The #pKa# of butyric acid #"HBut"# is 4.7. How do you calculate #K_b# for the butyrate ion #"But"^-#?
1 Answer
I got
To be pedantic,
Using the
Recall that
#10^(-"pKa") = color(green)(K_a = 10^(-4.7))#
Now, let's solve for
#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
#"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).#
#"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,
So, we know we wrote the expressions correctly, and we now see how butyric acid and butyrate dissociate in water.