$\ce{H2SO4}$ is a strong acid, which follows the reaction $$\ce{H2SO4 -> HSO4- + H+}$$ Not $$\ce{H2SO4 -> SO4^2- + 2 H+}$$


$$\ce{Ba(OH)2 -> Ba^2+ + 2OH-}$$

Is there a reason why these "polyprotic analogs" in bases completely dissociate but polyprotic acids do not?

Note: This is in context to titrations, where $\pu{0.10M}$ of $.\ce{Ba(OH)2}$ is $\pu{0.20M}$ of $\ce{OH-}$, and leads to different values of titrant needed.


In simplistic terms, Arrhenius acids are molecules while Arrhenius bases are ionic compounds. Much like $\ce{BaCl2}$ – another ionic compound – it is a good first rule of thumb to assume it completely dissociates into its respective ions: one $\ce{Ba^2+}$ and two single-charge anions. You could say that barium were never really connected to either hydroxide; ionic compounds are often shown as collections of balls simply in a certain arrangement (crystal structure) but easily separable.

Arrhenius acids, which are all molecules, are different. There is no inherent separation point; dissiociation in water actually requires the transfer of a proton, i.e. the breaking of an $\ce{X-H}$ covalent bond. Two such bonds don’t get broken simultaneously, instead in a first reaction one proton is removed – but then you end up with an anion. Removing another $\ce{H+}$ has to go against electrostatic attraction and that is a harder job to do so it happens less and the $\mathrm pK_\mathrm a$ values differ.

In actual fact, the picture I have drawn for Arrhenius bases is simplistic. Ionic bonds also involve orbital interactions; no bond is purely ionic or purely covalent. Cations and anions obviously also electrostatically attract each other. As barium is twice positively charged, it does tend to bind the hydroxide fragments rather tightly. Overall, upon dissolution it is much more likely to end up with what effectively is a coordination compound of the $\ce{[Ba(OH)(H2O)_n]+ + OH-}$ type.

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