A solution of the primary standard potassium hydrogen phthalate, KHC8H4O4, was p

Question

A solution of the primary standard potassium hydrogen phthalate, KHC8H4O4, was prepared by dissolving 0.4877 g of potassium hydrogen phthalate in about 50 mL of water. Titration with a KOH solution required 18.81 mL to reach a phenolphthalein end point. What is the molarity of the KOH solution?

Explanation / Answer

KHC8H4O4 + KOH ==>K2C8H4O4 + H2O mols phthalate - grams/molar mass = ? mols KOH = same (look at the coefficients in the balanced equation.) M KOH = mols KOH/L KOH molarity of potassium hydrogen phthalate = (0.4877/204.22)(1000/50) = 0.0478M MOlarity of KOH = (0.0478*50)(26.50) = 0.09 M

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