Question #d1319

1 Answer
Aug 18, 2017

Answer:

#K_a = 1.7 * 10^(-4)#

Explanation:

Methanoic acid is a weak acid, which implies that it will only partially ionize in aqueous solution to produce hydronium cations, #"H"_3"O"^(+)#, and methanoate anions, #"HCOO"^(-)#.

The ionization equilibrium can be written as

#"HCOOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCOO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)#

Now, notice that every mole of methanoic acid that ionizes produces #1# mole of methanoate anions and #1# mole of hydronium cations.

This means that, at equilibrium, you will have

#["HCOO"^(-)] = ["H"_3"O"^(+)]#

Moreover, you can say that the equilibrium concentration of the methanoic acid will be equal to

#["HCOOH"] = ["HCOOH"]_ 0 - ["H"_ 3"O"^(+)]#

This basically tells you that with every #1# mole of hydronium cations produced by the ionization of the acid, the initial concentration of the acid, #["HCOOH"]_0#, decreases by #1# mole.

The acid dissociation constant for the ionization of methanoic acid can be written as

#K_a = (["HCOO"^(-)] * ["H"_3"O"^(+)])/(["HCOOH"])#

Now, you know that the #"pH"# of the solution is equal to #2.4#. As you know, you have

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

which implies that

#["H"_3"O"^(+)] = 10^(-"pH")#

At equilibrium, the concentration of the methanoate anions is equal to that of the hydronium cations, so you can say that

#["HCOO"^(-)] = 10^(-"pH")#

The equilibrium concentration of the acid will be equal to--keep in mind that the initial concentration of the acid is #"0.10 M"#

#["HCOOH"] = 0.10 - 10^(-"pH")#

This means that the acid dissociation constant is equal to

#K_a = (10^(-"pH") * 10^(-"pH"))/(0.10 - 10^(-"pH")) = 10^(-2"pH")/(0.10 - 10^(-"pH"))#

Plug in the value you have for the #"pH"# of the solution to find

#K_a = 10^(-2 * 2.4)/(0.10 - 10^(-2.4)) = color(darkgreen)(ul(color(black)(1.7 * 10^(-4))))#

I'll leave the answer rounded to two sig figs.