A 52.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 26.0 mL of KOH at 25 ∘C.

The balanced equation for the above reaction is

HBr + KOH - - - > KBr + H₂O

stoichiometry of HBr to KOH is 1:1

HBr is a strong acid and KOH is a strong base and they both completely dissociate.

The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol

The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL = 0.013 mol

the number of H⁺ ions = number of OH⁻ ions

therefore complete neutralisation occurs.

Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.

Then pH of solution is 7