+ and OAc-) is x,
then the final concentrationof each species at equilibrium is
given by:
Example:
Calculate the pH and pOH of a
1.0 x 10-3 solution of acetic acid
Solution :
HOAc  H+ + OAcKa= [H+][OAc-] = 1.75 x 10-5
[HOAc]
Equation
HOAc
H+
OAc-
Initial (M)
Change (M)
Equilibrium (M)
1.0 x 10-3
-x
1.0 x 10-3 - x
0
+x
x
0
+x
x
x2
= 1.75 x 10-5
1.0 x 10-3 - x
Ka =[H+][OAc-] = 1.75 x 10-5
[HOAc]
x is smaller than C, neglect it, therefore,
x2
= 1.75 x 10-5 = 1.32 x 10-4M = [H+ ]
1.0 x 10-3
pH = -log 1.32 x 10-4 = 3.88
pOH = 14.00 – 3.88 = 10.12
Weak Bases


A weak base is a chemical base that only partially
ionize in water.
Refer Example 7.8 for more understand
Common Weak Bases
Common Weak Acids
Acid
Formula
Base Formula
Formic HCOOH
ammonia NH3
Acetic CH3COOH
trimethyl
N(CH3)3
ammonia
Trichloroacetic CCl3COOH
Hydrofluoric HF
Hydrocyanic HCN
Hydrogen
HS
sulfide 2
Water H2O
Conjugate acids
NH4+
of weak bases
pyridine C5H5N
ammonium
NH4OH
hydroxide
water H2O
HS- ion HSconjugate bases e.g.:
of weak acids HCOO-
Salts of Weak Acids and Bases



The salt of a weak acid for example NaOAc is
strong electrolyte, like all salt and completely
ionizes.
In addition, the anion of the salt of a weak acid is a
Brønsted base which will accept protons.
It partially hydrolyzed in water to form hydroxide
ion and the corresponding undissociated acid.
Salts of Weak Acids and Bases



The ionization constant for sodium acetate is equal
to basicity constant of the salt.
If the salt hydrolyzes that salt is consider as a weak
base.
The weaker the conjugate acid, the stronger the
conjugate base, that is, the more strongly the salt
will combine with a proton, as from the water , to
shift the ionization to the right.
Hydrolysis
constant


The value of Kb can be calculated from Ka of
acetic acid and Kw, if we multiply both the
numerator and denominator by [H+]:
The quantity inside the dashed line in Kw and the
remainder is 1/Ka, hence:


The product of Ka of any weak acid and Kb of its
conjugate base is always equal to Kw:
For any salt of weak acid HA that hydrolyzes in
water,


The pH of such a salt (Bronstead base) is calculated
as the same manner as for any other weak base
When the salts hydrolyzes, it forms an equal amount
of HA and OH- ,If the [] of A- is CA-, then,
The quantity x can be neglected , which will
generally be the case for such weakly ionizes bases

Example
Calculate the pH of a 0.10 M solution of sodium
acetate
Solution:

Write the equilibria:

Write the equilibrium constant
Since COAc > Kb , neglect x compared to COAc.
Then,


Similar equations can be derived for the cations of
salts of weak base.
Refer Example 7.10 for more understanding
WHAT IS BUFFERS???


Buffer is defined as a solution that resists change in
pH when a small amount of an acid or base is
added or when the solution is diluted.
To maintain the pH of the reaction at an optimum
value.


Buffer solution consists mixture of weak acid and its
conjugate base or weak base with it conjugate acid
at predetermined concentration or ratios.
That is mixture of a weak acid and its salt or a
weak base and its salt.
Buffer Solutions
21

A buffer solution can be:

a solution containing a weak acid and its conjugate base, or
CH3COOH (aq)
CH3COO‒ (aq) + H+ (aq)
weak acid
conjugate base
It is known as an acidic buffer solution and it maintains a pH
value that is less than 7.

a solution containing a weak base and its conjugate acid.
NH3 (aq) + H2O(l)
NH4+ (aq) + OH‒ (aq)
weak base
conjugate acid
It is known as a basic buffer solution and it maintains a pH
value that is greater than 7.
• Consider an acetic acid-acetate buffer.
• The acid equilibrium that governs the system is :


HOAc  H  OAc
• Now,
we have added a supply of acetate ions to the system
• The hydrogen ion [ ] is NO LONGER EQUAL to the acetate ion [ ]
• The hydrogen ion concentration is equal to :
HOAc 
H   K OAc 

a

• Taking the (-ve) logarithm of each of this equation :
  K
 log H

pH  pK a
a

HOAc 
 log
OAc 


HOAc 
 log
OAc 

• Inverting the last term, it becomes (+ve)

OAc 
pH  pK a  log
HOAc 


This form of the ionization constant equation is
called the Handerson-Hasselbalch equation
Useful for calculating the pH of weak acid solution
containing its salt

HA  H  A

Example :
Calculate the pH of a buffer prepared by adding
10 mL of 0.10 M acetic acid to 20 mL of 0.10 M
Sodium acetate.

