Calculate percent dissociation of a)0.0074 M butanoic acid (b) 0.0075 M butanoic acid containing 0.085M sodium butanoate?
1 Answer
Here's what I got.
Explanation:
!! LONG ANSWER !!
Before jumping in, make sure that you have a clear understanding of what exactly is percent dissociation.
As you know, weak acids do not dissociate (or ionize) completely in aqueous solution to form hydronium ions,
Instead, an equilibrium is established between the undissociated acid molecules and the two aforementioned ions.
When a problem asks you to determine percent dissociation or percent ionization, you must determine what percentage of the acid molecules have donated their acidic proton.
Simply put, you must figure out what the concentration of the two ions is in solution, and compare it with the initial concentration of the acid.
So, butanoic acid,
Use an ICE table to help you figure out the equilibrium concentration of hydronium and butanoate ions
#" " "C"_4"H"_8"O"_text(2(aq]) + "H"_2"O"_text((l]) rightleftharpoons "C"_4"H"_7"O"_text(2(aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#
The acid dissociation constant,
http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm
By definition, the acid dissociation constant is equal to
#K_a = (["H"_3"O"^(+)] * ["C"_4"H"_7"O"_2^(-)])/(["C"_4"H"_8"O"_2])#
Since the value of
#0.0074 - x ~~ 0.0074#
This means that you have
#1.5 * 10^(-5) = (x * x)/0.0074 = x^2/0.0074#
This will give you
#x = sqrt(1.5 * 0.0074 * 10^(-5)) = 3.33 * 10^(-4)#
Since
#["H"_3"O"^(+)] = ["C"_4"H"_7"O"_2^(-)] = 3.33 * 10^(-4)"M"#
So, for every molecule of butanoic acid that dissociates, one mole of hydronium ions and one mole of butanoate ions are formed in solution.
This means that the percent dissociation of the acid will be equal to
#"% dissociation" = (3.33 * 10^(-4) color(red)(cancel(color(black)("M"))))/(0.0074 color(red)(cancel(color(black)("M")))) xx 100 = color(green)("4.5%")#
This means that out of
Now for part (b).
This time, you're adding the acid to a solution that contains sodium butanoate,
You will thus be dealing with the common ion effect - because the solution already contains a significant amount of conjugate base molecules, the equilibrium will be pushed to the left, i.e. fewer acid molecules will release their acidic proton.
The ICE table will look like this
#" " "C"_4"H"_8"O"_text(2(aq]) + "H"_2"O"_text((l]) rightleftharpoons "C"_4"H"_7"O"_text(2(aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#
This time,
#1.5 * 10^(-5) = (x * (0.085 + x))/(0.0074 - x)#
This time, you can assume that
#0.0074 - x ~~ 0.0074" "# and#" "0.085 + x ~~ 0.085#
This will get you
#1.5 * 10^(-5) = 0.085/0.0074 * x implies x = 1.31 * 10^(-6)#
The percent dissociation will now be
#"% dissociation" = (1.31 * 10^(-6)color(red)(cancel(color(black)("M"))))/(0.0074color(red)(cancel(color(black)("M")))) xx 100 = color(green)("0.018%")#
This time, only
SIDE NOTE When making approximations like the one we made here
you need to make sure that the percent error you get is smaller than
#"% error" = (|(0.0074 - 3.33 * 10^(-4)) - 0.0074|)/0.0074 xx 100#
#"% error" = 4.5% < 5%#
This means that the approximation holds.