In: Chemistry
What is the pH at the quivalence point when 50.0 mL of 0.10 M ethylamine (C2H5NH2) is titrated with 0.20 M HCLO4? (Kb of C2H5NH2 = 5.6 x 10-4) (Hint: You will need to determine the volume of HCLO4 solution used) Be sure to show ICE table.
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
C2H5NH3+ + H2O <==> C2H5NH2 + H3O+
initial 0.067 excess - -
change -x excess +x +x
final 0.067-x excess x 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