Question

`40*mL` of sulfuric acid with a concentration of `0.05*mol*L^-1` was titrated with `50*mL` volume of `NaOH(aq)`. What was the concentration of `NaOH(aq)`?

Answer 1

We interrogate the chemical equation:

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

Explanation:

And we write this equation in order to establish the molar equivalence...

We use a volume of `40*mL` of `0.05*mol*L^-1` concentration with respect to sulfuric acid....

And this is a molar quantity of `40*mLxx10^-3*L*mL^-1xx0.05*mol*L^-1=2.0xx10^-3*mol`.

And so the stoichiometry requires that FOR EQUIVALENCE there were TWICE this molar quantity with respect to `NaOH`..

`i.` `[NaOH(aq)]=(2xx2.0xx10^-3*mol)/(0.050*L^-1)=0.080*mol*L^-1`...you can check on the description of `"normality"` versus `"molarity"`...here, for `NaOH`, BUT NOT for sulfuric acid, they are the same....

`ii.` Now in aqueous solution....`pH+pOH=14`...here...`pOH=-log_10(0.080)=1.10`..

...but `pH=14-pOH=14-1.10=12.90`...