# How do I show that for acetic acid and sodium ethanoate, K_w = K_a K_b if I don't know the K_a?

Feb 4, 2016

It ought to be known ahead of time that the $\text{pKa}$ of acetic acid is $4.74$. Or, perhaps it is known ahead of time that the ${\text{K}}_{a}$ of acetic acid is $1.8 \times {10}^{- 5} \text{M}$. One of these is in the back of your textbook in one of the appendices... I believe ${\text{K}}_{a}$ would be.

For the partial dissociation of acetic acid, we have the associated ${\text{K}}_{a}$, the acid dissociation constant.

Since you are actually working with sodium ethanoate, you are starting with ${\text{CH"_3"COO}}^{-}$.

$\text{CH"_3"COO"^(-) + "H"_2"O"(l) rightleftharpoons "OH"^(-)(aq) + "CH"_3"COOH} \left(a q\right)$

$\textcolor{b l u e}{\text{K"_a = \frac(["CH"_3"COO"^(-)]["H"^(+)])(["CH"_3"COOH"]) = 1.8xx10^(-5) "M}}$

"K"_b = \frac(["CH"_3"COOH"]["OH"^(-)])(["CH"_3"COO"^(-)]) = ?

(I don't always write units for equilibrium constants, though I should. It's because I tend to use them in logarithms, which gets rid of the units anyways.)

You can tell that's what the units are because the concentrations have units of $\text{M}$. Therefore, ("M"*cancel"M")/cancel"M" = "M". Note that $\text{1 dm"^3 = "1 L}$, which means that $\text{mol"cdot"dm"^(-3) = "mol"cdot"L"^(-1) = "M}$.

Whichever one you have in the back of your textbook, you should be able to interconvert fairly quickly on a test.

\mathbf("pKa" = -log("K"_a))

or

\mathbf("K"_a = 10^(-"pKa"))

As an example, you should be able to perform:

$\textcolor{b l u e}{\text{pKa}} = - \log \left(1.8 \times {10}^{- 5}\right) \approx \textcolor{b l u e}{4.74}$

$\textcolor{b l u e}{\text{K"_a) = 10^(-4.74) ~~ color(blue)(1.82xx10^(-5) "M}}$

Because you're in water, you can use ${K}_{w}$ to relate back to ${K}_{b}$. We could consider the ${\text{K}}_{w}$ (the water autoionization constant) and its relationship to ${\text{K}}_{b}$ (aptly named the base dissociation constant) and ${\text{K}}_{a}$:

$\setminus m a t h b f \left({\text{K"_w = "K"_a"K}}_{b}\right)$

Or, in terms of ${\text{pK}}_{X}$'s, we can start with:

$- \log \left({\text{K"_w) = -log("K"_a"K}}_{b}\right)$

Using the properties of logarithms:

-log("K"_w) = -(log("K"_a) + log("K"_b))

-log("K"_w) = -log("K"_a) + [-log("K"_b)]

And using the definition of "pK"_X = -log("K"_X):

$\textcolor{b l u e}{{\text{pK"_w = "pK"_a + "pK}}_{b}}$

Whichever one is easier to remember is fine. So, going back to the context of the problem:

${\text{K"_w = ["H"^(+)]["OH"^(-)] stackrel(?)(=) "K"_a"K}}_{b}$

stackrel(?)(=) \frac(cancel(["CH"_3"COO"^(-)])["H"^(+)])(cancel(["CH"_3"COOH"]))\frac(cancel(["CH"_3"COOH"])["OH"^(-)])(cancel(["CH"_3"COO"^(-)]))

$\textcolor{g r e e n}{\left[{\text{H"^(+)]["OH"^(-)] = ["H"^(+)]["OH}}^{-}\right]}$

There, we have shown that ${\text{K"_w = "K"_a"K}}_{b}$!