Answer:
The Molarity of KOH is
[tex]7,01.10^{-4}M[/tex]
Explanation:
The endpoint indicates the volume necessary to neutralize the moles of acid.
In other words, the point at which the moles of both solutions are the same.
[tex]M_{(HCl)}V_{HCl}=n\\ \\M_{(KOH)}V_{KOH}=n[/tex]
we match these equations and find the concentration of KOH
[tex]M_{(HCl)}. V_{(HCl)} =M_{(KOH)} .V_{(KOH)}\\ \\M_{(KOH)}=\frac{M_{(HCl)}. V_{(HCl)}}{V_{(KOH)}} \\\\M_{(KOH) =\frac{(25mL)(0,00124m)}{(44,25mL)}\\\\[/tex]
[tex]M_{(KOH)}=7,01.10^{-4}[/tex]
