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

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !