# Lactic acid HC3H5O3 has one acidic hydrogen. A 0.10 M solution of lactic acid has the concentration of hydronium ion of 0.00363 M. Calculate Ka for lactic acid?

Dec 8, 2015

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

#### Explanation:

All you have to do in order to find the value of the acid dissociation constant, ${K}_{a}$, is to write a balanced chemical equation for the equilibrium that is established when lactic acid, ${\text{HC"_3"H"_5"O}}_{3}$, is added to water.

Being a weak acid, lactic acid will not dissociate completely to form lactate, ${\text{C"_3"H"_5"O}}_{3}^{-}$, its conjugate base, and hydronium ions, ${\text{H"_3"O}}^{+}$, as you can see from the difference between the concentration of the acid and that of the hydronium ions.

Use an ICE table to help you find the equilibrium concentration of the lactic acid

${\text{ ""HC"_3"H"_5"O"_text(3(aq]) + "H"_2"O"_text((l]) rightleftharpoons "C"_3"H"_5"O"_text(3(aq])^(-) + "H"_3"O}}_{\textrm{\left(a q\right]}}^{+}$

color(purple)("I")" " " " " "0.10" " " " " " " " " " " " " " " " " "0" " " " " " " "0
color(purple)("C")" " " "(-x)" " " " " " " " " " " " " " " "(+x)" " " "(+x)
color(purple)("E")" " " "0.10-x" " " " " " " " " " " " " " " "x" " " " " " " "x

By definition, the acid dissociation constant will be equal to

${K}_{a} = \left(\left[{\text{C"_3"H"_5"O"_3^(-)] * ["H"_3"O"^(+)])/(["HC"_3"H"_5"O}}_{3}\right]\right)$

You know that the equilibrium concentration of hydronium ions is

["H"_3"O"^(+)] = x = "0.00363 M"

This means that the equilibrium concentration of the lactic acid will be

$\left[{\text{HC"_3"H"_5"O}}_{3}\right] = 0.10 - x$

["HC"_3"H"_5"O"_3] = "0.10 M" - "0.00363 M" = "0.09637 M"

This means that the value of the acid dissociation constant will be

${K}_{a} = \frac{0.00363 \cdot 0.00363}{0.09637} = 1.367 \cdot {10}^{- 4}$

Rounded to two sig figs, the number of sig figs you have for the molarity of the lactic acid solution, the answer will be

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