Question

A 25.0 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.

Answer 1

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