For the weak acid dissociation HA⇌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=[HA][H+]2
Substituting: 10−4=0.01[H+]2, which yields [H+]2=10−6, so [H+]=10−3 M.
Therefore, pH = −log(10−3)=3.