# The degree of ionisation $\alpha$

## Definition

The degree of dissociation of an acid or a base is the number of moles divided by the initial number of moles:

Degree of dissociation:
$\alpha$ $=$ $\frac{n_{dissoc.}}{n_{init.}}$

"Dissociated" means ionized or dissociated in the case of an acid, hydrolyzed in the case of a base.
"Initial" means the total number introduced into the medium before there was ionization.

Dividing by the volume of the solution, we have:
$\alpha$ $=$ $\frac{\frac{n_{dissoc.}}{V}}{\frac{n_{init.}}{V}}$
In the case of an acid $HB$:
$\frac{n_{dissoc.}}{V}$ $=$ $[H_3O^+]$
and then:

Acid:
$\alpha$ $=$ $\frac{[H_3O^+]}{c_{HB}}$

In the case of a base $B$:
$\frac{n_{dissoc.}}{V}$ $=$ $[OH^-]$
and then:

Base:
$\alpha$ $=$ $\frac{[OH^-]}{c_{B}}$

## Strong acids and bases

Strong acids and strong bases are completely dissociated, the number of moles dissociated equals the initial number of moles, so:

Strong acid or base:
$\alpha$ $=$ $1$

## Weak acids

Weak acids are not fully dissociated.
$\alpha $ is computed from $c_{HB}$ and $H_3O^+$ (pH calculation of a weak acid!)

Example:
A solution of a weak acid $0.10\frac{mol}{L}$ with $K_a$ $=$ $4.00\cdot 10^{-2}$ has a
$pH= 2.1$ (do the math):
$\alpha$ $=$ $\frac{[H_3O^+]}{c_{HB}}$ =
$\frac{10^{-2.1}}{0.1}$ $=$
$ 0.079$
We can say that this acid is dissociated up to $7.9\%$

## Weak bases

Weak bases are not fully hydrolyzed.
$\alpha$ is computed from $c_ {B}$ and $OH^-$ (pOH calculation of a weak base!)

Example:
A solution of a weak base $0,10\frac{mol}{L}$ with $K_b$ $=$ $4,00\cdot 10^{-2}$ has a
$pOH$ $=$ $ 2.1$ (do the math):
$\alpha$ $=$ $\frac{[OH^-]}{c_{B}}$ =
$\frac{10^{-2.1}}{0.1}$ $= $
$0.079$
We can say that this acid is dissociated up to $7.9\%$