Question

#32 Calculate the pH at the equivalent point for the titration of 0.10M NH3 by 0.10M perchloric acid. (For NH3, pKb=4.74, hint: consider the change in volume before and after reaction)

Answer 1

Pkb = -log Kb

4.74 = -log Kb

Kb = 1.8197 x 10-5

also

Ka = 10-14 / Kb

Ka = 10-14 / 1.8197 x 10-5

Ka = 5.495 x 10-10

the reaction is given by

Nh3 + HCl04 ----> Nh4Cl04

given

0.1 M each of Nh3 and HCl04

at equivalence point

Ma x Va = Mb x Vb

0.1 x Va = 0.1 x Vb

Va = Vb

now

final volume V = Va + Vb = 2Va = 2Vb

now consider

Nh3 + HCl04 ----> Nh4Cl04

from the above reaction

1 mole of NH3 on reaction with 1 mole HCl04 reacts to give 1 mole of NH4Cl04

so

moles of Nh4Cl04 formed = moles of NH3 reacted

we know that

moles = molarity x volume

so

moles of NH3 reacted = 0.1 x Vb = 0.1 Vb

so

moles of NH4CL04 formed = 0.1 Vb

now

final volume = 2Vb

molarity of Nh4CL04 = moles / volume = 0.1 Vb / 2Vb = 0.05 M

as NH4CL04 is a salt of strong acid and weak base

Nh4Cl04 undergoes cationinc hydrolysis in water

Nh4+ + H20   ---> NH3 + H30+

Ka = [NH3] [H30+] / [ Nh4+]

Ka = x *x / ( C-x)

as Ka ia very less

we can reduce it to the form

[H30+] = x = sqrt ( Ka x C )

= sqrt ( 5.495 x 10-10 x 0.05 )

[H30+] = 5.24 x 10-6

now

pH = -log [H30+]

pH = -log 5.24 x 10-6

pH = 5.28

so the pH at equivalence point is 5.28