Question #fe19f
1 Answer
Explanation:
You know that you're dealing with a weak acid, so you can write its ionization equilibrium like this
#"C"_ 3"H"_ 6"O"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "C"_ 3"H"_ 5"O"_ (3(aq))^(-) + "H"_ 3"O"_ ((aq))^(+)#
Now, the problem tells you in a
In other words, for every
This means that if you start with an initial concentration of
#["C"_3"H"_5"O"_3^(-)] = 3.7/100 * ["C"_3"H"_6"O"_3]_0#
Similarly, at equilibrium, the concentration of the lactic acid will be
#["C"_ 3"H"_ 6"O"_ 3] = overbrace(["C"_ 3"H"_ 6"O"_ 3]_ 0)^(color(blue)("what you start with")) - overbrace(3.7/100 * ["C"_3"H"_6"O"_3]_0)^(color(blue)("what dissociates"))#
#{"C"_ 3"H"_ 6"O"_ 3] =96.3/100 * ["C"_ 3"H"_ 6"O"_ 3]_ 0#
By definition, the acid dissociation constant,
#K_a = (["C"_ 3"H"_ 5"O"_ 3^(-)] * ["H"_ 3"O"^(+)])/(["C"_ 3"H"_ 6"O"_ 3])#
Since the dissociation of the acid produces lactate anions and hydronium cations in a
#["C"_ 3"H"_ 5"O"_ 3^(-)] = ["H"_ 3"O"^(+)] = 3.7/100 * ["C"_3"H"_6"O"_3]_0#
This means that the acid dissociation constant is equal to
#K_a = (3.7/color(red)(cancel(color(Black)(100))) * color(red)(cancel(color(black)(["C"_ 3"H"_ 6 "O"_ 3]_ 0))) * 3.7/100 * ["C"_ 3"H"_ 6"O"_ 3]_ 0 )/(96.3/color(red)(cancel(color(Black)(100))) * color(red)(cancel(color(black)(["C"_ 3"H"_ 6 "O"_ 3]_ 0))) )#
This is equivalent to
#K_a = 3.7^2/(100 * 96.3) * ["C"_3"H"_6"O"_3]#
Plug in the value you have for the initial concentration of the acid to get--I'll leave the answer without added units
#K_a = 3.7^2/9630 * 0.100 = color(darkgreen)(ul(color(black)(1.4 * 10^(-4))))#
The answer is rounded to two sig figs, the number of sig figs you have for the percent dissociation of the acid.
The value listed here for the acid dissociation constant of lactic acid is
#K_a = 1.38 * 10^(-4)#
so this is an excellent result.