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#?

Question

Cameron Andrews · Accepted Answer

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"^(-)]) = ?#

(Although I should, I don't always write equilibrium constants in units; this is because I usually use them in logarithms, which eliminates the units anyhow.)

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"#.

When taking an exam, you should be able to interconvert fairly quickly using the one you have in the back of your textbook.

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

or

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

For instance, you ought to be capable of carrying out:

#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)#

Making use of logarithmic properties:

#-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)#

Whatever is simpler to recall is acceptable. Now, let's return to the problem's context:

#"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#!

Naomi Aronson · Answer

undefined You can use the equation for the auto-ionization of water (Kw = [H+][OH-]) and the equilibrium constant expression for the reaction between acetic acid and sodium ethanoate (Ka = [CH3COOH][OH-]/[CH3COO-]) to derive Kw = KaKb.