A80.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 40.0 ml of koh at 25 ∘c.

Number of moles of HBr = 80.0 mL x(1L / 1000 mL) x (0.25 mol / L) = 0.02 mol
Number of moles of KOH = 40.0 mL x (1 L / 1000 mL) x (0.5 mol / L) = 0.02 mol
At this point all the acid is completely neutralized by base so; at the equivalence point pH of solution equal 7 (salt of strong acid and strong base KBr)