Question

A 3.458 g sample of KHP, a monoprotic acid, requires 45.71 mL of a KOH solution to reach the endpoint. What is the concentration of the KOH solution? The molar mass of KHP is 204.22 g/mol.

Answer 1

`M_(KOH)=0.3704M`

Step-by-step 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:

`n_(KHP)=n_(KOH)`

In such a way, we first calculate the moles of KOH given the mass and molar mass of KHP:

`n_(KHP)=n_(KOH)=3.458g*(1mol)/(204.22g)=0.0169mol`

Next, since we have the volume of KOH, we first take it to liters (0.04571 L) to that we obtain the following concentration:

`M_(KOH)=(0.0169mol)/(0.04571L)\\\\M_(KOH)=0.3704M`

Regards!