A 96.0 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