Question

the hydronium ion concerntration of an aqueous solutin of 0.375 M pyridine (a weak base with the formula C5H5N) is...

[H3O+]=

Answer 1

pyridine is a weak base . In aqueous solution it is

py + H2O <-------> pyH⁺  + OH-

Since it is basic inaqueous solution , the [OH-] > [H3O+] and we can calculate the [H3O+] as

[H3O+] = Kw / [OH-]

Now let us calculate the [OH-] in solution .

The pKb of pyridine = 1.6949x10^-9

given [py] = 0.375 M

For a weak base   py + H2O <-------> pyH⁺  + OH-

c - 0 0 initail concentrations

c(1-x) cx cx at equilibrium

Since Kb = cx.cx /c(1-x) and x <<<unity , the equation becomes

Kb = cx²

Thus x = square root of Kb/c and [OH-] = square root of Kb.C

Thus we can calculate [OH-] = square root of [1.6949x10^-9 x 0.375] = 2.52 x10^-5 M

and the hydronium ion [H3O+] = 1.0x10^-14 / 2.52x10^-5

= 3.968x10^-10