Question

A 92.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 46.0 ml of koh at 25 ∘c.

Answer 1

In neutralization reactions, a base and an acid react to form an ionic salt and water. There is a rule that when a strong base and a strong acid react, the pH of their salt is always neutral which is at pH 7. However, this is only true if equal amounts of acid and base are consumed and that there is no excess. Otherwise, the excess acidity or basicity will adjust the total pH.

Strong acids are the following: HCl, HBr, HI, HClO4, HClO3, HNO3 and H2SO4. Strong bases are KOH, LiOH, NaOH, Ca(OH)2, Sr(OH)2 and Ba(OH)2. Therefore, we can already establish that both of the reactants are strong. The complete reaction is

HBr + KOH ⇒ KBr + H₂O

So, 1 mole of HBr would require 1 mol of KOH, and vice versa. Let'scompute the amount of the initial reactants:

mol HBr: (0.25 mol/L)*(0.92 L) = 0.23 mol HBr
mol KOH: (0.5 mol/L)*(0.46 L) = 0.23 mol KOH

There are equal amounts of acid and base. Thus, pH of the KBr solution is neutral at pH 7.