A weak acid or base does not ionize completely, so its pH must be found by combining the initial concentration with its ionization equilibrium.

Consider a $0.100,\mathrm M$ solution of a weak monoprotic acid $\mathrm{HA}$ with

$$K_a=1.8\times10^{-5}.$$

The reaction is

$$\mathrm{HA+H_2O\rightleftharpoons H_3O^+ + A^-}.$$

Let $x$ be the amount that ionizes, expressed as a concentration. Starting with negligible product concentrations from the acid gives

Substitution into the acid-ionization expression gives

$$1.8\times10^{-5}=\frac{x^2}{0.100-x}.$$

For a weak acid, $x$ may be much smaller than $0.100$. Trying the approximation

$$0.100-x\approx0.100$$

gives

$$x\approx\sqrt{(1.8\times10^{-5})(0.100)} \approx1.34\times10^{-3},\mathrm M.$$

The fraction ionized is

$$\frac{x}{0.100}\times100%\approx1.34%,$$

so neglecting $x$ in the denominator is self-consistent. Therefore

$$[\mathrm{H_3O^+}]\approx1.34\times10^{-3},\mathrm M$$

and

$$\mathrm{pH}\approx2.87.$$

If the calculated ionization is not small compared with the initial concentration, the approximation must be abandoned and the equilibrium equation solved without dropping $x$.

Weak bases are handled in the same way. For

$$\mathrm{B+H_2O\rightleftharpoons HB^+ + OH^-},$$

$K_b$ determines the equilibrium $[\mathrm{OH^-}]$; pOH is found first and then converted to pH using $pK_w$.

The crucial distinction from a strong acid or base is that the analytical concentration is not automatically equal to $[\mathrm{H_3O^+}]$ or $[\mathrm{OH^-}]$. Equilibrium determines what fraction ionizes.