Question

calculate the pH at the equilivalence point when 12.35 mL of a 0.100M codeine soultion (Kb=1.6x10^-6) is titrated with 0.100 M hydrochloric acid.

Answer 1

At equivalence point

Moles of codeine = moles of HCl.

As given molarity is equal for codeine and HCl then at equivalence point

Volume of Codeine = Volume of HCl = 12.35 mL.

As equal moles of codeine reacts with equal moles of HCl , then moles of salt = moles of codeine

=( 12.35*0.100)*10^-3 = 1.235*10^-3.

Now after reaction total volume =( 12.35+12.35 )= 24.70 mL.

= 24.70*10^-3 L.

Then concentration of salt = (1.235*10^-3)/(24.70*10^-3)

= 0.05 M.

The salt is formed by weak base and strong acid

Hence , pH = 7 - ( pKb + logC)

= 7 - ( -log1.6*10^-6 + log 0.05)

= 7 - ( 5.795 - 1.30)

= 7 - 2.247

= 4.753