A25.0 ml sample of 0.723 m hclo4 is titrated with a 0.27 m koh solution. the h3o+ concentration after the addition of 80.0 ml of koh is m. a 25.0 ml sample of 0.723 m hclo4 is titrated with a 0.27 m koh solution. the h3o+ concentration after the addition of 80.0 ml of koh is m. 3 × 10-13 0.7 0.4 4 × 10-2 1 × 10-7

 the balanced equation for the acid base reaction is 
KOH + HClO₄ ---> KClO₄ + H₂O
stoichiometry of KOH to HClO₄ is 1:1
The number of HClO₄ moles - 0.723 M / 1000 mL/L x 25.0 mL = 0.0181 mol
Number of KOH moles - 0.27 M/ 1000 mL/L x 80.0 mL = 0.0216 mol
KOH is a strong acid and HClO₄ is a strong base therefore complete dissociation takes place.
acid reacts with base in 1:1 molar ratio, there's excess base remaining.
excess OH⁻ ions - 0.0216 - 0.0181 = 0.0035 mol
concentration is calculated as number of moles/volume
volume of solution - 25.0 + 80.0 = 105.0 mL 
[OH⁻] = 0.0035 mol / 0.105 L = 0.033 M
pOH = -log [OH⁻]
pOH = -log(0.033 M)
pOH = 1.48
pH can be calculated by knowing pOH
pH + pOH = 14
pH = 14 - 1.48 = 12.52
pH = -log [H₃O⁺]
[H₃O⁺] = antilog(-12.52)
[H₃O⁺]  = 3.0 x 10⁻¹³ M
Answer is 3 x 10⁻¹³ M