Acid–Base Titration

Report
Clinical Pharmacy
Code: 105
Acid/Base Titration
Dr. Hisham Ezzat Abdellatef
Prof. of Analytical Chemistry
Electrolyte and the theory of
electrolytic Dissociation
• Electrolytes
• conduct the electric current
• mineral acid, caustic alkalies and salts
• non–electrolytes
• non–conducting solutions
• cane sugar, glycerin and ethyl acetate.
Acid–Base Titration
•
Pure water is a bad conductor of
electricity
• acid as HCl
Dissolved
• base as KOH
in water
• salt as Na2SO4
•
Molecule
Dissociation
H+ ClK+ OHNa+ SO42ions
Strong and weak electrolytes
•
•
•
•
NaCl ⇌ Na+ + Cl–
K2SO4 ⇌
2K+ + SO42–
Na2HPO4 ⇌ 2Na+ + H+ + PO43
Arrhenius therefore introduced a quantity "a",
called "the degree of dissociation“
• "a“ = 1 "strong electrolyte“
• “a” very far from unity.= weak electrolyte
Law of mass action
• "The rate of a chemical
reaction is proportional to the
active masses of the reacting
substances."
diluted
solution
concentration
Law of mass action
• Vf = [A].[B]. Kf
• Vb = [C]. [D]. Kb
• Kf [A]. [B] = Kb [C]. [D]
The dissociation of water
• H2O ⇌ H+ + OH–
• [H+] . [OH]  = Kw………..(2)
• "The ionic product of water"
The dissociation of water
• 25oC; the value of Kw
[H+] . [OH] = 1014
[H+]2 = Kw = 1 x 1014 ………(3)
Hydrogen ion exponent (pH):
10–6, 10–5
[H+]
1 x 107
Acid
10–8 , 10–9
base
Neutral
Procedure for Titration
• pH is defined as equal to the
logarithm of the hydrogen ion
concentration with a negative sign.
pH = –log
+
[H ]
The pH scale
1
2
3 4
acid
5
6 7 8 9 10 11 12 13 14
neutral
base
Stomach juice: pH = 1.0 – 3.0
Human blood: pH = 7.3 – 7.5
Lemon juice: pH = 2.2 – 2.4
Seawater: pH = 7.8 – 8.3
Vinegar: pH = 2.4 – 3.4
Ammonia: pH = 10.5 – 11.5
Carbonated drinks: pH = 2.0 – 4.0
Orange juice: pH = 3.0 – 4.0
0.1M Na2CO3: pH = 11.7
1.0M NaOH: pH = 14.0
Figure : The pH
scale and pH values
of some common
substances
Figure: A pH meter
Figure: Indicator paper
being used to measure
the pH of a solution
p
pKw = pH + pOH
 pH = pKw – pOH
or pH = 14 – pOH
or pOH = 14 – pH
Acids and bases:
• Arrhenius theory
• Acid
•
• Bases,
[H+]
when dissolved in water
[OH–]
Acid and base

H 2SO 4  H  HSO
HSO

 H  SO
4
(H )(HSO-4 )

 K1  1
(H2SO 4 )
4
(H )(SO 24- )
2


K
2

3
x
10
(HSO -4 )
24
Also


4

2
4
H 3 P O4  H  H 2 P O

4
H 2 P O  H  HP O
2
4
HP O

3
4
 H  PO
(H )(H2 P O4 )

 K1  1.1x10-2
(H3 P O4 )
(H )(HP O24 )
-7


K

2.0x10
2
(H2 P O4 )
(H )(P O34 )
-13


K

3.6x10
3
(HP O24 )
pH calculations
1. Solution of strong acids
and strong bases
[H+] or [OH–] =concentration
Example 1
• Calculate the pH value of a solution of a
completely ionised 1.0 N solution of acid; or
base. ?
• Solution:
[H+] = 1M
pH = –log 1 = 0 (zero)
similarly, in a completely ionised 1.0 N solution of base
[OH–] = 1 M
pOH = –log 1 = 0 (zero)
Example 2
Calculate the [H+] and pH of 0.009 N
hydrochloric acid?
solution
[H+] = 0.009 N
pH =– log (9.0 X 10–3) = – log 9.0 – log 10–3
pH = – 0.95 + 3.00 = 2.05
Example 3
• Calculate the pH values of a solution of
sodium hydroxide whose [OH–] is 1.05 x
10–3?
solution
pOH = – (log 1.05 + log 10–3)
pOH = – (0.02 –3 ) = 2.98
pH = 14 – 2.98 = 11.02
Example 4
• Calculate the hydrogen ion concentration
of a solution of pH 5.3?
solution
pH = – log [H+]
5.3 = –log [H+]
[H+] = 5.01 x 10–6 M
Example 5
• Calculate the hydroxyl ion concentration
of a solution of pH 10.75 ?
solution
pOH = 14 – 10.75 = 3.25
[OH–] = the antilog of –3.25
[OH–] = 5.62 x 10–4 M
pH calculations
• 2. Solution of weak acids
and bases
• A) Calculation of pH of
solution of weak acids
pH = ½ (pKa + pCa)
• B) Calculation of pH of
solution of weak bases:
pH = pKw – ½ (pKb + pCb)
pH of salts
• A) Salts of strong acids or
bases
KCl  K+ + ClK+ + OH- ⇌ KOH
Cl- + H+ ⇌ HCl
The equilibrium between the hydrogen and hydroxyl ions
in water
H2O ⇌ H+ + OH–
• B) Salts of weak acids or
bases (hydrolysis)
• Salts of weak acids (or bases) react with
water to give basic (or acidic) solutions
• Hydrolysis reverse of neutralization
• i) Salts of weak acids and
strong bases:
H2O + CH3COO ⇌ CH3COOH + OH
• pH = ½ (pKw – pCa + pKa)
y
h
CS
• ii) Salts of weak bases and
strong acids:
NH4+
+ H2O ⇌ NH4OH + H+
pH = ½ (pKw + pCs – pKb)
y
h
CS
Hydrolysis of NH4NO3
NH4
++H
2O
 NH3+H3
+
O
• The SCA ion of the salt
NH4NO3 reacts with water to
produce hydronium ions which
causes the soln to be basic
• iii) Salts of weak acids and
weak bases:
pH = ½ pKw + ½ pKa – ½pKb
Buffer solutions
• Resist changes in pH caused
by addition of small amounts
of acid or base; or upon
dilution.
Types of buffers

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