Download Q1) Discuss the following briefly: (a) The effect of hydrogen bond on

Q1) Discuss the following briefly:
(a) The effect of hydrogen bond on solubility of solute in water
The ability of the solute to form hydrogen bonds is a far more significant factor than is the polarity thus water also dissolves phenols, alcohols, aldehydes,
ketones, amines, and other oxygen and nitrogen containing compounds that can form hydrogen bonds with water.
(b) Semipolar compounds can act as intermediate solvents.
semipolar compounds can act as intermediate solvents to bring about miscibility of polar and nonpolar liquids. Accordingly, acetone increases the
solubility of ether in water. Propylene glycol has been shown to increase the mutual solubility of water and peppermint oil and of water and benzyl
benzoate.
(c) Solutes are often precipitated from solutions by the addition of an electrolyte.
solutes are often liberated from solutions in which they are dissolved by the introduction of an electrolyte such as sodium chloride and sometimes
by a nonelectrolyte such as sucrose. This phenomenon is known as salting out. The resultant precipitation is due to the attraction of the salt ions or
the highly polar nonelectrolyte for the water molecules, which reduces the density of the aqueous environment adjacent to the solute molecules.
Salting out can also occur in solutions of gas in liquids and liquids in liquids.
(d) The effect of critical temperature on the solubilities of partially miscible liquids.
The solubility of some liquid pairs, can increase as the temperature is lowered, and the system will exhibit a lower critical temperature, below
which the two liquids are soluble in all proportions and above which two separate layers form. Another type, involving a few mixtures such as
nicotine and water, shows both an upper and a lower critical temperature with an intermediate temperature region in which the two liquids are only
partially miscible. A final type exhibits no critical solution temperature; the pair ethyl ether and water, for example, has neither an upper nor a lower
critical temperature and shows partial miscibility over the entire temperature range at which the mixture exists.
(e) The effect of pH on the solubility of weak electrolyte drugs.
Weak electrolytes can behave like strong electrolytes or like nonelectrolytes in solution. When the solution is of such a pH that the drug is entirely
in the ionic form, it behaves as a solution of a strong electrolyte, and solubility does not constitute a problem. However, when the pH is adjusted to
a value at which un-ionized molecules are produced in sufficient concentration to exceed the solubility of this form, precipitation occurs.
(f) The effect of dilution of strong electrolytes on specific and equivalent conductance.
As the solution of a strong electrolyte is diluted, the specific conductance  decreases because the number of ions per unit volume of solution is
reduced. Conversely, the equivalent conductance  of a solution of a strong electrolyte steadily increases on dilution. The increase in  with
dilution is explained as follows: The quantity of electrolyte remains constant at 1 gram equivalent according to the definition of equivalent
conductance; however, the ions are hindered less by their neighbors in the more dilute solution and hence can move faster.
Q2) Representing the free acid form of phenobarbital as HP and the soluble ionized form as P–, the
equilibria in a saturated solution of this slightly soluble weak electrolyte is written as:
HPsolid  HPsol
HPsol  H 2 O  H 3O   P 
Derive an equation to compute the pH below which the salt of sodium Phenobarbital begins to
precipitate from aqueous solution.
Because the concentration of the un-ionized form in solution, HPsol is essentially constant, the equilibrium constant for the equation is:
S   [ HP ]sol
Where
Ka 
S is molar or intrinsic solubility. The constant for the acid-base equilibrium is:
[ H 3O  ][ P  ]
or
[ HP ]
[P ]  Ka
[ HP ]
...........1
[ H 3O]
The total solubility, S of phenobarbital consists of the concentration of the undissociated acid, [HP], and that of the conjugate base or ionized form,
[P–]:
S  [ HP]  [ P  ].............2
Substituting
S for [HP] from equation (1) and the expression from equation (2) for [P–] yields:
S  S  K a
S
[ H 3O  ]
(S  S )  K a
S
[ H 3O  ]
log( S  S  )  log K a  log S   log[ H 3O  ]
pH p  pK a  log
S  S
S
Where pHp is the pH below which the drug separates from solution as the undissociated acid.
Q3) Below what pH will free phenobarbital begin to separate from a solution having an initial
concentration of 5% (w/v)? The molar solubility, S of phenobarbital is 0.005 and the pKa is 7.41 at
25°C. The secondary dissociation of Phenobarbital can be disregarded. The molecular weight of
sodium phenobarbital is 254.
The molar concentration of salt initially added is:
5 gm 

   50 gm / Liter
0.1L 1L
Molarity 
Weight
50

 0.1968mole / Liter
Molecular .Weight 254
pH p  pK a  log
pH p  7.41  log
S  S
S
0.1968  0.005
 7.41  log 38.36  7.41  158  5.826
0.005
Q4) The equivalent conductance of a 5.9 × 10–3 M solution of acetic acid is 14.4 ohm cm2/Eq, and
the degree of dissociation of acetic acid at this concentration is 0.037. Calculate the equivalent
conductance of acetic acid at infinite dilution.

c

14.4 ohm cm 2 / Eq
 0  c 
 390.7ohmcm2 / Eq
0

0.037