What is the pH?

Question

A 975.0 mL sample of 0.18 M HClO₄ is titrated with 0.54 M KOH. Determine the pH of the solution after the addition of 325.0 mL of KOH.

Answer 1

This is a titration of a strong acid, perchloric acid (HCLO4) with a strong base potassium hydroxide (KOH).

There are seven strong acids: H2SO4, HNO3, HCLO3, HCLO4, HCL, HBr, HI.

The bases made from Group 1A like KOH are all essentially strong.

When we say strong with regards to Bronsted-Lowry acids and bases it means they completely dissociate.

Equation HCLO4 + KOH >>>>H2O + KCLO4 which is already balanced and it goes to completion

You have 0.975 L of a 0.18 M solution of HCLO4 titrated with 0.325 L of a 0.54 M solution of KOH.

You would use a B (before), C (change) and A (after table) converting the HCLO4 and KOH to moles since the volumes are not equal for each compound, But lets see first if you need to use this table and its calculations.

You start with determining the Equivalence Point where the number of moles of acid equal the number of moles of base. You can use the dilution formula to easily determine this:

M1V1(acid) = M2V2 (base) where M is the Molarity and V is the Volume.

We are going to solve for the volume of base needed to reach the equivalence point

(0.18)(0.975) = (0.54)(Vb)

solving for Vb = 0.325 L which is exactly the volume of KOH you are titrating the HCLO4 with

To confirm this determine, the # moles of HCLO4 and KOH

moles HCLO4 = 0.975 L x 0.18 M = 0.1755 moles

moles KOH = 0.325 L X 0.54 M = 0.1755 moles

you can see they are equal, and you have reached the equivalence point with this titration.

Since your at the equivalence, point the HCLO4 is completely neutralized by the KOH leaving you with H20 and KCLO4, a salt of a strong base and a strong acid (therefore there is no hydrolysis reaction)

There is no need as well to actually use a BCA table as no further calculations are needed.

Main point, at the equivalence point with the titration of a strong acid with a strong base the

pH will ALWAYS = 7.