Question

A 75.00-mL portion of a solution that is 0.0460 M in HClO4 is treated with 150.00 mL of 0.0230 M KOH(aq). Is the pH of the resulting mixture greater than, less than, or equal to 7.0? Explain.

Answer 1

we know that

moles = molarity x volume (L)

so

moles of HCl04 taken = 0.046 x 75 x 10-3

moles of HCl04 taken = 3.45 x 10-3

moles of KOH taken = 0.023 x 150 x 10-3

moles of KOH taken = 3.45 x 10-3

now

consider the reaction

HCl04 + KOH ---> KCl04 + H20

we can see that

moles of HCl04 reacted = moles of KOH added = 3.45 x 10-3

so

all the HCl04 is reacted with all the KOH

so

only KCl04 remains in the solution

we know that

HCl04 is a very strong acid and KOH is a very strong base

so

their salt KCl04 will be a neutral one

we know that

pH of neutral solution = 7

so

pH of KCl04 = 7

as the resulting mixture only contains KCl04

the pH of the resulting mixture is equal to 7