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

Apr 16, 2018

See below

Explanation:

As we are finding the ${K}_{b}$ we want to know the $p O H$ of the solution as a first step:

Using
$p O H + p H = 14$ (assuming standard conditions)
$p O H = 4.91$
So $O {H}^{-} = {10}^{- 4.91}$

${K}_{b}$ for this species would be (I assume):

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

(But ${H}_{2} O$ is excluded)

Because ${C}_{6} {H}_{5} N + {H}_{2} O r i g h t \le f t h a r p \infty n s O {H}^{-} + {C}_{6} {H}_{5} N {H}^{+}$

So setting up an ICE table:
${C}_{6} {H}_{5} N + {H}_{2} O r i g h t \le f t h a r p \infty n s O {H}^{-} + {C}_{6} {H}_{5} N {H}^{+}$
I: 0.1/-/0/0

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

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

But from finding the $p O H$, we found the $O {H}^{-}$ and know it's concentration: ${10}^{- 4.91}$, so this must be the value of $x$.

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

${K}_{b} = {\left({10}^{- 4.91}\right)}^{2} / 0.0999876$

${K}_{b} \approx 1.51 \times {10}^{-} 9$