Question

Question #d1319

Answer 1

`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.