A solution of the primary standard potassium hydrogen phthalate, KHC8H4O4, was prepared by dissolving 0.4877 g...

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A solution of the primary standard potassium hydrogen phthalate, KHC8H4O4, was prepared by dissolving 0.4877 g...

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 36.21 mL to reach a phenolphthalein end point. What is the molarity of the KOH solution?

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Answer 1

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Answer #1

\(\mathrm{KHC} 8 \mathrm{H} 404+\mathrm{KOH}=\mathrm{K}_{2} \mathrm{C} 8 \mathrm{H} 4 \mathrm{O} 4+\mathrm{H} 2 \mathrm{O}\)

Moles of \(\mathrm{KOH}=\) moles of \(\mathrm{KHC} 8 \mathrm{H} 404\) = mass \(/\) molar mass of \(\mathrm{KHC} 8 \mathrm{H} 404\)

\(=0.4877 / 204.22=0.0023881 \mathrm{~mol}\)

Molarity of \(\mathrm{KOH}=\) moles \(/ \mathrm{volume}\) of \(\mathrm{KOH}\)

\(=0.0023881 / 0.03621\)

\(=0.06595 \mathrm{M}\)

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Answer 2

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