Question

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`?

Answer 1

It ought to be known ahead of time that the `"pKa"` of acetic acid is `4.74`. Or, perhaps it is known ahead of time that the `"K"_a` of acetic acid is `1.8xx10^(-5) "M"`. One of these is in the back of your textbook in one of the appendices... I believe `"K"_a` would be.

For the partial dissociation of acetic acid, we have the associated `"K"_a`, the acid dissociation constant.

Since you are actually working with sodium ethanoate, you are starting with `"CH"_3"COO"^(-)`.

`"CH"_3"COO"^(-) + "H"_2"O"(l) rightleftharpoons "OH"^(-)(aq) + "CH"_3"COOH"(aq)`

`color(blue)("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 `"M"`. Therefore, `("M"*cancel"M")/cancel"M" = "M"`. Note that `"1 dm"^3 = "1 L"`, which means that `"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:

`color(blue)("pKa") = -log(1.8xx10^(-5)) ~~ color(blue)(4.74)`

`color(blue)("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 `"K"_w` (the water autoionization constant) and its relationship to `"K"_b` (aptly named the base dissociation constant) and `"K"_a`:

`\mathbf("K"_w = "K"_a"K"_b)`

Or, in terms of `"pK"_X`'s, we can start with:

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

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)`:

`color(blue)("pK"_w = "pK"_a + "pK"_b)`

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

`"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"^(-)]))`

`color(green)(["H"^(+)]["OH"^(-)] = ["H"^(+)]["OH"^(-)])`

There, we have shown that `"K"_w = "K"_a"K"_b`!