Question

A 3.458 g sample of KHP, a monoprotic acid, requires 45.71 mL of a KOH solution to reach the endpoint. What is the concentration of the KOH solution

Answer 1

Answer:

0.3704M is the concentration of KOH

Explanation:

The reaction of the monoprotic acid, KHP, with KOH is:

KHP + KOH → H₂O + K₂P

An acid reacts with a base to produce water and a salt

The molar mass of KHP is 204.22g/mol, that means in 3.458g of KHP you have:

3.458g * (1mol / 204.22g) = 0.01693 moles KHP.

At the endpoint:

Moles KHP = Moles KOH. That means moles KOH Are 0.01693 moles. These moles are in 45.71mL = 0.04571L, the molar concentration is:

0.01693 moles KOH / 0.04571L =

0.3704M is the concentration of KOH

Answer 2

The concentration of the KOH solution needed for the reaction is 0.370 M.

We'll begin by calculating the number of mole in 3.458 g of KHP. This can be obtained as follow:

NOTE:

KHP => Potassium hydrogen phthalate

Mass of KHP = 3.458 g

Molar mass of KHP = 204.22 g/mol

Mole of KHP =?

Mole = mass / molar mass

Mole of KHP = 3.458 / 204.22

Mole of KHP = 0.0169 mole

• Next, we shall determine the number of mole of KOH needed to react with 0.0169 mole of KHP. This can be obtained as follow:

KHP + KOH → H₂O + K₂P

From the balanced equation above,

1 mole of KHP reacted with 1 mole of KOH.

Therefore,

0.0169 mole of KHP will also react with 0.0169 mole of KOH to the end point.

• Finally, we shall determine concentration of the KOH solution. This can be obtained as follow:

Mole of KOH = 0.0169 mole

Volume = 45.71 mL = 45.71 / 1000 = 0.04571 L

Concentration of KOH =?

Concentration = mole / Volume

Concentration of KOH = 0.0169 / 0.04571

Concentration of KOH = 0.370 M

Therefore, the concentration of the KOH solution needed for the reaction is 0.370 M

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