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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

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    $\begingroup$ 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
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    $\begingroup$ Please have a look at the mhchem extension of MathJax for correct typography. $\endgroup$ Commented Mar 8, 2022 at 14:50
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    $\begingroup$ 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

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