A56.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 28.0 ml of koh at 25 ∘c.

Reaction equation :
         HBr + KOH → KBr + H₂O
1 mole of HBr reacts with 1 mole of KOH. So, mole ratio between them is 1:1
mmol of HBr   = 56.0 ml x 0.25 M = 14 mmol
mmol of KOH = 28.0 ml x 0.50 M = 14 mmol
So after the mixing neither HBr nor KOH remains and the solution contains only KBr and water which are both neutral so the pH of the solution will be 7.0

The angle of reflection is 30°