A96.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 48.0 ml of koh at 25 ∘c.

KOH is a strong base and HBr is a strong acid and completely dissociates.
The balanced equation for the reaction is;
KOH + HBr ---> KBr + H₂O
Stoichiometry of acid to base is 1:1
The number of KOH moles reacted - 0.50 M / 1000 mL/L x 48.0 mL = 0.024 mol
number of HBr moles reacted - 0.25 M/ 1000 mL/L x 96.0 mL  = 0.024 mol
the number of H⁺ ions are equal to number of OH⁻ ions. 
Then the solution is neutral.
pH of neutral solutions at 25 °C is 7. 
Therefore pH is 7

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.