Consider the dissociation equilibrium of the following weak acid HA...

For the weak acid dissociation $HA \rightleftharpoons H^+ + A^-$ with pKa = 4
We have Ka = 10⁻⁴.
For a 10 mM (0.01 M) solution where the degree of dissociation is negligible with respect to unity, the concentration of HA remains approximately 0.01 M.
The equilibrium expression gives
$Ka = \frac{[H^+]^2}{[HA]}$
Substituting: $10^{-4} = \frac{[H^+]^2}{0.01}$, which yields $[H^+]^2 = 10^{-6}$, so $[H^+] = 10^{-3}$ M.
Therefore, pH = $-\log(10^{-3}) = 3$.