Question

A NaOH solution is standardized using KHP and the molarity of the NaOH solution is determined to be 0.4150 M. Since the acid in vinegar is the monoprotic acid, acetic acid, the concentration of the acetic acid can easily be determined by titration. If 90.72 mL of this solution is required to titrate 10.32 mL of vinegar to the Phenolphthalein endpoint, what is the concentration of acetic acid in the vinegar?

_______________ M

95.71 mL of NaOH solution is required to titrate 0.6718 g of potassium hydrogen phthalate (KHP; FW=204.224) to the Phenolphthalein endpoint. How much of this NaOH solution would be required to titrate 0.5222 g of potassium hydrogen phthalate (KHP) to the Phenolphthalein endpoint?

_______________ mL

Answer 1

no. of mole = molarity X volume of solution in liter

no. of mole of NaOH required = 0.4150 X 0.09072 = 0.0376488 mole

NaOH react with acetic acid as given below

NaOH + CH₃COOH   CH₃COO^-Na⁺ + H₂O

According to reaction NaOH react with acetic acid in equimolar proportion therefore to react with 0.0376488 mole NaOH acetic acid = 0.0376488 mole

molarity = no. of mole / volume of solution in liter

molarity of vinegar = 0.0376488 / 0.01032 = 3.648 M

Ans - 3.648 M

95.71 ml of NaOH solution required to titrate 0.6718 gm KHP then to titrate 0.5222 gm KHP required NaOH =

0.5222 X 95.71 / 0.6718 = 74.397 ml

Ans - 74.397 ml