# Calculation of volume required for titration with dibasic acid

Here is a problem:

Titrate $1 M$ sulfuric acid with $\pu{50 ml}$ of $1 M$ sodium hydroxide solution. What volume of sulfuric acid will be required for neutralization.

This is a simple problem. My reasoning is since both reactants are strong we will have $\pu{0.05 L}$ $\ce{OH-}$ ions and for neutralization we need an equal number of $\ce{H+}$ ions. Since $\ce{H2SO4}$ is a strong acid and ignoring the second dissociation which is weak, we get an equal number of $\ce{H+}$ as $\ce{H2SO4}$ ions. Since concentrations are equal, the volume is simply $\pu{50 mL}$.

Apparently this is wrong. I think the source of confusion is what is meant by "required for neutralization". If you write out the molecular equation for this reaction and do stoichiometry you get a different volume.

• "Neutralized" usually means "will not react upon further addition of strong acid/base". What would this mean in your case? – Nicolau Saker Neto May 2 '15 at 0:59
• You should get different volume. The only way you should get equal volume is if the ratio between protons and hydroxide are equal. Are they equal in this case? – Diehardwalnut May 2 '15 at 1:30
• We'll by definition that's what you would need for neutralization, so yes. Unless neutralization means something different. – Joshua Benabou May 2 '15 at 1:34
• Actually, I don't think we need to do stoichiometry at the ionic level, that is to say we don't need to compare ions during stoichiometry, only the compounds. – Diehardwalnut May 2 '15 at 2:28

We first observe the balanced equation. $$\ce {H2SO4 + 2NaOH <=> Na2SO4 + 2H2O}$$ we see the ratio of $\ce {NaOH}$ and $\ce {H2SO4}$ is 2:1. Given $0.050~\mathrm{L}$ of $\ce {1M~NaOH}$ we find moles of $\ce {NaOH}$ and use stoichiometry to find how many moles of $\ce {H2SO4}$ is needed. We find the volume to be $25~\mathrm{ml}$.