Question #2506e

In order to be able to provide a numerical answer, you must know the value of the acid dissociation constant, #K_a#, for lactic acid, #"C"_3"H"_6"O"_3#, which you can find listed here

#K_a = 1.38 * 10^(-4)#

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

So, what does degree of ionization mean?

In simple terms, the problem wants you to determine how many molecules of lactic acid will donate their acidic proton. Keep in mind that lactic acid is a monoprotic acid, i.e. it only has one acidic proton to donate.

Since lactic acid is a weak acid, you can expect the majority of lactic acid molecules to remain unionized in aqueous solution.

The problem tells you that your starting solution ha a concentration of #"0.05 F"#, or #"0.05 formal"#. I won't go into detail about this, but in this context you can assume it to be equal to #"0.05 M"#, or #"0.05 molar"#.

Now, your tool of choice here will be an ICE table. The equilibrium reaction that describes the ionization of lactic acid looks like this

#" " "C"_3"H"_6"O"_text(3(aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" ""C"_3"H"_5"O"_text(3(aq])^(-) + "H"_3"O"_text((aq])^(+)#

#color(purple)("I")" " " "color(white)(a)0.05" " " " " " " " " " " " " " " " " " " " "0" " " " " " " " " "color(white)(a)0#
#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " " " " " "(+x)" " " " "(+x)#
#color(purple)("E")" "color(white)(a)0.05-x" " " " " " " " " " " " " " " " " "color(white)(a)x" " " " " " " " "x#

By definition, the acid dissociation constant for this equilibrium will be

#K_a = (["C"_3"H"_5"O"_3^(-)] * ["H"_3"O"^(+)])/(["C"_3"H"_6"O"_3])#

Now, notice that every molecule of lactic acid that ionizes will produce one molecule of lactate ions, #"C"_3"H"_5"O"_3^(-)#, and one molecule of hydronium ions, #"H"_3"O"^(+)#.

This is why the concentration of the products increases by the same amount, #x#, as the concentration of the reactant decreases.

You will thus have

#K_a = (x * x)/(0.05 - x) = x^2/(0.05 - x) = 1.38 * 10^(-4)#

Rearrange to form a quadratic equation

#x^2 = 1.38 * 10^(-4) * (0.05 - x)#

#x^2 + 1.38 * 10^(-4)x - 6.9 * 10^(-6) = 0#

This quadratic will produce two values for #x#, one positive and one negative. Since #x# represents concentration, the only valid solution will be the positive one.

#x = 2.56 * 10^(-3)#

So, the equilibrium concentrations of the two ions are

#["C"_3"H"_5"O"_3^(-)] = ["H"_3"O"^(+)] = 2.56 * 10^(-3)"M"#

Now, the degree of ionization, #alpha#, is calculated by dividing the concentration of ionized molecules by the total concentration of molecules of acid.

#color(blue)(|bar(ul(color(white)(a/a)alpha = "concentration of ionized molecules"/"total concentration of the acid"color(white)(a/a)|)))#

In your case, you have a total of #"0.05 M"# acid molecules, out of which only #2.56 * 10^(-3)#"M"# are ionized. This means that you have

#alpha = (2.56 * 10^(-3)color(red)(cancel(color(black)("M"))))/(0.05color(red)(cancel(color(black)("M")))) = color(green)(|bar(ul(color(white)(a/a)0.053color(white)(a/a)|)))#

To get the percent ionization of the acid, simply multiply the degree of ionization by #100#.

#color(blue)(|bar(ul(color(white)(a/a)%alpha = alpha xx 100color(white)(a/a)|)))#

In your case, you will have

#%alpha = 0.053 xx 100 = color(green)(|bar(ul(color(white)(a/a)5.3%color(white)(a/a)|)))#