# The venom of biting ants contains formic acid, "HCOOH". What is the "pH" of a "0.0700-M" solution of formic acid?

## As the feedback and solution both indicate, the difference in calculated $\text{pH}$ when solving with the quadratic equation and the assumption that $x$ is small enough differ by only $0.01$ $\text{pH}$ units, so using the $x$-is-small assumption is recommended. K_a = 1.8 * 10^(–4) at ${25}^{\circ} \text{C}$.

Apr 12, 2018

$\text{pH} = 2.450$

#### Explanation:

Formic acid is a weak acid, so you know that it will only partially ionize in aqueous solution to produce formate anions and hydronium cations.

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

Notice that every $1$ mole of formic acid that ionizes produces $1$ mole of formate anions and $1$ mole of hydronium cations.

Now, let's say that $x$ $\text{M}$ is the concentration of formic acid that ionizes. Since formic acid is a weak acid, you know for a fact that

$x \quad \text{M" < "0.0700 M}$

In other words, only a fraction of the initial concentration of the acid will actually ionize. According to the balanced chemical equation, the solution will contain

["HCOO"^(-)] = ["H"_ 3"O"^(+)] = x quad "M"

When $x$ $\text{M}$ of formic acid ionizes, you get $x$ $\text{M}$ of formate anions and $x$ $\text{M}$ of hydronium cations.

So at equilibrium, you can express the concentration of the formate anions and the concentration of the hydronium cations in terms of the concentration of formic acid that ionizes. At equilibrium, the resulting solution will also contain

["HCOOH"] = (0.0700 - x) quad "M"

When $x$ $\text{M}$ of formic acid ionizes, the initial concentration of the acid decreases by $x$ $\text{M}$.

By definition, the acid dissociation constant that describes the ionization equilibrium of formic acid is equal to

${K}_{a} = \left(\left[\text{HCOO"^(-)] * ["H"_ 3"O"^(+)])/(["HCOOH}\right]\right)$

In your case, this will be equal to

${K}_{a} = \frac{x \cdot x}{0.0700 - x}$

which is

$1.8 \cdot {10}^{- 4} = {x}^{2} / \left(0.0700 - x\right)$

Now, you know that using the approximation

$0.0700 - x \approx 0.0700$

is recommended here, so rewrite the expression of the acid dissociation constant as

$1.8 \cdot {10}^{- 4} = {x}^{2} / 0.0700$

This will get you

$x = \sqrt{0.0700 \cdot 1.8 \cdot {10}^{- 4}} = 0.003550$

Since $x$ represents the equilibrium concentration of hydronium cations, you can say that the solution will have

["H"- 3"O"^(+)] = "0.003550 M"

As you know, the $\text{pH}$ of the solution is given by

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

Plug in your value to find

$\text{pH} = - \log \left(0.003550\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.450}}}$

The answer is rounded to three decimal places because you have three sig figs for the initial concentration of formic acid.