Question

What will be the pH of the resulting solution when 70mL of a 0.312M KOH solution is added to 25mL of a 0.723M HClO4 solution?

Answer 1

mmol of KOH = 0.312 M x 70 mL = 21.84 mmol

mmol o fHCLO4 = 0.723 M x 25 mL = 18.075 mmol

let see the balanced equation

KOH + HClO4 -----> KClO4 + H2O

from the balanced equation

1 mmol of HClO4 react with 1 mmol of KOH accordingly

18.075 mmol of HClO4 react with 18.075 mmol of KOH

mmoles of KOH remaining = 21.84 - 18.075 = 3.765 mmol

total volume = 70+25 = 95 mL

concentration of KOH remaining = 3.765 mmol / 95 mL = 0.0396 M

pOH = -log(0.0396) = 1.4

pH = 14-pOH = 14 - 1.4 = 12.6