Question

When 4.15g of potassium hydrogen phthalate (KHPh M=204.2 g/mol) is titrated with KOH solution and it takes 46.42 mL of the base to reach the endpoint, calculate the concentration (M) of the KOH solution.

I'm confused about the wording of the problem and how to approach it.

Does titrated mean just a normal chemical equation problem like KHP + KOH -> ???, or am I suppose to do something special?

Also, what does it mean to reach the endpoint? Or is it just a complicated way of saying the final products.

Answer 1

We can solve given problem in following steps.

Step 1 : Calculation of moles of KHP

We have, no. of moles = Mass /Molar mass

No. of moles of KHP = 4.15 g / ( 204.2 g /mol ) = 0.02032 mol

Step 2 : Calculation of moles of KOH .

Consider reaction, HC₈H₄KO₄ ( KHP ) + KOH KC₈H₄KO₄ + H₂O

According to reaction , 1 mol KHP reacts with 1 mol KOH.

Therefore, 0.02032 mol KHP will react with 0.02032 mol KOH.

Hence, moles of KOH used in the titration = 0.02032 mol.

Step 3 : Calculation of molar concentration of KOH

We know that, [ KOH ] = No. of moles of KOH / volume of solution in L

[ KOH ] = 0.02032 mol / 0.04642 L

[ KOH ] = 0.4377 M

ANSWER : Molar concentration of KOH : 0.438 M