 Diprotic acid

A diprotic acid can donate two protons per molecule through successive acid-base steps. The name counts ionizable protons in the stated chemistry, rather than all hydrogen atoms in a molecular formula. The intermediate after the first loss is itself able to act as an acid in the second step. Carbonic acid and sulfuric acid provide examples with different dissociation strengths.^[ref]

For ${\mathrm{H}}_2\mathrm{A}$ in water, write the equilibria as ${\mathrm{H}}_2\mathrm{A}\rightleftharpoons {\mathrm{H}}^{+}+\mathrm{H}{\mathrm{A}}^{-}$ and $\mathrm{H}{\mathrm{A}}^{-}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{2-}$, suppressing the explicit water/hydronium notation. There are two acid dissociation constants, $K_1$ and $K_2$. In an ideal dilute approximation let $h$ be hydrogen-ion concentration divided by the standard concentration, soSquare brackets denote concentrations. With total acid concentration $C_T=[{\mathrm{H}}_2\mathrm{A}]+[\mathrm{H}{\mathrm{A}}^{-}]+[{\mathrm{A}}^{2-}]$, the fractions in the three forms areTheir sum is one by construction; this is a species balance, not a claim that every molecule has already released both protons.

For illustrative constants $K_1=10^{-3}$ and $K_2=10^{-6}$, a maintained $h=10^{-4}$ gives fractions approximately $(0.0901,0.9009,0.00901)$. The singly deprotonated form dominates in this chosen regime. These are stipulated parameters, not measured constants for the named example acids.

If the acid is placed in initially pure water, $h$ is generally unknown: species balance must be combined with charge neutrality, water ionization and any other ions. Holding pH fixed, as in the example, is a different boundary condition. Likewise, two available protons do not imply two resolved features in every acid-base titration: the separation of the two dissociation steps and the experimental conditions determine how distinguishable they are.