We need (i) pK_B values for methylamine, alternatively pK_a values for H_3CNH_3^+. This site gives pK_b=3.66.
And (ii) we need a stoichiometric equation.........
H_3CNH_2(aq) + H_2O(l)rightleftharpoonsH_3CNH_3^(+) +HO^-
And by definition, this equilibrium is governed by the quotient.....
K_b=10^(-pK_b)=10^(-3.66)=([H_3CNH_3^+][HO^-])/([H_3CNH_2]).
Now if initially, [H_3CNH_2]=0.470*mol*L^-1, and we say the amount of dissociation is x, then we can write:
K_b=([H_3CNH_3^+][HO^-])/([H_3CNH_2])=((x)xx(x))/(0.470-x)=x^2/(0.470-x)=10^(-3.66).
This is a quadratic in x, which we could solve exactly, but because chemist are lazy folk, we make the approximation, that 0.366">>"x, and that x^2/(0.470-x)=10^(-3.66)~=x^2/(0.470).
And thus x_1=sqrt(10^(-3.66)xx0.470)=1.01xx10^-2, and if we recycle this first approximation back into the equation, we gets.....
x_2=sqrt(10^(-3.66)xx(0.470-1.01xx10^-2))=1.00xx10^-2*mol*L^-1
x_3=sqrt(10^(-3.66)xx(0.470-1.00xx10^-2))=1.00xx10^-2*mol*L^-1
Because the approximations have converged, we are willing to accept this value. But x=[HO^-] by definition; and so pOH=-log_10(1.00xx10^-2)=+2
And since we know (or should know) that pOH+pH=14, pH=12.
