# Question 38939

Feb 15, 2018

$\text{pH} = 2.2$

#### Explanation:

Lactic acid will only partially ionize in aqueous solution to produce lactate anions and hydronium cations as described by the equilibrium equation

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

Notice that for every mole of lactic acid that ionizes, you get $1$ mole of lactate anions and $1$ mole of hydronium cations.

This means that if you take $x$ $\text{M}$ to be the concentration of the acid that ionizes, you can say that, at equilibrium, the solution will contain

["C"_ 2"H"_ 5"OCOO"^(-)] = ["H"_ 3"O"^(+)] = x quad "M"

["C"_ 2"H"_ 5"OCOOH"] = (0.06 - x) quad "M"

This basically means that in order for the reaction to produce $x$ $\text{M}$ of lactate anions and $x$ $\text{M}$ of hydronium cations, the initial concentration of the acid must decrease by $x$ $\text{M}$.

By definition, the acid dissociation constant for this reaction looks like this

${K}_{a} = \left(\left[\text{C"_ 2"H"_ 5"OCOO"^(-)] * ["H"_ 3"O"^(+)])/(["C"_ 2"H"_ 5"OCOOH}\right]\right)$

This will be equivalent to

$8.4 \cdot {10}^{- 4} = \frac{x \cdot x}{0.06 - x}$

$8.4 \cdot {10}^{- 4} = {x}^{2} / \left(0.06 - x\right)$

${x}^{2} + 8.4 \cdot {10}^{- 4} \cdot x - 0.06 \cdot 8.4 \cdot {10}^{- 4} = 0$

This quadratic equation will produce two solutions, one positive and one negative. Since $x$ represents concentration, you can discard the negative value to get

$x = 0.0066917$

Since $x$ $\text{M}$ represents the equilibrium concentration of the hydronium cations, you can say that you have

["H"_3"O"^(+)] = "0.0066917 M"

Consequently, the $\text{pH}$ of the solution, which is calculated using the equation

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

will be equal to

$\text{pH} = - \log \left(0.0066917\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.2}}}$

The answer must be rounded to one decimal place, the number of significant figure you have for the initial concentration of the acid.