# Calculating acid dissociation constant for polyprotic acids

I'm confused with $K_{a1}$ and $K_{a2}$ for a polyprotic acid and how they are found, consider the following:

A solution of $1 M$ of $\ce{H2SO4}$ would yield $1 M$ of $\ce{H3O+}$ and $\ce{HSO4-}$ since $K_{a1}$ is very large.

In the case of the second ionization, $\ce{HSO4-}$ would give $1+x$ M of $\ce{H3O+}$ and $x$ M of $\ce{SO4^2-}$ , if $x$ is known how do we calculate K$_{a2}$

Do we use first concentration + second concentration as following

• $K_{a2} = (x+1)(x)/(1-x)$

Or only second concentrations

• $K_{a2} = (x)(x)/(1-x)$

Which one is correct and why?

• It should be $1+x$ because the equilibrium accounts for all hydrogen ions in solutions. That's only if you choose to start with $\ce{H2SO4}$ for that. You can also just start with something like $\ce{NaHSO4}$. Since bisulfate is a very weak base, there's no concern with the first dissociation, and there's no extra hydrogen floating around from the first dissociation. – Zhe Nov 26 '16 at 14:44
• @Zhe so Ka2 for bisulfate would be different for sulfuric acid? – alend ahmed Nov 26 '16 at 14:50
• Bisulfate's Ka is sulfuric acid's Ka2. – Zhe Nov 26 '16 at 14:52