Question

A student has a 0.00124 M HCl (aq) solution and she titrates 25.00 mL of this solution against an unknown potassium hydroxide solution. The acid requires 44.25 mL of base to reach the end point, what is the molarity of the KOH (aq) solution?

Answer 1

Answer:

The Molarity of KOH is

$7,01.10^{-4}M$

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.

$M_{(HCl)}V_{HCl}=n\\ \\M_{(KOH)}V_{KOH}=n$

we match these equations and find the concentration of KOH

$M_{(HCl)}. V_{(HCl)} =M_{(KOH)} .V_{(KOH)}\\ \\M_{(KOH)}=\frac{M_{(HCl)}. V_{(HCl)}}{V_{(KOH)}} \\\\M_{(KOH) =\frac{(25mL)(0,00124m)}{(44,25mL)}$

$M_{(KOH)}=7,01.10^{-4}$