Question

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH.

Answer 1

neutrilization reaction between HBr and KOH is

HBr + KOH KBr + H₂O

no. of mole = molarity volume of soluution in liter

no. of mole of HBr = 0.15 M 0.050 L = 0.0075 mole

no. of mole of KOH = 0.25 M 0.013 L = 0.00325 mole

According to reaction HBr and KOH react in equimolar proportion therefore 0.00325 mole of KOH react with 0.00325 mole of HBr

mole of HBr remain in solution = 0.0075 - 0.00325 = 0.00425 mole

total volume of solution = 50 + 13 = 63 ml = 0.063 liter

Molarity = no. of mole / volume of solution in liter

Molarity of HBr = 0.00425 / 0.063 = 0.0675 M

HBr is strong acid and dissociate completely therefore [ HBr] = [H₃O⁺] = 0.0675 M

pH = -log(H₃O⁺) = -log(0.0675) = 1.17

pH = 1.17