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.
We first observe the balanced equation. $$\ce {H2SO4 + 2NaOH &lt;=&gt; Na2SO4 + 2H2O}$$ 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 $25~\mathrm{ml}$.