Calculate percent dissociation of a)0.0074 M butanoic acid (b) 0.0075 M butanoic acid containing 0.085M sodium butanoate?

1 Answer
Dec 30, 2015

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, #"H"_3"O"^(+)#, and their respective conjugate base.

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, #"C"_4"H"_8"O"_2#, will react with water to form hydronium ions and the butanoate anion, #"C"_4"H"_7"O"_2^(-)#, the acid's conjugate base.

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])^(+)#

#color(purple)("I")" " " "0.0074" " " " " " " " " " " " " " " " "0" " " " " " " " " " " " " 0#
#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x) " " " " " " " "(+x)#
#color(purple)("E")" "0.0074-x" " " " " " " " " " " " " "x " " " " " " " " " " " "x#

The acid dissociation constant, #K_a#, for butanoic acid is equal to #1.5 * 10^(-5)# at room temperature

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 #K_a# is relatively small compared with the initial concentration of the acid, you can say that

#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 #x# represents the equilibrium concentration of both the hydronium and of the butanoate ions, you can say that

#["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 #1000# butanoic acid molecules, only #45# will release their acidic proton in solution. The rest will remain undissociated.

Now for part (b).

This time, you're adding the acid to a solution that contains sodium butanoate, #"C"_4"H"_7"O"_2"Na"#. Sodium butanoate is soluble in aqueous solution, which means that it will dissociate completely to form sodium cations, which are of no interest to you, and butanoate anions.

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])^(+)#

#color(purple)("I")" " " "0.0074" " " " " " " " " " " " " " " " "0" " " " " " " " " " " " 0.085#
#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x) " " " " " " " "(+x)#
#color(purple)("E")" "0.0074-x" " " " " " " " " " " " " "x " " " " " " " " " " "0.085+x#

This time, #K_a# will be equal to

#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 #18# out of #10^6# butanoic acid molecules will release their acidic proton. The rest will remain undissociated.

SIDE NOTE When making approximations like the one we made here

#0.0074 - x ~~ 0.0074#

you need to make sure that the percent error you get is smaller than #5%#. In this case, you would have

#"% error" = (|(0.0074 - 3.33 * 10^(-4)) - 0.0074|)/0.0074 xx 100#

#"% error" = 4.5% < 5%#

This means that the approximation holds.