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?

1 Answer
Dec 8, 2015

Answer:

#1.4 * 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, #"HC"_3"H"_5"O"_3#, is added to water.

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

http://ceaccp.oxfordjournals.org/content/6/3/128/F2.expansion

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

#" ""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"_text((aq])^(+)#

#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 = ( ["C"_3"H"_5"O"_3^(-)] * ["H"_3"O"^(+)])/(["HC"_3"H"_5"O"_3])#

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

#["HC"_3"H"_5"O"_3] = 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 = ( 0.00363 * 0.00363)/(0.09637) = 1.367 * 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 = color(green)(1.4 * 10^(-4))#