What is the pH of a 0.470 M solution of methylamine?

1 Answer

Answer:

#pH=12#

Explanation:

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#.