# Lactic acid has one acidic hydrogen. A 0.1 M solution of lactic acid has a pH of 2.44. Calculate Ka for lactic acid?

Dec 27, 2015

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

#### Explanation:

You don't actually need to know the molecular formula for lactic acid in order to solve this problem.

You are told that lactic acid has one acidic proton, which means that you can represent it as $\text{HA}$.

So, you know that the pH of a solution that is $\text{0.10 M}$ lactic acid is equal to $2.44$. As you know, a solution's pH is simply a measure of the concentration of hydronium ions, ${\text{H"_3"O}}^{+}$.

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

This means that you can use the given pH to determine the concentration of hydronium ions in this solution

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

["H"_3"O"^(+)] = 10^(-2.44) = 3.63 * 10^(-3)"M"

As you know, a weak acid does not dissociate completely in aqueous solution to form hydronium ions and the conjugate base of the acid.

Instead, an equilibrium is established between the unprotonated molecules and the two ions.

Use an ICE table to help you determine the acid dissociation constant, ${K}_{a}$, for this weak acid. Keep in mind that the equilibrium concentration of hydronium ions must be equal to $3.63 \cdot {10}^{- 3} \text{M}$, since that corresponds to the pH of the solution.

${\text{ " "HA"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "A}}_{\textrm{\left(a q\right]}}^{-}$

color(purple)("I")" " " " "0.10" " " " " " " " " " " " " " " " " " "0" " " " " " " " " "0
color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " "(+x)
color(purple)("E")" "0.10-x" " " " " " " " " " " " " " " "x" " " " " " "3.63 * 10^(-3)

Since you have

$x = 3.63 \cdot {10}^{- 3}$

you can say that the equilibrium concentrations of the weak acid and of the conjugate base will be

["A"^(-)] = 3.63 * 10^(-3)"M"

["HA"] = 0.10 - 3.63 * 10^(-3) = "0.09637 M"

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

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

Plug in your values to get - I'll leave the acid dissociation constant unitless

${K}_{a} = \frac{3.63 \cdot {10}^{- 3} \cdot 3.63 \cdot {10}^{- 3}}{0.09637} = \textcolor{g r e e n}{1.4 \cdot {10}^{- 4}}$

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

The listed value for lactic acid's acid dissociation constant is

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

so this is an excellent result.