Question

Calculate the pH at the equivalence point when you titrate 1.6M of CH_3NH_2 and 1.6M of HCl.

Answer 1

VF = V1+V2 = 2V1

then, concnetration will be halved, that is 1.6/2 = 0.8 M of CH3NH2

CH3NH3Cl will be formed

then

CH3NH3+ + Cl- in aquous solution

CH3NH3+ + H2O <--> CH3NH2 + H3O+

this is slightly acidic so

Ka = [CH3NH2][H3O+]/[CH3NH3+]

Ka = (Kw/Kb) = (10^-14)/(4.4*10^-4) = 2.27*10^-11

then

[CH3NH2]= x= [H3O+]

[CH3NH3+] = M-x = 0.08-x

Ka = [CH3NH2][H3O+]/[CH3NH3+]

2.27*10^-11 = (x*x)/(0.08-x)

x = H3O+ = 1.34*10^-6

p H= -log(H) = -log( 1.34*10^-6 = 5.872