For $y$ $=$ $[H_3O^+]$ or $y$ $=$ $[OH^-]$ and $K$ $=$ $K_a$ or $K$ $=$ $K_b$, we have seen: $\alpha$ $=$ $\frac{y}{c}$ $y^2$ $+$ $K\;y$ $-$ $K\;c$ $=$ $0$ so:
$\alpha^2c$ $+$ $K\alpha$ $-$ $K$ $=$ $0$
Example 1 A $0.10\frac{mol}{L}$ solution of a weak acid with $K_a$ $=$ $4.00\cdot 10^{-2}$ obeys the equation: $0.10\alpha^2$ $+$ $4.00\cdot 10^{-2}\alpha$ $-$ $4.00\cdot 10^{-2}$ $=$ $0$, so: $\alpha$ $=$ $0.079$ We can say that this acid is dissociated to $7.9\%$
Example 2 A $0.10\frac{mol}{L}$ weak base solution with $K_b$ $=$ $4.00\cdot 10^{-2}$ obeys the equation: $0.10\alpha^2$ $+$ $4.00\cdot 10^{-2}\alpha$ $-$ $4.00\cdot 10^{-2}$ $=$ $0$, so: $\alpha$ $=$ $\frac{10^{-2.1}}{0.1}= 0.079$ We can say that this base is dissociated to $7.9\%$
From the previous equation, we deduce: $\frac{\alpha^2}{1-\alpha}$ $=$ $\frac{K}{c}$ If $c$ decreases, this fraction should increase, which is possible only if the numerator $\alpha^2$ increases and the denominator $1-a$ decreases.
If the dilution increases, $\alpha $ increases. If $c\rightarrow 0$ then $\alpha \rightarrow 1$ (complete dissociation)