Question

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

Answer 1

I got `K_b = 5.0xx10^(-10)`.

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To be pedantic, `"Bu"` is a butyl group, so I would write `"BuOOH"` and `"BuOO"^(-)`.

Using the `"pKa"` of butyric acid (butanoic acid), we can calculate the `K_a`, and then go from there.

Recall that `"pKa" = -log(K_a)`. Thus,

`10^(-"pKa") = color(green)(K_a = 10^(-4.7))`

Now, let's solve for `K_b`. Recall that `K_aK_b = K_w = 10^(-14)`.

`color(blue)(K_b) = (K_w)/(K_a)`

`= (10^(-14))/(10^(-4.7))`

`= 10^(-14 - (-4.7))`

`= 10^(-9.3)`

`= color(blue)(5.0xx10^(-10))`

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If you wish to see the context, and to see how we know that `K_aK_b = K_w`, the dissociation of the water-miscible butyric acid in water is:

`"BuOOH"(l) stackrel("H"_2"O"(l)" ")(rightleftharpoons) "BuOO"^(-)(aq) + "H"^(+)(aq)`

And for this process,

`color(green)(K_a = (["BuOO"^(-)]["H"^(+)])/(["BuOOH"])) = 10^(-4.7).`

`K_b` for the butyrate ion, `"BuOO"^(-)`, is for the following reaction, where we have acceptably utilized sodium butyrate:

`"BuOONa"(s) + "H"_2"O"(l) rightleftharpoons "BuOOH"(aq) + "NaOH"(aq),`

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

`color(green)(K_b = (["BuOOH"]["OH"^(-)])/(["BuOO"^(-)]))`

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

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

`= (cancel(["BuOO"^(-)])["H"^(+)])/(cancel(["BuOOH"]))*(cancel(["BuOOH"])["OH"^(-)])/(cancel(["BuOO"^(-)]))`

`= color(highlight)(["H"^(+)]["OH"^(-)] = K_w = 10^(-14))`

Well, that's nice. So, `\mathbf(K_aK_b = K_w)`. 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.