1
$\begingroup$

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?

$\endgroup$
2
  • 3
    $\begingroup$ This is the typical pattern. Roughly speaking, there are no strong polyprotic acids. $\endgroup$ Feb 23 '18 at 9:59
  • 1
    $\begingroup$ @IvanNeretin Actually some complex acid would qualify... But roughly speaking... :D $\endgroup$
    – Mithoron
    Feb 23 '18 at 15:06
4
$\begingroup$

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.


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 $\ce{H2SO4}$.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.