Pharmaceutical Analytical Chemistry (PHCM223-SS16)
Lecture 1
ACID- BASE EQUILIBRIUM-II
“pH of salts”
Dr. Rasha Hanafi
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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LEARNING OUTCOMES
By the end of this session the student should be able to:
1. Determine ka- kb relationship, and vice versa.
2. Calculate the pH of a salt solution.
3. Compare acid base properties of salts.
4. Comprehend the effect of common ion on dissociation of
polyprotic acids.
5. Calculate the equilibrium concentrations of acids and bases
containing a common ion.
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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SALTS: ACIDIC/ BASIC/ NEUTRAL?
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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SALTS: ACIDIC/ BASIC/ NEUTRAL?
SALTS
Ionic compounds
Dissociate in water
Acidic
Basic
Neutral
pH<7
pH>7
pH7
Salts of weak
bases (ex: NH3)
Salts of weak acids
(ex: CH3COOH)
Ex: NH4 Cl
Ex: CH3COONa
Salts made of
strong acids and
strong bases
Ammonium Chloride
Sodium acetate
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
Ex: KCl, NaNO3
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NEUTRAL SALTS
Strong acid + water  weak conjugate base + H+
HCl

HNO3 
Cl-
NO3-
No affinity for
H+
No effect on
pH
Strong bases + water  weak conjugate acid + OHNaOH 
Na+
KOH 
K+
No ability to
produce H+
No effect on
pH
Salts made of cations of strong bases and the anions of strong
acids have no effect on [H+] when dissolved in water.
Aqueous solutions of salts such as KCl, NaCl, NaNO3 and KNO3
are neutral ( pH= 7)
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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BASIC SALTS
SALTS OF WEAK ACIDS IN WATER GIVE BASIC SOLN.
NaC2H3O2 + H2O  Na+ + H2O + C2H3O2salt dissociates
HC2H3O2(aq) + OH-(aq)
strong base
affinity for H+ of water
The pH of this solution will be determined by the C2H3O2-ion as it
is the base.
C2H3O2-(aq) + H2O(l)
HC2H3O2(aq) + OH-(aq)
xM
Same conc.
of salt (M)

[ HC 2 H 3O2 ][OH ]
Kb 
[C2 H 3O2 ]
xM
Kb of a conjugate base (acetate ion) is not
found in textbooks, yet it can be obtained
from ka of the acid (acetic acid) which is
common in any data base.
Kw=ka x kb
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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pKa , pKb RELATIONSHIP
HA +H2O
[ H  ][ A ]
ka 
[ HA]
A- + H2O
[ HA][OH  ]
kb 
[ A ]
H+ + A[ HA]
[A ] 
ka

[H ]

HA + OH-

[ HA]
[ HA][OH ]
ka 

[H ]
kb

[
HA
][
OH
]

[A ] 
kb
k a kb  [ H  ][OH  ]  k w  10 14
pk a  pkb  pk w  14
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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Test yourself
Calculate the pH of a 0.30 M NaF solution. The Ka
value for HF is 7.2x10-4.
NaF+ H2O  Na++ F-+ H2O
HF + OH-
The Fluoride ion is a base and its dissociation has a Kb that can be obtained
from Ka of its corresponding hydrofluoric acid.
Kb = Kw / Ka = 1.0x10-14/7.2x10-4 = 1.4x10-11
Kb = 1.4x10-11 = [HF][OH-]/[F-] = (x)(x)/0.3
x = 2.0x10-6 = [OH-] pOH = 5.69
pH = 14.00 -5.69 = 8.31
The solution is basic, as stated:
“the salt of a weak acid yields a basic soln.”
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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ACIDIC SALTS
SALTS OF WEAK BASES IN WATER GIVE ACIDIC SOLN.
NH4Cl + H2O  Cl- + NH4+ + H2O
salt
dissociates
acid
Same conc.
of salt (M)
NH3 + H3O+ (or H+)
xM
xM
The pH of this solution will be determined by the NH4+ as it is the conjugate
acid of a weak base (NH3).
[ H  ][ NH 3 ]
ka 
[ NH 4 ]
Ka of a conjugate acid (ammonium ion) is not
found in textbooks, yet it can be obtained
from kb of the base (ammonia) which is
common in any data base.
Kw=ka x kb
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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Test yourself
Calculate the pH of a 0.01 M NH4Cl solution. The Kb of
NH3 is 1.8x10-5.
Ka x Kb = Kw
Ka (NH4+) = Kw / Kb(NH3) = 1.0x10-14/1.8x10-5 = 5.6x10-10
5.6x10-10 = Ka = [H+][NH3]/[NH4+] = (x)(x)/(0.01)
x= [H+] = 7.5x10-6M
pH = 5.31
The solution is acidic, as stated:
“the salt of a weak base yields an acidic soln.”
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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EFFECT OF COMMON ION ON DISSOCIATION OF
ACIDS/BASES
If HF and NaF are present together in soln.,
would the pH be different if only HF was present?
NaF(aq) + H2O
HF(aq) + H2O

