..........or at least you need #K_a# for methyl ammonium ion.
From this site, we learn that #K_b# for methylamine is #4.4xx10^-4#. I don't know whether the site is right, I will check later.
Now in aqueous solution, methylamine undergoes the acid base reaction:
#H_3CNH_2(aq) + H_2O(l) rarr H_3CNH_3^+ + HO^-#
And we write the equilibrium equation in the usual way:
#K_b=([H_3CNH_3^+][HO^-])/([H_3CNH_2(aq)])=4.4xx10^-4#
So if #x# moles of methylamine associate, we can put in some numbers:
#K_b=(x^2)/(0.26-x)=4.4xx10^-4#
As is normal, this is a quadratic in #x#, the which I can solve precisely. But since chemists are simple folk, we can make the approximation that #(0.26-x)~=0.26#. We have to justify this approximation later.
And so #x_1=sqrt(4.4xx10^-4xx0.26)=0.0107#.
#x_1#, our first approx. was indeed small compared to #0.26*mol*L^-1#, but we can recycle this approximation to see how good it was, #x_2=0.0105#, and #x_3=0.0105#. Since #x_3# has converged, this is equal to the solution we would have got by the quadratic equation.
So, #[HO^-]=0.0105*mol*L^-1#; #pOH=-log_10[HO^-]=#
#-log_10(0.0105)=1.98#
But #pH+pOH=14# in aqueous solution, and thus #pH=12.0#
Now I admit this does seem like a lot of work (it was more work for me, because I had to format the equations on this editor!). But I can assure you that you can get very proficient at these sorts of problems, especially if you can work your calculator competently. Remember the approach, approximate and then justify. And then recycle the approximation. Good luck.