Need to calculate the [ ] of the acid and salt in the
solution. The final volume is 30 mL:
So,

M1  mL1  M 2  mL2
For HOAc,
0.10 mmol/mL X 10 mL = MHOAc X 30 mL
MHOAc = 0.033 mmol/mL
For OAcˉ ,
0.10 mmol/mL X 20 mL = MOAcˉ x 30 mL
MOAcˉ = 0.067 mmol/mL
pH   log K a  log
 protonacceptor
 protondonor 
pH   log( 1.75 10 5 )  log
= 4.76 + log 2.0
= 5.06
0.067 mmol / mL
0.033mmol / mL

Refer to example 7.12 for more understanding
BUFFERING MECHANISM
For a mixture of weak acid and its salt, it can be
explain as follows.
 The pH is governed by the logarithm of the ratio of
the salt and acid
pH = constant + log [A⁻]
[HA]
*if solution is diluted the ratio remains constant
So, the pH of the solution does not change.

•
•
•
•
If small amount of strong acid added it will combined
with an equal amount of the A⁻ to convert it to HA.
HA
H⁺ + A⁻
Le Chatelier’s principle dictates added H⁺ will
combined with A⁻ to form HA.
The change in ratio [A⁻]/[HA] is small and hence the
change in pH is small.
If acid added in unbuffered solution (NaCl solution) the
pH will decreased markedly.



If small amount of strong base is added it will
combined with part of HA to form an equivalent
amount of A⁻. Again, change in ratio is small.
Buffering capacity : amount of acid or base that can
be added without causing a large change in pH.
This is determine by the concentrations of HA and
A⁻.
•
•
•
•
↑ concentrations ,↑ acid/base can tolerate
Buffer capacity of a solution is defined as
β = dCBOH / dpH
= - dCHA / dpH
dCBOH and dCHA represents the number of moles per liter of
strong base or acid.
For weak acid or conjugate base buffer solution of greater
than 0.001 M the buffer capacity is approximate by
  2.303
C HAC A
C HA  C A


The CHA and CAˉ represent the analytical [ ] of the
acid and its salt respectively.
If we have a mixture of 0.10 mol/L acetic acid and
0.10 mol/L sodium acetate, the buffer capacity is :
0.10  0.10
  2.303
 0.050mol / LperpH
0.10  0.10

If we add solid sodium hydroxide until it becomes
0.0050 mol/L, the change in pH is :
dpH  dCBOH /   0.0050 / 0.050  0.10  pH


In addition to [ ], the buffering capacity is governed
by the ratio of HA to Aˉ.
It is maximum when the ratio is unity that is the pH =
pKa
1
pH  pK a  log  pK a
1
Example
A buffer solution is 0.20 M in acetic acid and
in sodium acetate. Calculate the change in pH
upon adding 1.0 mL of 0.10M hydrochloric acid
to 10 mL of this solution.

Solution :
mmol HOAc = 2.0 + 0.1 = 2.1 mmol
mmol OAcˉ = 2.0 – 0.1 = 1.9 mmol
1.9mmol / 11mL
pH  4.76  log
2.1mmol / 11mL
1.9mmol
 4.76  log
 4.76  log 0.90
2.1mmol
 4.76  (0.95  1)  4.71
The change in pH is -0.05.

A buffer can resist a pH change even when there is
added an amount of strong acid or base greater
than the equilibrium amount of H⁺ or OHˉ in the
buffer.


Same goes for the weak base and its salt.
Consider the equilibrium between the base B and its
conjugate (BrØnsted) acid :
BH   B  H 
Ka


B H  


BH 

Kw
Kb
The logarithmic Henderson-Hasselbalch form is
derived exactly as above:
H   K

 log H

BH   K . BH 
.


a

w
B 
   log K
Kb
B 
BH    log K
 log

a
B 
BH 
 log

w
Kb
B 
pH  pK a  log
B
BH 

 ( pK w  pK b )  log
B
BH 



protonacceptor 
protonacceptor 
pH  pK a  log
 ( pK w  pK b )  log
 protondonor 
 protondonor 
Since pOH =pKw -pH, we can also write , the above equation form pKw
pOH  pKb

BH 
 log
 pK

B
b
 log
 protondonor 
 protonacceptor
Example :
Calculate the volume of concentrated ammonia
and the weight of ammonium chloride you
would have to take to prepare 100 mL of a
buffer at pH 10.00 if the final concentration of
salt is said to be 0.200 M
We want 100 mL 0f 0.200 M NH4Cl.
Therefore :
mmol NH4Cl = 0.200 mmol/mL × 100 mL
= 20.0 mmol
mg NH4Cl = 20.0 mmol × 53.5 mg/mmol
= 1.07 × 10³ mg
So, 1.07 g NH4Cl. Calculate [ ] of NH3 by
pH  pK a  log
 protonacceptor
 protondonor 
 (14.00  pK b )  log
NH 3 
NH 
10  (14.00  4.76)  log

4
NH 3 
0.200mmol / mL
log
NH 3 
0.200mmol / mL
NH 3 
0.200mmol / mL
 0.76
 10 0.76  5.8
NH 3   (0.200)(5.8)  1.16mmol / mL
The molarity of concentrated ammonia is 14.8 M
100mL 1.16mmol / mL  14.8mmol / mL  mL( NH3 )
mL( NH 3 )  7.8

Refer example 7.15 for ,ore understanding