Question

Determine the pH of the solution after 25.00 mL of 0.2397 M HClO₄ has been titrated with 13.74 mL of 0.5381 M KOH.

Answer 1

Since HClO₄ is a strong acid and KOH is a strong base, they react completely as per the reaction

HClO₄(aq) + KOH(aq) -----> KClO₄(aq) + H₂O(l)

moles of KOH present = molarity*volume of solution in litres = 0.5381*0.01374 = 0.0074

moles of HClO₄ present = molarity*volume of solution in litres = 0.2397*0.025 = 0.006

Clearly, HClO₄ & KOH reacts in the molar ratio of 1:1

Thus, KOH is in excess and the excess amount is 0.0014

Now, [OH^-] = left [KOH] = 0.0014/(0.025+0.01374) = 0.0361

Thus, pOH = -log[OH^-] = 1.442

pH = 14 - pOH = 12.558