


We start with the fundamental acid–base equilibrium:
mathrm{HA + H_2O ightleftharpoons H_3O^+ + A^-}
Key idea:
Acid strength = how much HA wants to become A⁻
Ka values are:
Instead of using Ka directly, we usually use pKa:
oxed{mathrm{pKa = -log(Ka)}}
This inverse relationship is critical and comes up constantly in biochemistry exams.
For the conjugate base (A⁻):
For a conjugate acid–base pair in water:
oxed{mathrm{pKa + pKb = 14}}
This means:
You cannot have both strong at the same time.
The balance between pH and pKa determines whether the molecule is mostly:
This equation connects everything:
oxed{mathrm{pH = pKa + logleft(rac{A^-}{HA} ight)}}
🧠 Think: “Low pH = hold onto protons”
mathrm{HA = A^-}
This point is special:
Let’s apply this to a physiological setting.
We rearrange Henderson–Hasselbalch:
mathrm{pH - pKa = logleft(rac{A^-}{HA} ight)}
1.5 - 2.5 = -1
rac{A^-}{HA} = 10^{-1} = 0.1
➡️ Some dissociation ➡️ Moderate amount of A⁻
1.5 - 4.5 = -3
rac{A^-}{HA} = 10^{-3} = 0.001
➡️ Almost no dissociation ➡️ Mostly protonated (HA)
Even a 2-unit difference in pKa leads to a 100-fold difference in protonation state.
That’s enormous in biochemistry and pharmacology.