This is a base in water so, let the base be Pyridine = "B" and Pyridine H+ = HB+ the protonated base "HB+"
there are free OH- ions so, expect a basic pH
B + H2O <-> HB+ + OH-
The equilibrium Kb:
Kb = [HB+][OH-]/[B]
initially:
[HB+] = 0
[OH-] = 0
[B] = M
the change
[HB+] = x
[OH-] = x
[B] = - x
in equilibrium
[HB+] = 0 + x
[OH-] = 0 + x
[B] = M - x
Now substitute in Kb
Kb = [HB+][OH-]/[B]
Kb = x*x/(M-x)
x^2 + Kbx - M*Kb = 0
x^2 + (1.7*10^-9)x - (0.08)(1.7*10^-9) = 0
solve for x
x = 1.166*10^-5
substitute:
[HB+] = 0 + x = 1.166*10^-5 M
[OH-] = 0 + x = 1.166*10^-5M
pH = 14 + pOH = 14 + log(1.166*10^-5) = 9.066
pH = 9.066
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