Question

What volume of `0.5*mol*L^-1` `"sulfuric acid"` is required for equivalence with a `20*mL` volume of `0.5*mol*L^-1` `NaOH(aq)`?

Answer 1

We need a `2.0*mL` volume.......

Explanation:

We need (i) a stoichiometric equation.....

`2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) + 2H_2O(l)`

And (ii) we need equivalent quantities of each reagent....

`"Moles of NaOH"=20*mLxx10^-3*L*mL^-1xx0.1*mol*L^-1=2xx10^-3*mol.`

And we thus need a `1xx10^-3*mol` quantity with respect to the DIACID `H_2SO_4`....

`V=n/C=(1xx10^-3*mol)/(0.5*mol*L^-1)xx10^3*mL*L^-1=2.0*mL`

Clearly, we should titrate the base with a LESS concentrated acid.....