# Titration of H2SO4 w NaOH: Solving for the molarity of H2SO4?

I am given $\ce{H2SO4}$ in a reaction vessel of about $50~\mathrm{mL}$. I need to solve for the molarity of $\ce{H2SO4}$. A $10~\mathrm{mL}$ sample of $\ce{H2SO4}$ is removed and then titrated with $33.26~\mathrm{mL}$ of standard $0.2643\ \mathrm{M}\ \ce{NaOH}$ solution to reach the endpoint. How do I solve for titration of the $50~\mathrm{mL}$ sample?

Obviously I can use the formula: $$M_i \times V_i = M_f \times V_f$$

Which brings me to

$$M_i \times 10~\mathrm{mL} = 0.2643~\mathrm{M} \times 33.26~\mathrm{mL}$$

Thus:

$$M_i = (0.2643~\mathrm{M} \times 33.26~\mathrm{ml}) / (10~\mathrm{mL})$$

This will find the molarity of the $10~\mathrm{mL}$ sample of $\ce{H2SO4}$.

Now, how do I find the molarity of the $50~\mathrm{mL}$ sample of $\ce{H2SO4}$ from this? Would I just do five times the $10~\mathrm{mL}$ sample's molarity?

• Actually, 𝑀𝑖×𝑉𝑖=𝑀𝑓×𝑉𝑓 is for dilution. It seems to be like you're dealing with a stoichiometry problem. This would require dimensional analysis to determine the molarity of the final product... but I could be wrong. – lilythegreat Dec 7 '19 at 4:08

The molarity would be the same whether you have $5~\mathrm{mL}$ of $\ce{H2SO4}$ or a swimming pool full of it.
However, if you wanted to solve for moles of $\ce{H2SO4}$ in $50~\mathrm{mL}$, you would have to multiply the number of moles in the $10~\mathrm{mL}$ sample by $5$.