The pH of aquous 0.10M pyridine (C6H5N) ion is 9.09. What is the Kb for this base?

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Answer 1 · 2018-04-16T09:22:10

See below

As we are finding the $K_b$ we want to know the $pOH$ of the solution as a first step:

Using
$pOH+pH=14$ (assuming standard conditions)
$pOH=4.91$
So $OH^(-)=10^(-4.91)$

$K_b$ for this species would be (I assume):

$K_b=([OH^-] times [C_6H_5NH^+])/([C_6H_5N] times [H_2O]$

(But $H_2O$ is excluded)

Because $C_6H_5N + H_2O rightleftharpoons OH^(-) + C_6H_5NH^(+)$

So setting up an ICE table:
$C_6H_5N + H_2O rightleftharpoons OH^(-) + C_6H_5NH^(+)$
I: 0.1/-/0/0

C:-x/-/+x/+x/

E:(0.1-x)/-/x/x

But from finding the $pOH$ , we found the $OH^-$ and know it's concentration: $10^(-4.91)$ , so this must be the value of $x$ .

So
$K_b=([x^2])/([1-x]$

$K_b=(10^(-4.91))^2/0.0999876$

$K_b approx 1.51 times 10^-9$