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

Question

A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.06 g potassium hydrogen phthalate KHCBH O, treat this as a monoprotic acid). It requires 33.3 m of potassium hydroxide to reach the endpoint. A. What is the molarity of the potassium hydroxide solution? This potassium hydroxide solution is then used to titrate an unknown solution of hydroiodie acid. B. If 193 mlL of the potassium hydroxide solution is required to neutralize 19.3 mL of hydroiodic acid, what is the molarity of the hydroiodic acid solution? Retry Entire Group 4 more group attempts remaining

Explanation / Answer

A) The reaction between KHP and KOH is

KHP + KOH - - - - - > K2P + H2O

Stoichiometrically, 1mole of KOH reacting with one mole of KHP

given mass of KHP = 1.06g

molar mass of KHP = 204.22g/mol

No of moles of KHP = 1.06g/204.22g/mol = 0.00519mol

0.00519 moles of KHP react with 0.00519moles of KOH

Volume of KOH solution consumed= 33.3ml

Molarity of KOH solution =( 0.00519mol/33.3ml)×1000ml = 0.1559M

B) The reaction between potassium hydroxide and Hydrogen iodide is

KOH + HI - - - - > KI + H2O

stoichiometrically, 1mole of KOH react with 1mole of HI

Therefore,

V1 × M1 = V2 × M2

M1 = V2 × M2/V1

M1 = 19.3ml × 0.1559M/19.3ml

M1 = 0.1559M

therefore,

Molarity of hydriodic acid solution = 0.1559M

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