What does first, second, third etc. Acid dissociation constants mean? They are written as $pK_{a_{1}}, pK_{a_2}$ etc. If any one would explain it that would be good.


The first, second, third, etc... acid dissociation constants refer to the equilibrium constant for loss of the first, second, third, and so on proton. For example, $H _2SO_4$ can lose one proton to make HSO4–, which can then lose another proton to generate SO42–. Ka1 and Ka2 would be the equilibrium constants for these reactions. pKa1 and pKa2 are, of course, -ln(Ka1) and -ln(Ka2).


Acid dissociation constants in general are the equilibrium constants that describe acid dissociation reactions. So for:

$\ce{HA + H2O <-> H3O+ + A-}$

the equilibrium constant would be

$K_a = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}$

This is enough for monoprotic acids, compounds that can donate a single proton. However, some acids, like the common inorganic acids $\ce{H2SO4}, \ce{H3PO4},$ and many, many organic acids can donate more than one proton. They are called polyprotic. The issue with their protons is that they are not all donated at the same time, or under the same conditions.

So what we do not see is:

$\ce{H2A + 2H2O <-> 2H3O+ + A^{2-}}$

with an equilibrium constant of

$K_a = \frac{[\ce{H3O+}]^2[\ce{A2-}]}{[\ce{H2A}]}$

Instead there are two (or more, if there are more protons to be donated) equilibria, each of which has its own equilibrium constant.

Dissociation 1:

$\ce{H2A + H2O <-> H3O+ + HA^{-}}$

$K_{a1} = \frac{[\ce{H3O+}][\ce{HA-}]}{[\ce{H2A}]}$


Dissociation 2:

$\ce{HA- + H2O <-> H3O+ + A^{2-}}$

$K_{a2} = \frac{[\ce{H3O+}][\ce{A2-}]}{[\ce{HA-}]}$

Each of these acid dissociation constants can be acted on by the p operator such that:

$pK_{a1} = -log_{10}(K_{a1})$


$pK_{a2} = -log_{10}(K_{a2})$

Note that there are polyprotic bases as well, substances that can accept more than one proton, and they are described using $K_{b1}, K_{b2}$, etc, where $K_b$ is the equilibrium constant for the base reacting with water to form hydroxide and a conjugate acid.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.