# Question 23e08

Jan 4, 2018

${K}_{a} = 1.4 \cdot {10}^{- 4}$

#### Explanation:

If you represent lactic acid as a generic monoprotic weak acid $\text{HA}$, you can write its ionization equilibrium as

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

Now, notice that every mole of lactic acid that dissociates produces $1$ mole of hydronium cations and $1$ mole of lactate anions, the conjugate base of the lactic acid, which is represented here as ${\text{A}}^{-}$.

This means that if you know the equilibrium concentration of hydronium cations, you also know the equilibrium concentration of the lactate anions and the concentration of lactic acid that dissociated.

To find the equilibrium concentration of hydronium cations, use the $\text{pH}$ of the solution.

"pH" = - log(["H"_3"O"^(+)])#

so you can say that

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

This means that, at equilibrium, the solution contains

$\left[\text{A"^(-)] = ["H"_3"O"^(+)] = 10^(-"pH}\right)$

Similarly, you have

$\left[\text{HA"] = ["HA"]_0 - 10^(-"pH}\right)$

This basically means that in order for the reaction to produce $\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$, the initial concentration of lactic acid must decrease by ${10}^{- \text{pH}}$.

The acid dissociation constant, ${K}_{a}$, is defined as

${K}_{a} = \left(\left[\text{A"^(-)] * ["H"_3"O"^(+)])/(["HA}\right]\right)$

which, in your case, is equal to

${K}_{a} = \left({10}^{- \text{pH") * 10^(-"pH"))/(0.10 - 10^(-"pH}}\right)$

${K}_{a} = \left({10}^{- 2 \cdot \text{pH"))/(0.10 - 10^(-"pH}}\right)$

Since you know that

$\text{pH} = 2.44$

you can say that the acid dissociation constant for lactic acid is equal to

${K}_{a} = {10}^{- 2 \cdot 2.44} / \left(0.10 - {10}^{- 2.44}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.4 \cdot {10}^{- 4}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the initial concentration of the acid.