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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?

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  • $\begingroup$ Actually, ๐‘€๐‘–×๐‘‰๐‘–=๐‘€๐‘“×๐‘‰๐‘“ 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. $\endgroup$ – lilythegreat Dec 7 '19 at 4:08
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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 $5$.

Note: Make sure you're working with molarity and not moles. I'm in analytical chem right now and often we're multiplying the number of moles in our sample by the total volume of the volumetric flask from which the sample was drawn, so we're doing calculations similar to this.

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A couple things to think about:

  • How many protons can one molecule of sulfuric acid give? Will this affect the amount of NaOH it takes to neutralize a given amount of sulfuric acid?

  • Molarity is the number of moles in a Litre of solution. If I double the volume, it doubles the number of moles. Does this change the ratio of moles to litres?

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