Na = 23 x 2.40 = 55.2
O = 16 x 2.40 = 38.4
H = 1 x 2.40 = 2.40
55.2 + 38.4 + 2.4 = 96
2.40 mol of NaOH = 96 amu
We are given the base dissociation constant, Kb, for Pyridine (C5H5N) which is 1.4x10^-9. The acid dissociation constant, Ka for the Pyridium ion or the conjugate acid of Pyridine is to be determined. We know from our chemistry classes that:
Kw = Kb * Ka
where Kw is always equal to 1x10^-14
so, to solve for Ka of Pyridium ion, substitute Kb to the equation together with Kw and solve for Ka:
1x10^-14 = 1.4x10^-9 * Ka
solve for Ka
Ka = 7.14x10^-6
Therefore, the acid dissociation constant of Pyridinium ion is 7.14x10^-6.
<span />
