Question

Calculate the pH at the equivalence point for the titration of .100M methylamine (CH3NH2) with .100 M HCL. The Kb of methylamine is 5.0x10^-4

Answer 1

The reaction involved in the titration

HCl + CH₃NH₂CH₃NH₃⁺Cl^-

HCl and methylamine reacts in equimolar quantity, therefore at equivalence point

Moles of HCl = Moles of CH3NH2

At equivalence point suppose 1L HCl reacts with 1L CH₃NH₂. Total volume becomes 2 litres.

Molarity of CH₃NH₃⁺ = no. of moles / Volume of solution = 0.1 / 2 = 0.05 M

At equivalence point pH is calculated due to dissociation of salt formed

CH₃NH₃⁺CH₃NH₂ + H+

pH = 7 - 1/2 (pKb + log c) (Here c is concentration of salt formed in the titration)

pKb = -Log Kb = -log (5 x 10-4) = -(log5 - 4log10) = -(log5 - 4) = 4 - log5 = 4 - 0.69 = 3.30

pH = 7 - 1/2(3.30 + log 0.05)

pH = 6.0