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 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 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 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 $7.9\%$