Question

A 21.8 mL sample of 0.223 M trimethylamine, (CH3)3N, is titrated with 0.207 M hydrobromic acid. At the equivalence point, the pH is

Answer 1

in equivalence point:

mol of acid = mol of base

mol of base = 21.8*0.223 = 4.8614 mmol of base

mmol of acid = 4.8614

V = mmol/M = 4.8614/(0.207) = 23.48 ml

VT = V1+V2 = 23.48+21.8 = 45.28 ml

conjugate =

[(CH3)3NH+] = 4.8614 /(45.28 ) = 0.107363 M

HB + H2O <--> B + H3O+

Ka = [(CH3)3N][H3O+]/[(CH3)3NH+]

KB = 6.3*10^-5

Ka = (10^-14)/(6.3*10^-5) = 1.5873*10^-10

Ka = [(CH3)3N][H3O+]/[(CH3)3NH+]

1.5873*10^-10 = (x*x)/(0.107363 -x)

x = [H+] = 4.12*10^-6

pH = -logx =-log( 4.12*10^-6) =

pH = 5.3851