Question

A `43.6*mL` volume of `0.125*mol*L^-1` `KOH` was titrated with a `25.0*mL` volume of hydrochloric acid. What was `[HCl]`?

Answer 1

`[HCl]=0.218*mol*L^-1`

Explanation:

We assess the reaction....

`HCl(aq) + KOH(aq) rarr KCl(aq) + H_2O(l)`

....which is stoichiometrically balanced with respect to mass and charge.....and there is thus 1:1 equivalence between the acid and base....and we can assess the molar quantity of the base we used....

`"Moles of KOH"=43.6*mLxx10^-3*L*mL^-1xx0.125*mol=5.45xx10^-3*mol`.

There was thus an equimolar quantity of acid, and this was dissolved in a volume of `25.0*mL`....and we gets a concentration of....

`(5.45xx10^-3*mol)/(25.0*cancel(mL)xx10^-3*L*cancel(mL^-1))=0.2180*mol*L^-1`