Question

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

Answer 1

See below

Explanation:

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`