# Question fe19f

Oct 28, 2017

$1.4 \cdot {10}^{- 4}$

#### Explanation:

You know that you're dealing with a weak acid, so you can write its ionization equilibrium like this

${\text{C"_ 3"H"_ 6"O"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "C"_ 3"H"_ 5"O"_ (3(aq))^(-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

Now, the problem tells you in a $\text{0.100-M}$ solution of lactic acid, the acid is only 3.7% dissociated. This basically means that for every $1000$ molecules of lactic acid present in the solution, only $37$ will dissociate to produce lactate, the conjugate base of lactic acid, and hydronium cations.

In other words, for every $100$ moles of lactic acid present in this solution, only $3.7$ moles will dissociate.

This means that if you start with an initial concentration of ${\left[{\text{C"_3"H"_6"O}}_{3}\right]}_{0}$, you can say that at equilibrium, the solution will contain

${\left[{\text{C"_3"H"_5"O"_3^(-)] = 3.7/100 * ["C"_3"H"_6"O}}_{3}\right]}_{0}$

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"))#

${\left\{{\text{C"_ 3"H"_ 6"O"_ 3] =96.3/100 * ["C"_ 3"H"_ 6"O}}_{3}\right]}_{0}$

By definition, the acid dissociation constant, ${K}_{a}$, is equal to

${K}_{a} = \left(\left[{\text{C"_ 3"H"_ 5"O"_ 3^(-)] * ["H"_ 3"O"^(+)])/(["C"_ 3"H"_ 6"O}}_{3}\right]\right)$

Since the dissociation of the acid produces lactate anions and hydronium cations in a $1 : 1$ mole ratio, you can say that, at equilibrium, you have

${\left[{\text{C"_ 3"H"_ 5"O"_ 3^(-)] = ["H"_ 3"O"^(+)] = 3.7/100 * ["C"_3"H"_6"O}}_{3}\right]}_{0}$

This means that the acid dissociation constant is equal to

${K}_{a} = \left(\frac{3.7}{\textcolor{red}{\cancel{\textcolor{B l a c k}{100}}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left[{\text{C"_ 3"H"_ 6 "O"_ 3]_ 0))) * 3.7/100 * ["C"_ 3"H"_ 6"O"_ 3]_ 0 )/(96.3/color(red)(cancel(color(Black)(100))) * color(red)(cancel(color(black)(["C"_ 3"H"_ 6 "O}}_{3}\right]}_{0}}}}\right)$

This is equivalent to

${K}_{a} = {3.7}^{2} / \left(100 \cdot 96.3\right) \cdot \left[{\text{C"_3"H"_6"O}}_{3}\right]$

Plug in the value you have for the initial concentration of the acid to get--I'll leave the answer without added units

${K}_{a} = {3.7}^{2} / 9630 \cdot 0.100 = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.4 \cdot {10}^{- 4}}}}$

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

${K}_{a} = 1.38 \cdot {10}^{- 4}$

so this is an excellent result.