Question

Calculate the pH at the equivalence point for the titration of 0.100 M methylamine (CH3NH2) with 0.100 M HCl. The Kb of methylamine is 5.0

Answer 1

at equivalance point

CH3NH2 + HCl ------------------> CH3NH3^+Cl^-

0.1 xv           0.1 xv                       0                   before titration

0                    0                               0.1 xv               after titration

CH3NH3^+Cl^ moles = 0.1 xv  

CH3NH3^+Cl^ concentration = 0.1 xv   /total volume

[CH3NH3^+Cl^] = 0.1 v/2v

                       C       = 0.05M

salt is strong acid weak base type on hydrolysis of this salt

pKb = -logKb= -log(5 x10^-4) = 3.30

pH= 7- 1/2 (pKb + logC)

    = 7- 1/2 (3.30 + log(0.05))

   pH = 6.00