# 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? Commented May 2, 2015 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? Commented May 2, 2015 at 1:30
• We'll by definition that's what you would need for neutralization, so yes. Unless neutralization means something different. Commented May 2, 2015 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. Commented May 2, 2015 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 $25~\mathrm{ml}$.