SATURATED SOLUBILITY OF BENZOIC ACID

SATURATED
SOLUBILITY
OF
BENZOIC ACID
Solubility: It is the concentration of
a solute in a saturated solution at a
certain temperature
Classification of solutions (Based on
amount of solute):
I)
Unsaturated Solutions
II)
Saturated Solutions
III)
Super--Saturated Solutions
Super
There are three types of solutions
(Based on nature of solute and solvent)
Gas/ liquid
Liquid/ liquid
Solid/ liquid
Solubility is affected by:
Temperature
Type of Solvent
pH
Particle size
Molecular Structure
Crystal Structure
Added Substance:
Substance: e.g. electrolytes.
Salting out Effect
Determination of Solubility
This comes in two steps:
1. Preparation of saturated solution
2. Analysis of saturated solution according
to the nature of the drug
Evaporation
method
Volumetric
method
Gravimetric
method
Instrumental
method
Experiment 9
• Determination of Solubility of Benzoic acid and effect of added salt
50 ml.
Saturated
solution of
benzoic acid
•Vigorous shake for 2
Take 10 ml
minutes.
filterate
•Set aside
for 10
minutes.
•Filter.
•Take 10ml of the
filtrate.
•Add 5 dps phph.
•Titrate10 ml against
0.05 N NaOH
•End Point : PERSISTENT
PINK color.
Xss benzoic acid
filtrate
0
10
20
30
40
50
25 ml
Titrate against 0.1N NaOH
25 ml
25
ml
PERSISTENT PINK color
25 ml
25 ml
Calculation of % of NaCl added:
• NaCl was added to a volume of 25 ml of the
filtrate.
e.g. Conc. Of NaCl= 1X100/25= 4%
1 gm
?
25 ml
100ml
Calculation:
1. Determination of solubility of
benzoic acid:
N1.V1 = N2.V2
?
Normality of
benzoic acid
Volume of
sample(10ml)
Normality of NaOH
In absence of NaCl:
N2 = 0.1N
mls of titrant
(E.P.)
In presence of NaCl:
N2 = 0.05N
Solubility of benzoic acid (gm/L)=
N1.E
Equivalent weight of
benzoic acid= 122
TO convert solubility (grams/l) into (gm/100ml)
solubility in gm
1000 ml
?
100 ml
=..............gm/100ml
solubility (%)
Table :
Amount of
Electrolyte
added (g)
% of electrolyte
added
Volume of
Solution
Titrated
(ml)
0
0
10
1
4
10
2
8
10
3
12
10
4
16
10
5
20
10
Volume of
NaOH
Used
(Av. Of 2
readings)
End point
Solubility
(% w/v)
Benz. ac. Dissolved (g/100ml)
Plot % dissolved benzoic acid against
% of added NaCl
The graph (best fitting line)
must intercept the Y axis.
axis.
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
5
10
% w/v NaCL
15
20
PRECAUTIONS:
Titration with the correct Titrant:
1. 0.1 N NaOH for benzoic acid in absence of
electrolyte.
2. 0.05 N NaOH for benzoic acid in presence of
electrolyte.
The graph (best fitting line) must intercept the Y
axis.
•Effect of pH on Solubility:
• Weak acids better dissolve at high pH (basic
medium), e.g. phenobarbital
• Weak bases better dissolve at low pH (acidic
medium), e.g. Atropine
• For weak electrolytes in Sat .Aq. Solution & @Eq.:
HP
undissolved
(molecular)
HP
dissolved
(molecular)
H + + Pdissolved
(ionized)
Equations of weak acids and bases:
Weak base:
Weak acid:
ܵ − ܱܵ
‫ ܽ݇݌ = ܪ݌‬+ ݈‫݃݋‬
ܱܵ
ܱܵ
‫ ݓ݇݌ = ܪ݌‬− ‫ ܾ݇݌‬+ ݈‫݃݋‬
ܵ − ܱܵ
S or St= Total solubility (Initial molar conc. Or Solubility
of ionized+ unionized forms.)= at certain pH not 7
S0 or Ks= water or molar solubility (unionized form).
Ka= dissociation constant of free acid.
Kb= dissociation constant of free base.
Kw=dissociation constant of water.
Knowing that:
pKw = 14
pH = -log H+
pKa = -log Ka
pKb = -log Kb
Problems:
[1] What must be the pH of an aqueous formulation be in
order to maintain in solution 10 mg per ml. of a weakly
acidic drug M. Wt = 200
200,, Ka = 1 x l0
l0-5 and Ks = 0.00
00ll M/L.
Solution:
Desired molar concentration (S)
= 10 mg/ml = 10 gm/liter
= 10
10//200 = 0.05 mol/liter = S
ܵ − ܱܵ
‫ ܽ݇݌ = ܪ݌‬+ ݈‫݃݋‬
ܱܵ
0.05 − 0.001
‫ = ܪ݌‬5 + ݈‫݃݋‬
= 6.69
0.001
Below what pH will free Phenobarbital begin to separate
from a solution having an initial concentration of 1 g of
sodium Phenobarbital per 100 ml at 25oC. The molar
solubility So of Phenobarbital is 0.005 and the pKa = 7.41 at
25oC. The molecular weight of sodium Phenobarbital is 254.
Solution:
The molar concentration (S) of salt initially added is:
1gm
100 ml
x gm
1000 ml
X = 10 gm/l = 10/254 = 0.039 mole/l = S
ܵ − ܱܵ
‫ ܽ݇݌ = ܪ݌‬+ ݈‫݃݋‬
ܱܵ
0.039 − 0.005
‫ = ܪ݌‬7.41 + ݈‫݃݋‬
= 8.24
0.005