# Question d1319

Aug 18, 2017

${K}_{a} = 1.7 \cdot {10}^{- 4}$

#### Explanation:

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

The ionization equilibrium can be written as

${\text{HCOOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCOO"_ ((aq))^(-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

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

$\left[{\text{HCOO"^(-)] = ["H"_3"O}}^{+}\right]$

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

$\left[{\text{HCOOH"] = ["HCOOH"]_ 0 - ["H"_ 3"O}}^{+}\right]$

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, ${\left[\text{HCOOH}\right]}_{0}$, decreases by $1$ mole.

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

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

Now, you know that the $\text{pH}$ of the solution is equal to $2.4$. As you know, you have

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

which implies that

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

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

$\left[\text{HCOO"^(-)] = 10^(-"pH}\right)$

The equilibrium concentration of the acid will be equal to--keep in mind that the initial concentration of the acid is $\text{0.10 M}$

$\left[\text{HCOOH"] = 0.10 - 10^(-"pH}\right)$

This means that the acid dissociation constant is equal to

${K}_{a} = \left({10}^{- \text{pH") * 10^(-"pH"))/(0.10 - 10^(-"pH")) = 10^(-2"pH")/(0.10 - 10^(-"pH}}\right)$

Plug in the value you have for the $\text{pH}$ of the solution to find

${K}_{a} = {10}^{- 2 \cdot 2.4} / \left(0.10 - {10}^{- 2.4}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.7 \cdot {10}^{- 4}}}}$

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