please calculate the pH of an aqueous solution of pyridine (pKb=8.77) of the following concenteration:(a)1.0M?

Answer 1

I've already showed you how to do this. Pyridine is C5H5N
You have

.. .. .. .. .. C5H5N +H2O <=> C5H5NH(+) + OH-
Initial .. .. .. .. 1
Dissociate .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. x
At Equil .. .. 1-x .. .. . .. .. .. .. .. .. .. x .. .. .. .. .. x

Kb=x^2/(1-x) = 10^-8.77
x^2 +(10^-8.77)x- 10^-8.77 =0

x =0.5*(-10^-8.77 +SquareRoot( (10^-8.77)^2 +4*10^-8.77))=>
x=4.12*10^-5
pOH=-logx= 4.39
pH=14-4.39= 9.61