Question

A 31.0 mL sample of 0.624M perchloric acid is titrated with a 0.258M sodium hydroxide solution.

Answer 1

Answer:

0.0922 M

Explanation:

The problem first states that the titration is made using NaOH, and later asks about the addition of KOH. I'm going to assume NaOH was used throughout the whole problem. The result does not change if it was KOH instead.

The reaction that takes place is:

• HClO₄ + NaOH → NaClO₄ + H₂O

First we calculate how many HClO₄ moles are there in the sample, using the given molarity and volume:

• 0.624 M * 13.0 mL = 8.11 mmol HClO₄

Then we calculate how many NaOH moles were added:

• 0.258 M * 15.0 mL = 3.87 mmol NaOH

Now we calculate how many HClO₄ remained after the reaction:

• 8.11 - 3.87 = 4.24 mmol HClO₄

As HClO₄ is a strong acid, 4.24 mmol HClO₄ = 4.24 mmol H⁺

Finally we calculate the molarity of H⁺, using the calculated number of moles and final volume:

• Final volume = 31.0 mL + 15.0 mL = 46.0 mL

• 4.24 mmol / 46.0 mL = 0.0922 M