moles of ethylamine = 0.1 M x 50 ml = 5 mmol
At equivalence point moles of base = moles of acid added
So moles of HClO4 = 5 mmol
Volume of HClO4 added = 5 mmol/0.2 M = 25 ml
C2H5NH2 + HClO4 ---> C2H5NH3ClO4
concentration of salt formed [C2H5NH3+] = 5 mmol/75 ml = 0.067
M
         Â
C2H5NH3+ + H2O <==> C2H5NH2 + H3O+
initial     0.067         excess
             -              Â
-Â Â Â Â
change   Â
-x          Â
excess           Â
+x           Â
+x
final    Â
0.067-x      Â
excess            Â
x             Â
x
Ka = [C6H5NH2][H3O+]/[C6H5NH3+]
Kw/Kb = 1 x 10^-14/5.6 x 10^-4 = x^2/0.067 - x
let x be a small amount and can be ignored in the
denominator
x = [H3O+] = 1.10 x 10^-6 M
pH = -log[H+] = 5.96
