### Ionic Equilibrium Past Papers Collection

Ionic Equilibrium Past Papers Collection
MCQ and Structured Question 1983 - 2014

## Ionic Equilibrium

H2O is the medium of biological systems one must consider the role of this molecule in the dissociation of ions from biological molecules. Water is essentially a neutral molecule but will ionize to a small degree. This can be described by a simple equilibrium equation:

H2O ↔ H+ + OH

This equilibrium can be calculated as for any reaction:
Since the concentration of H2O is very high (55.5M) relative to that of the [H+] and [OH], consideration of it is generally removed from the equation by multiplying both sides by 55.5 yielding a new term, Kw:
Kw = [H+][OH]
This term is referred to as the ion product. In pure water, to which no acids or bases have been added:
Kw = 1 x 10–14 M2
As Kw is constant, if one considers the case of pure water to which no acids or bases have been added:
[H+] = [OH-] = 1 x 10–7 M
This term can be reduced to reflect the hydrogen ion concentration of any solution. This is termed the pH, where:
pH = –log[H+]

Acids and bases can be classified as proton donors and proton acceptors, respectively. This means that the conjugate base of a given acid will carry a net charge that is more negative than the corresponding acid. In biologically relevant compounds various weak acids and bases are encountered, e.g. the acidic and basic amino acids, nucleotides, phospholipids etc.

Weak acids and bases in solution do not fully dissociate and, therefore, there is an equilibrium between the acid and its conjugate base. This equilibrium can be calculated and is termed the equilibrium constant = Ka.

This is also sometimes referred to as the dissociation constant as it pertains to the dissociation of protons from acids and bases. In the reaction of a weak acid:
HA ↔ A + H+

the equilibrium constant can be calculated from the following equation:
As in the case of the ion product:
pKa = –logKa

Therefore, in obtaining the –log of both sides of the equation describing the dissociation of a weak acid we arrive at the following equation:
Since as indicated above –logKa = pKa and taking into account the laws of logarithms: