A student is asked to standardize a solution of potassium hydroxide . He weighs

Question

A student is asked to standardize a solution of potassium hydroxide. He weighs out 0.950 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).

It requires 31.8 mL of potassium hydroxide to reach the endpoint.

A. What is the molarity of the potassium hydroxide solution?  M

This potassium hydroxide solution is then used to titrate an unknown solution of perchloric acid.

B. If 16.7 mL of the potassium hydroxide solution is required to neutralize 27.5 mL of perchloric acid, what is the molarity of theperchloric acid solution?  M

Explanation / Answer

Mass of Potassium hydrogen phthalate (KHP) = 0.950 g

Molar mass of KHP = 204.22 g/mol

Moles of KHP = Mass / Molar mass = 0.950 g/204.22 g/mol = 0.004652 moles

Volume of KOH = 31.8 ml = 0.0318 L (1000 ml = 1L)

At equivalent point , Moles of acid = moles of base

Molarity of KOH = Moles Of KOH /Litre =0.004652moles / 0.0318 L= 0.1463 Moles/L

Molarity of KOH = 0.1463 Moles/L

Volume of KOH = 16.7 ml

Volume of perchloric acid = 27.5 ml

At eqvt point volume of acid x molarity of acid = Volume of base x molarity of base

27.5 ml x Molarity of acid = 16.7ml x 0.1463 M

Molarity of acid = (16.7ml x 0.1463M ) / 27.5ml =0.0888 M

Molarity of perchloric acid = 0.0888 M

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