Question #38939

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

#"C"_ 2"H"_ 5"OCOOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "C"_ 2"H"_ 5"OCOO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)#

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# #"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# #"M"# of lactate anions and #x# #"M"# of hydronium cations, the initial concentration of the acid must decrease by #x# #"M"#.

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

#K_a = (["C"_ 2"H"_ 5"OCOO"^(-)] * ["H"_ 3"O"^(+)])/(["C"_ 2"H"_ 5"OCOOH"])#

This will be equivalent to

#8.4 * 10^(-4) = (x * x)/(0.06 - x)#

#8.4 * 10^(-4) = x^2/(0.06 - x)#

Rearrange to quadratic equation form

#x^2 + 8.4 * 10^(-4) * x - 0.06 * 8.4 * 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# #"M"# represents the equilibrium concentration of the hydronium cations, you can say that you have

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

Consequently, the #"pH"# of the solution, which is calculated using the equation

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

will be equal to

#"pH" = - log(0.0066917) = color(darkgreen)(ul(color(black)(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.