Na+(aq) +
H+(aq) +
F-(aq)
F-(aq)
The major species in soln. are HF, Na+, F- and H2O.
F- is a common ion produced by both hydrofluoric acid and
sodium fluoride. Following Le Chatelier’s principle, the
dissociation equilibrium for HF would be driven to the left, the
extent of dissociation of HF will be lower in presence of NaF
 less acidic solution than in case of HF alone.
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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EFFECT OF COMMON ION ON DISSOCIATION OF
POLYPROTIC ACIDS
The production of protons by the first dissociation step greatly
inhibits the succeeding dissociation steps due to common ion
effect of H+.
 the hydrogen ion source will be only the first dissociation
step.
H2CO3 is stronger acid than HCO3-
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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Test yourself
If the equilibrium concentration of H+ in a 1.0M HF solution is 2.7x10-2M and
the percent dissociation of HF is 2.7%. Calculate [H+] and the percent
dissociation of HF in a solution containing 1.0M HF (Ka = 7.2x10-4) and 1.0M
NaF.
NaF
1M

Na+
+
1M
F1M
HF
H+
1M
xM
+
FxM
HF soln.
HF soln. + NaF
[HF] = 1.0 M
[HF] = 1.0 M
[NaF] = 1.0 M
[H+]= 2.7x10-2 M
[H+]= ??
2.7% of HF dissociates ??% of HF dissociates
-4
Ka of HF = 7.2x10
Ka = [H+][F-] / [HF] = 7.2x10-4
[F-] = 1.0M in addition to (x) resulting from dissociation of HF
K a  7.2x10
4
[H ][F ] (x)(1.0  x) (x)(1.0)



[HF]
(1.0  x)
(1.0)
x= [H+] = 7.2X10-4 M
PHCM223,SS16
pH =- log 7.2X10-4 = 3.14
Lecture 2, Dr. Rasha Hanafi.
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Test yourself
Cont.,
The percent dissociation of HF in this solution is:
[H  ]
7.2x10 4 M
x 100 
x 100  0.072%
[HF]0
1.0M
In presence of NaF, HF dissociated in 0.072%, while in absence of
HF it dissociated in 2.7%.
A shift in the equilibrium to the left by the presence of the a
common ion from the dissolved salt.
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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SUMMARY
Strong acid
• An acid shows lower acidity in presence of one of its salts due to common
ion effect.
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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REFERENCES
1.
2.
3.
S.S. Zumdahl, S.A. Zumdahl, Chemistry 6th Ed., Houghton Mifflin
Company, ISBN 0-618-22156-5, (Chapters 14, 15)
D. A. Skoog, D.A. West, F.J. Holler, S.R. Crouch, Analytical Chemistry, an
introduction, 7th Edition, ISBN 0-03-020293-0. (Chapters 4, 8, 10, 11, 12,
13, 14, 15, 18)
Lecture 2, PHCM223, by Prof. Rasha Elnashar, GUC, SS 2014.
PHCM223,SS16
Lecture 2, Dr. Rasha Hanafi.
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