Question

How many milliliters of .45 M `HCl` will neutralize 25.0 mL of 1.00 M `KOH`?

Answer 1

Approx. `60*mL` of hydrochloric acid are required.

Explanation:

With all problems of this type, a BALANCED chemical equation is required to establish the equivalence:

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

Given the 1:1 equivalence:

`"Moles of KOH"=25xx10^-3*Lxx1.00*mol*L^-1=25xx10^-3*mol`

Thus we require an equivalent quantity of `0.45*mol*L^-1` hydrochloric acid:

`=` `(25xx10^-3*mol)/(0.45*mol*L^-1)xx10^3*mL*L^-1` `=` `??mL`