Answer:
[tex]K_{b}[/tex] is 0.000438 and [tex]pK_{b}[/tex] is 3.36
Explanation:
Methylamine is a monoprotic base.
For a monoprotic base, [tex]K_{b}=\frac{ca^{2}}{(1-a)}[/tex]
where, c is concentration of base in molarity and a is it's degree ionization
Here [tex]a=\frac{6.4}{100}=0.064[/tex] and c = 0.100 M
So, [tex]K_{b}=\frac{(0.100)\times (0.064)^{2}}{(1-0.064)}=0.000438[/tex]
We know, [tex]pK_{b}=-logK_{b}[/tex]
Hence, [tex]pK_{b}=-log(0.000438)=3.36[/tex]
