Acid Base Titration

Question

A 0.2650g sample of KHP required 13.42mL of Ca(OH)₂ solutions reach the phenolphthalein endpoint. What is the molarity of the calcium hydroxide solution?

my Final answer I got is 9.69 · 10^-2M. Please feel free to correct me if I am wrong.

Answer 1

I think your answer is off because you probably used 1 mole Ca(OH)₂ per mole KHP, but in fact, it takes 2 moles KHP to react with 1 mole Ca(OH)₂. See my solution below.

KHP = potassium hydrogen phthalate, a mono protic acid (has 1 titratable hydrogen)

2KHP + Ca(OH)₂ ==> 2H₂O + Ca(KP)₂ ... balanced equation

Note that 1 mol Ca(OH)₂ reacts with TWO mols KHP

molar mass KHP = 204.2 g

mols KHP used = 0.2650 g x 1 mol / 204.2 g = 1.2977x10^-3 mols KHP

mols Ca(OH)₂ present = 1.2977x10^-3 mols KHP x 1 mol Ca(OH)₂ / 2 mols KHP = 6.4887x10^-4 mols Ca(OH)₂

Molarity of Ca(OH)₂ = 6.4887x10^-4 mols Ca(OH)₂ / 13.42 ml x 1000 ml / L = 0.04835 M