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 0.965 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).

It requires 25.9 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 hydrochloric acid.

B. If 21.5 mL of the potassium hydroxide solution is required to neutralize 27.7 mL of hydrochloric acid, what is the molarity of the hydrochloric acidsolution? ________M

Please explain :)

Explanation / Answer

first find out the no of moles potassium hydrogen phthalate

moles of potassium hydrogen phthalate = weight / molar mass

= 0.965 / 204.2212

= 0.004725 moles

if you look at the balanced reaction

KHC8H4O4 + KOH --------> K2C8H4O4 + H2O

one mole of KHC8H4O4 required one mole of KOH

so no o fmoles of KOH = no of moles KHC8H4O4 =0.004725 moles

PartA

now we know the moles of KOH = 0.004725 andvolume of KOH = 25.9mL

now Molarity of KOH = no of moles of KOH / volume of KOH in Liters

= 0.004725 / 0.0259 L

= 0.1824M

PartB

abive solution used means molarity of above KOH solution is 0.1824

volume of KOH used is 21.5 mL

find out the moles using these two

moles of KOH = Molarity x Volume in liters

= 0.1824 x 0.0215 L

= 0.003922 moles

KOH + HCl ----> KCl + H2O

from this balanced equation it is clear that on emole of KOH required on emole of HCl

so no of moles of HCl = 0.003922 moles

now we know the volume of HCl and moles of HCl = 27.7 mL = .0277 L

Molarity = 0.003922 / 0.0277

Molarity = 0.1416M

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