Question

A 0.108-M aqueous solution of CH3NH2 (methylamine) has a pH of 11.8. Calculate the pH of a buffer solution that is 0.108 M in CH3NH2 and 0.308 M in CH3NH3+

Answer 1

if this is a buffer

this must be basic, since we are talking about amines

use buffer equation for basic solution

pOH = pKb + log(CH3NH3+/CH3NH2)

pOH = 14-pH = 14-11.8 = 2.2

[OH-] = 10^-2.2 = 0.00630957

Kb methyl amine = 4.38 x 10-4

pKb = -log(Kb) = -log(4.38*10^-4) =3.35852

... prove that with info given

CH3NH2 + H2O <--> CH3NH3+ +OH-

Kb = [CH3NH3+][OH-]/[CH3NH2]

then

Kb = (0.00630957)(0.00630957)/(0.108-0.00630957) = 0.0003914

Kb = 3.914*10^-4

pKb = -log(KB) = -log(0.0003914) = 3.40737

which is pretty near to that in databases...

then

pOH = pKb + log(CH3NH3+/CH3NH2)

pOH = 3.40737 + log(0.308/0.108 )

pOH = 3.8624

pH = 14-pOH = 14-3.8624= 10.1376

pH = 10.1376