# The pKa of butyric acid "HBut" is 4.7. How do you calculate K_b for the butyrate ion "But"^-?

May 27, 2016

I got ${K}_{b} = 5.0 \times {10}^{- 10}$.

To be pedantic, $\text{Bu}$ is a butyl group, so I would write $\text{BuOOH}$ and ${\text{BuOO}}^{-}$.

Using the $\text{pKa}$ of butyric acid (butanoic acid), we can calculate the ${K}_{a}$, and then go from there.

Recall that $\text{pKa} = - \log \left({K}_{a}\right)$. Thus,

${10}^{- \text{pKa}} = \textcolor{g r e e n}{{K}_{a} = {10}^{- 4.7}}$

Now, let's solve for ${K}_{b}$. Recall that ${K}_{a} {K}_{b} = {K}_{w} = {10}^{- 14}$.

$\textcolor{b l u e}{{K}_{b}} = \frac{{K}_{w}}{{K}_{a}}$

$= \frac{{10}^{- 14}}{{10}^{- 4.7}}$

$= {10}^{- 14 - \left(- 4.7\right)}$

$= {10}^{- 9.3}$

$= \textcolor{b l u e}{5.0 \times {10}^{- 10}}$

If you wish to see the context, and to see how we know that ${K}_{a} {K}_{b} = {K}_{w}$, the dissociation of the water-miscible butyric acid in water is:

${\text{BuOOH"(l) stackrel("H"_2"O"(l)" ")(rightleftharpoons) "BuOO"^(-)(aq) + "H}}^{+} \left(a q\right)$

And for this process,

$\textcolor{g r e e n}{{K}_{a} = \left(\left[\text{BuOO"^(-)]["H"^(+)])/(["BuOOH}\right]\right)} = {10}^{- 4.7} .$

${K}_{b}$ for the butyrate ion, ${\text{BuOO}}^{-}$, is for the following reaction, where we have acceptably utilized sodium butyrate:

$\text{BuOONa"(s) + "H"_2"O"(l) rightleftharpoons "BuOOH"(aq) + "NaOH} \left(a q\right) ,$

which has an analogous base dissociation expression (ignoring the spectator sodium):

$\textcolor{g r e e n}{{K}_{b} = \left(\left[{\text{BuOOH"]["OH"^(-)])/(["BuOO}}^{-}\right]\right)}$

As an experiment, notice what happens if we do this:

color(highlight)(K_a*K_b) = ???

$= \left(\cancel{\left[{\text{BuOO"^(-)])["H"^(+)])/(cancel(["BuOOH"]))*(cancel(["BuOOH"])["OH"^(-)])/(cancel(["BuOO}}^{-}\right]}\right)$

$= \textcolor{h i g h l i g h t}{\left[{\text{H"^(+)]["OH}}^{-}\right] = {K}_{w} = {10}^{- 14}}$

Well, that's nice. So, $\setminus m a t h b f \left({K}_{a} {K}_{b} = {K}_{w}\right)$. We expected that, right? Good.

So, we know we wrote the expressions correctly, and we now see how butyric acid and butyrate dissociate in water.