# How much calcium hydroxide is needed to raise the pH of an aqueous sulphuric acid solution?

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?

• 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 $\ce{HSO4^-}$. Commented Mar 8, 2022 at 12:47