Why is the second acid dissociation constant for sulphuric acid much lower than the first in water?

In aqueous solution, $\ce{H2SO4}$ ionises in two steps :


$$\ce{H2SO4(aq) -> H+(aq) + HSO4-(aq)}; \quad K\_\mathrm{a\_1}=1\times10^3$$
$$\ce{HSO4-(aq) -> H+(aq) + SO4^2-(aq)}; \quad K\_\mathrm{a\_2}=1.2\times10^{-2}$$


The value of $K\_{\mathrm{a}\_2}$ caught my attention. I thought that $\ce{H2SO4}$ was a strong dibasic acid, but a very low value of $K\_{\mathrm{a}\_2}$ implies that the second proton is not that easily donated. Why is the value of $K\_{\mathrm{a}\_2}$ so less?




For the second deprotonation you are trying to remove a proton from something that is already negatively charged, a process far more difficult than removing a proton from something that is neutral.


Alternatively you can think that the $-2$ charged $\ce{SO4^2-}$ anion attracts really strongly the $\ce{H+}$ resulting in the equilibrium being towards $\ce{HSO4^-}$, making it a weak acid compared to $\ce{H2SO4}$.


It is similar thinking to why the second ionization energies are always much higher than the first.




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There can always be more than one ways to explain an observed phenomenon. With this in mind, one might be tempted to believe that there is **suppression of dissociation** of the $\ce{HSO4^-}$ ion due to the **common ion effect.** The reasoning goes along the lines of:



> 
> Note that even before $\ce{HSO4^-}$ dissociates, there is an equal concentration of $\ce{H+}$ ions *already* present in the solution. We know, by Le Chatelier's principle, that the increased concentration of the products favors the reactants. Hence for the second dissociation, since $\ce{H+}$ is one of the products, and is in large concentration, the second dissociation is actually favored in the *reverse* direction. Thus, deprotonation of $\ce{HSO4-}$ is not favorable.
> 
> 
> Hence, the $K\_\mathrm {a\_2}$ of $\ce{H2SO4}$ is significantly lower than its $K\_\mathrm {a\_1}$.
> 
> 
> 


However, this is a **wrong reasoning**. Note that $K\_\mathrm{a}$ values are always a *constant*. The $K\_\mathrm{a}$ value for the $\ce{HSO4-}$ ion would be the same whether we took $\ce{H2SO4}$ or $\ce{NaHSO4}$. However, in the case of $\ce{NaHSO4}$, there are **no pre-existing $\ce{H+}$ ions in solution when $\ce{HSO4-}$ deprotonates**. Hence, it would be *wrong* to invoke the common ion effect here.


Therefore, the common ion effect should **not** be invoked when explaining the low $K\_\mathrm a$ of $\ce{HSO4-}$ ion as compared to $\ce{H2SO4}$.


 5

The value of 2nd dissociation constant is less than 1st because in 2nd equation there is anion $\ce{HSO4-}$ which will not allow to release protons,but we know that $\ce{H2SO4}$ is strong dibasic acid.


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