Question

What is the concentration of sulfuric acid if a `50*cm^3` volume reacts with `50*cm^3` of `2*mol*dm^-3` `NaOH`?

Answer 1

`"Concentration"~=1.0*mol*dm^-3`

Explanation:

`"Concentration"-="Moles of stuff"/"Volume of solution"`

And here we gots.........

`50*cm^3xx10^-3*dm^3*cm^-3xx2*mol*dm^-3=0.10*mol` with respect to `NaOH`.

And given the stoichiometry expressed by the following equation.........

`2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) + 2H_2O`..
....there were `0.050*mol` with respect to sulfuric acid in the original solution.

And thus `C_(H_2SO_4)=(0.050*mol)/(50xx10^-3*dm^3)=1.0*mol*L^-1`