Lets say dissociation constants $K_{\mathrm{a}1}$ & $K_{\mathrm{a}2}$ of $\ce{H_2SO_4}$ in water are $1000$ and $0.012$ respectively. So the first dissociation step happens completely: $$\ce{ H_2SO_4 (l) + H_2O (l) \rightleftharpoons HSO_4^- (aq) + H_3O^+(aq)}$$ Where: $\ce{[HSO_4^-] = [H_2SO_4]_0)}$, let those be $c_0$. And the second step: $$\ce{ HSO_4^-(aq) + H_2O(l) \rightleftharpoons SO_4^{2-}(aq) + H_3O^+(aq)}$$ So the IE table for the second step would look something like this:

$\ce{[HSO_4^-]}$ $\ce{[SO_4^{2-}]}$ $\ce{[H_3O^+]}$
Initial $c_0$ 0 $c_0$
Equilibrium $c_0(1-\alpha)$ $c_0\alpha$ $c_0(1+\alpha)$

So at equilibrium, $K_{\mathrm{a2}} = \frac{[\ce{SO_4^{2-}][H_3O^+][HSO_4^-]}} {(1 - \alpha)} = \frac{c_0\alpha(1+\alpha)}{(1-\alpha)}$. Which can be solved for $\alpha$: $c_0\alpha^2+(c_0+K_{\mathrm{a2}})\alpha-K_{\mathrm{a2}}=0$

Then, $\alpha$ can be used to obtain the $p$H of the solution at equilibrium:

$p$H = - $log\ce{[H_3O^+]}$)= - $log$ $ { \pu{c_o( 1+\alpha)}}$)

So, if this solution is to be neutralized with $\ce{Ca(OH)_2}$ to a certain pH, lets say $11$, how would I calculate the amount of lime needed to increase the $p$H by that much?

Thank for your time

  • 1
    $\begingroup$ Why choosing $\ce{Ca(OH)2}$ and not $\ce{NaOH}$? Neutralization sulfuric acid with $\ce{Ca(OH)2}$ produces $\ce{CaSO4}$, which is not very soluble in water ($0.0047$ M, from solubility product). And measured solubility is much higher : $0.018$ M (Journal of Chemical Education 77, 12, Dec. 2000, p. 1558). This is due to other ions like $\ce{[Ca(OH)]^+}$ and $\ce{HSO4^-}$. $\endgroup$
    – Maurice
    Commented Mar 8, 2022 at 12:47
  • 3
    $\begingroup$ Please have a look at the mhchem extension of MathJax for correct typography. $\endgroup$ Commented Mar 8, 2022 at 14:50
  • 1
    $\begingroup$ If you wanted pH = 11 then acid would be irrelevant and the basic properties of hydroxide would be needed for calculation. $\endgroup$
    – Mithoron
    Commented Mar 9, 2022 at 15:49


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.