Answer:
[tex]M_{KOH}=0.3704M[/tex]
Explanation:
Hello there!
In this case, since the titration of bases when using monoprotic acids like KHP, occurs in a 1:1 mole ratio, it is possible to use the following equation, because at the endpoint the moles of the KHP and KOH get equal:
[tex]n_{KHP}=n_{KOH}[/tex]
In such a way, we first calculate the moles of KOH given the mass and molar mass of KHP:
[tex]n_{KHP}=n_{KOH}=3.458g*\frac{1mol}{204.22g}=0.0169mol[/tex]
Next, since we have the volume of KOH, we first take it to liters (0.04571 L) to that we obtain the following concentration:
[tex]M_{KOH}=\frac{0.0169mol}{0.04571L}\\\\M_{KOH}=0.3704M[/tex]
Regards!
