# Solubility: Calculating solubility of weak electrolytes as influenced by pH, influence of solvents on the solubility of drugs and MCQs for GPAT, NIPER, Pharmacist and Drug Inspector exam

## Solubility: Calculating solubility of weak electrolytes as influenced by pH, influence of solvents on the solubility of drugs and MCQs for GPAT, NIPER, Pharmacist and Drug Inspector exam

Calculating the Solubility of Weak Electrolytes as Influenced by pH:

From what has been said about the effects of acids and bases on solutions of weak electrolytes, it becomes evident that the solubility of weak electrolytes is strongly influenced by the pH of the solution. For example, a 1% solution of phenobarbital sodium is soluble at pH values high in the alkaline range. The soluble ionic form is converted into molecular phenobarbital as the pH is lowered, and below 9.3, the drug begins to precipitate from solution at room temperature. On the other hand, alkaloidal salts such as atropine sulfate begin to precipitate as the pH is elevated. To ensure a clear homogeneous solution and maximum therapeutic effectiveness, the preparations should be adjusted to an optimum pH. The pH below which the salt of a weak acid, sodium phenobarbital, for example, begins to precipitate from aqueous solution is readily calculated in the following manner. Representing the free acid form of phenobarbital as HP and the soluble ionized form as P-, we write the equilibria in a saturated solution of this slightly soluble weak electrolyte as

HPsolid ↔ HPsol                                                                                                                                               (1)

HPsol + H2O ↔ H3O+ + P–                                                                                                                                                    (2)

Because the concentration of the un-ionized form in solution, HPsol, is essentially constant, the equilibrium constant for the solution equilibrium, equation (1), is

S0 = HPsol                                                                                                                                                          (3)

where So is molar or intrinsic solubility. The constant for the acid–base equilibrium, equation (2), is

Ka = [H3O+] [P]/HP                                                                                                                                       (4)

Or

[P] = Ka [HP]/ [H3O+]                                                                                                                                   (5)

where the subscript ―sol has been deleted from [HP]sol because no confusion should result from this omission.

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]                                                                                                                                                 (6)

Substituting So for [HP] from equation (3) and the expression from equation (5) for [P-] yields

S = S0 + Ka S0/[H3O+]                                                                                                                                   (7)

S = S0(1+ Ka/[H3O+]                                                                                                                                      (8)

When the electrolyte is weak and does not dissociate appreciably, the solubility of the acid in water or acidic solutions is So = [HP], which, for phenobarbital is approximately 0.005 mole/liter, in other words, 0.12%.

The solubility equation can be written in logarithmic form, beginning with equation (7). By rearrangement, we obtain

(S – S0) = Ka S0/[H3O+]

log(S – S0) = log Ka + log S0– log[H3O+]

and finally

pHp = pKa + log S – S0/ S0                                                                                                                    (9)

where pHp is the pH below which the drug separates from solution as the undissociated acid.

An analogous derivation can be carried out to obtain the equation for the solubility of a weak base as a function of the pH of a solution. The expression is

pHp = pKw – pKb + log  S0/   S – S0                                                                                                   (10)

where S is the concentration of the drug initially added as the salt and So is the molar solubility of the free base in water. Here pHp is the pH above which the drug begins to precipitate from solution as the free base.

The Influence of Solvents on the Solubility of 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 serious 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. In this discussion, we are now interested in the solubility of nonelectrolytes and the undissociated molecules of weak electrolytes. The solubility of undissociated phenobarbital in various solvents is discussed here because it has been studied to some extent by pharmaceutical investigators. Frequently, a solute is more soluble in a mixture of solvents than in one solvent alone. This phenomenon is known as cosolvency, and the solvents that, in combination, increase the solubility of the solute are called cosolvents. Approximately 1 g of phenobarbital is soluble in 1000 mL of water, in 10 mL of alcohol, in 40 mL of chloroform, and in 15 mL of ether at 25°C.

Multiple choice questions:

1.From what has been said about the effects of acids and bases on solutions of weak electrolytes, it becomes evident that the solubility of weak electrolytes is strongly influenced by the pH of the solution.

a)true

b)false

2.In S0 = HPsol   S0     is

a)molar solubility

b)intrinsic solubility

c)both of these

d)none of these

3.When the electrolyte is weak and does not dissociate appreciably, the solubility of the acid in water or acidic solutions is So = [HP], which, for phenobarbital is approximately

a)0.002 mole/liter

b)0.005 mole/liter

c)0.05 mole/liter

d)0.02 mole/liter

4.When the electrolyte is weak and does not dissociate appreciably, the solubility of the acid in water or acidic solutions is So = [HP], which, for phenobarbital is approximately(in percentage)

a)0.10%

b)0.11%

c)0.12%

d)0.15%

5.pHp = pKa + log S – S0/ S0   In this equation pHp   is

a)pH below which the drug separates from solution as the undissociated acid

b)pH above which the drug separates from solution as the undissociated acid

c)pH below which the drug separates from solution as the dissociated acid

d)all of the above

6.pHp = pKw – pKb + log  S0/   S – S0    In this equation S is

a)concentration of the drug initially added as the salt

b)molar solubility of the free base in water

c)intrinsic solubility

d)all of the above

7.pHp = pKw – pKb + log  S0/   S – S0    In this equation So is

a)concentration of the drug initially added as the salt

b)molar solubility of the free base in water

c)intrinsic solubility

d)all of the above

8.pHp = pKw – pKb + log  S0/   S – S0    In this equation pHp is

a)pH below which the drug separates from solution as the undissociated acid

b)pH above which the drug begins to precipitate from solution as the free base

c)pH below which the drug begins to precipitate from solution as the free base

d)all of these

9.How weak electrolytes can behave in solution?

a) electrolytes

b)nonelectrolytes

c)only a

d)both of these

10.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 serious problem.

a)true

b)false

11.When the pH is adjusted to a value at which ionized molecules are produced in sufficient concentration to exceed the solubility of this form, precipitation occurs.

a)true

b)false

12.Frequently, a solute is more soluble in a ____ than in one solvent alone.

a)mixture of solvents

b)basic solvent

c)acidic solvent

d)water alone

13.Cosolvency is

a)a solute is less soluble in a mixture of solvents than in one solvent alone

b)a solute is more soluble in a mixture of solvents than in one solvent alone

c)a solute is more soluble in one solvent alone than in mixture of solvents

d)all of these

14.The solvents that, in combination, increase the solubility of the solute are called

a)nonelectrolytes

b)electrolytes

c)cosolvents

d)dissociated ions

15.Approximately 1 g of phenobarbital is soluble in (at temperature 25°C)

a)1000 mL of water

b)10 mL of alcohol

c)40 mL of chloroform

d)all of the above

Solutions:

1. a)true
2. c)both of these
3. b)0.005 mole/liter
4. c)0.12%
5. a)pH below which the drug separates from solution as the undissociated acid
6. a)concentration of the drug initially added as the salt
7. b)molar solubility of the free base in water
8. b)pH above which the drug begins to precipitate from solution as the free base
9. d)both of these
10. a)true
11. b)false
12. a)mixture of solvents
13. b)a solute is more soluble in a mixture of solvents than in one solvent alone
14. c)cosolvents
15. d)all of the above

References:

1. Martins Physical Pharmacy, 6th edition 2011, page no. 341-343.

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