A44.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 22.0 ml of koh at 25 ∘c.

The reaction between KOH and HBr is as follows ;
KOH + HBr ---> H₂O + KBr
Stoichiometry of base to acid is 1:1 molar ratio
Both are strong acid and strong base therefore complete ionization takes place
The number of KOH moles added - 0.50 M / 1000 mL/L x 22 mL = 0.011 mol
the number of HBr moles - 0.25 M /1000 mL/L x 44 mL = 0.011 mol
the number of H⁺ ions and OH⁻ ions are equal therefore the whole amount of acid has been completely neutralised by base.
No remaining acid nor base, therefore solution is neutral.
pH = 7 
thats the pH value for a neutral solution

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