In writing net ionic equations involving strong acids, one needs to know how strong acids dissociate. For example, the dissociation of $\ce{H2SO4}$:

\begin{align}\ce{ H2SO4 &-> 2H+ + SO4^{2-}\\ H_SO4 &-> H+ + HSO4^{-}\\ }\end{align}

The reason I ask is because apparently the molecular equation

$$\ce{H2SO4 + 2KOH -> 2H2O + K2SO4}$$

has the net-ionic form $$\ce{H+ + OH- -> H2O}$$

It would be helpful if you could describe in general how I should write the dissociation of the strong acids in the net ionic equation.


This is a little tricky. I would argue that only the first proton dissociates completely and should be considered "strong" - using the rules for writing net ionic equations, where we only dissociate strong electrolytes, this would meant that we would write:

$\ce{H2SO4 + 2KOH -> 2H2O + K2SO4}$

$\ce{H+ + HSO4- + 2K+ + 2OH- -> 2H2O + 2K+ + SO_4^{2-}}$ - here only the first proton dissociates completely

Canceling out spectator ions gives:

$\ce{H+ + HSO4- + 2OH- ->2H2O + SO_4^{2-}}$

Your teacher (or whoever is giving you that answer) is dissociating the second proton as well, to give:

$\ce{2H+ + SO_4^{2-} + 2OH- ->2H2O + SO_4^{2-}}$

Which becomes:

$\ce{H+ + OH- ->H2O}$

There are a few reasons why they might do this, but I would argue that the 1st approach (which only dissociates the first proton) is best because it more accurately reflects what is happening, in the sense that $\ce{HSO4-}$ is a weak electrolyte. This will become important later when you study acid-base equilibria.

In general, you dissociate protons from acids as if they were regular metal cations - just "pull them off" and don't forget the positive charge. For polyprotic acids, the first proton is strongly acidic, or dissociates completely, but the other protons are not. There might be exceptions to this rule but I can't think of any off the top of my head.

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  • $\begingroup$ Is the presence of excess $\ce{OH-}$ ions maybe forcing the $\ce{HSO4-}$ ions to dissociate as well? Kinda like how a weak acid behaves like a strong one in a basic medium. $\endgroup$ – Papul Jan 13 '15 at 6:54
  • $\begingroup$ @Papul - yes, that is probably the reason, but it would only be true for an excess of KOH, so leaving bisulfate out is not, in my opinion, the best answer. $\endgroup$ – thomij Jan 13 '15 at 13:58

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