A25.0 ml sample of 0.723 m hclo4 is titrated with a 0.273 m koh solution. the h3o concentration after the addition of 50.0 ml of koh is m.

This is an acid base reaction and the chemical equation for the above reaction is as follows;
KOH  + HClO₄ ---> KClO₄ + H₂O
the stoichiometry of acid to base is 1:1
KOH is a strong base and HClO₄ is a strong acid therefore they both ionize completely into their respective ions
Number of KOH moles - 0.723 M/1000 mL/L x 25.0 mL = 0.018 mol
Number of HClO₄ moles - 0.273 M/1000 mL/L x 50 mL = 0.013 mol 
since acid and base react completely, 0.013 mol of acid reacts with 0.013 mol of base.
The excess base remaining is - 0.018 - 0.013 = 0.005 mol 
total volume of solution = 25.0 mL + 50.0 mL = 75.0 mL 
[OH⁻] = 0.005 mol/0.075 L = 0.067 M 
pOH = -log[OH⁻]
pOH = -log(0.067 M)
pOH = 1.17
pOH + pH = 14
Therefore pH = 14 - 1.17 = 12.83
by knowing pH we can calculate the [H₃O⁺]
pH = -log [H₃O⁺]
[H₃O⁺] = antilog[-12.83]
[H₃O⁺]= 1.47 x 10⁻¹³ M

The answer is c fish