A52.0 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