Final answer:
The mass of KHP required to reach an endpoint of 25.00 mL when titrated with 0.093 M NaOH is 0.4748 g. This is calculated using the stoichiometry of the reaction and the molarity and volume of NaOH.
Step-by-step explanation:
To calculate the mass of KHP that will result in an endpoint of 25.00 mL when titrated against 0.093 M NaOH, we first need to know the stoichiometry of the reaction. KHP (potassium hydrogen phthalate) reacts with NaOH in a 1:1 molar ratio as follows: KHP + NaOH → KP + NaP + H₂O. Knowing this, we can determine the number of moles of NaOH that reacted, and therefore the number of moles of KHP that were present at the equivalence point.
The number of moles of NaOH is calculated by multiplying the volume in liters by the molarity (mol/L):
# mol NaOH = 0.02500 L × 0.093 M = 0.002325 mol NaOH.
Since the molar ratio of KHP to NaOH is 1:1, the moles of KHP is also 0.002325 mol. To find the mass of KHP, we multiply the number of moles by the molar mass of KHP (204.22 g/mol):
Mass of KHP = # mol KHP × molar mass of KHP= 0.002325 mol × 204.22 g/mol = 0.4748 g
Therefore, 0.4748 g of KHP is the mass needed for the titration.