A 44.0 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