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

1 Answer
May 27, 2016

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


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))#


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.