Question

A 370 mL sample of 0.180 M HClO4 is titrated with 0.542 M LiOH. Determine the pH of the solution after the addition of 125 mL of LiOH.

Answer 1

The neutralization reaction can be written as

HClO₄ (aq) + LiOH (aq) LiClO₄ (aq) + H₂O (l)

It is an example for strong acid and strong base titration. They dissociate completely into ions.

The no of moles of H⁺ion (HClO₄) = Molarity x volume (in litres) = 0.180 M x 0.370 lit = 0.0666 moles

The no of moles of OH^- ion in solution = 0.542 M x 0.125 litres = 0.06775 moles

Total volume in litres = 0.370 lit + 0.125 lit = 0.495 litres

Note: no of moles OH^- is more hence the solution will be basic the remaing OH^- ion concentration

[OH^-] = moles/volume (litres)

= (0.06775 - 0.0666)/0.495 (Note: The remaing moles after neutralization with H⁺)

= 0.00115/0.495 M

= 0.002323 M

[OH^-] = 2.3 x 10^-3 M

P^H = 14 - P^OH

= 14 - (- log(0.002323))

= 14 + log(0.002323)

= 14 - 2.6339

= 11.37

Hence P^H = 11.37