calculating pH?

Question

How would you calculate the pH of a 1.377 M CH3NH3+ solution? Kb=0.00036

well, i calculated the pOH, since CH3NH3+ + OH- --> CH3NH3OH, [OH-] = 1.377 M as well,
putting that in -log[OH-] gives me -0.1389 as pOH....
what would be the pH of this?!?!?

Answer 1

CH3NH3+ is the conjugate acid of the base CH3NH2.
Thus the solution is ACIDIC

For the conjugate acid Ka=Kw/Kb =10^-14 / 0.00036 = 2.78*10^-11

.. .. .. .. .. .. .. CH3NH3+ <=> CH2NH2 + H+
Initial .. .. .. .. .. 1.377
Dissociate .. .. ..x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. ..x
At Equil. .. .. 1.377-x .. .. .. .. .. .. ..x .. .. .. ..x

Ka = [CH3NH2][H+] / [CH3NH3+] = x^2/(1.377-x)

Let's assume that x<<1.377 so that 1.377-x=1.377
Then the equation is simplified to

Ka=x^2/1.377 =>
x=squareroot (1.377*Ka) =SQRT(1.377* 2.78*10^-11)= 6.187*10^-6 M which is <<1.377, thus our assumption is valid.

pH= -log[H+]=-logx =-log(6.187*10^-6) =5.21

Answer 2

pOH + pH = pKw, where Kw is the ionic product of water. the value at standard temp in 10^-14, so use this value if its not specified, giving pKw = 14, hence,
-0.1389 + pH = 14
pH = 14 + 0.1389
pH = 14.1389