how many grams of potassium hydrogen phthalate will be required to titrate 50 ml of 0.101 M NaOH solution

Answer 1

Potassium hydrogen phthalate (KHP) = C₈H₅KO₄

Molar mass = 204 g / mol

Note that KHP has ONE ionizable hydrogen

NaOH + C₈H₅KO₄ ==> C₈H₄KNaO₄ + H2O

moles NaOH present = 50 ml x 1 L / 1000 ml x 0.101 mol / L = 0.00505 moles NaOH

moles KHP needed = 0.00505 mols NaOH x 1 mol KHP / mol NaOH = 0.00505 mols KHP needed

grams KHP needed = 0.00505 mols KHP x 204 g / mol = 1.03 g KHP required