What concentration of formic acid will result in a solution with "pH" = 1.90? The value of K_a for formic acid is 1.8 * 10^(−4)

Oct 20, 2017

$\text{0.89 M}$

Explanation:

The idea here is that the $\text{pH}$ of the solution will give you the equilibrium concentration of hydronium cations in this formic acid solution.

Formic acid is a weak acid, which means that it only partially ionizes to produce formate anions, the conjugate base of formic acid, and hydronium cations. For the sake of simplicity, I'll use $\text{HA}$ to denote the formic acid and ${\text{A}}^{-}$ to denote the formate anions.

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

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

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

This implies that the concentration of hydronium cations is equal to

["H"_3"O"^(+)] = 10^(-"pH")color(white)(.)"M"

Now, notice that in order for the ionization of the weak acid to produce $1$ mole of hydronium cations and $1$ mole of conjugate base, $1$ mole of formic acid must ionize.

If you take $\left[\text{HA}\right]$ to be the equilibrium concentration of formic acid, you can say that the initial concentration of the acid is equal to

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

This is equivalent to saying that in order to get an equilibrium concentration of $\left[{\text{H"_3"O}}^{+}\right]$ in the solution, the initial concentration of the acid must decrease by $\left[{\text{H"_3"O}}^{+}\right]$.

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

Now, by definition, the acid dissociation constant is equal to

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

Since you know that, at equilibrium, you have

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

you can rewrite the expression you have for the acid dissociation constant as

${K}_{a} = \left(\left[{\text{H"_3"O"^(+)]^2)/(["HA"]_0 - ["H"_3"O}}^{+}\right]\right)$

which is, of course, equivalent to

${K}_{a} = \left({10}^{- \text{pH"))^2/(["HA"]_0 - 10^(-"pH}}\right)$

Rearrange to solve for the initial concentration of the formic acid

$\left[\text{HA"]_0 * K_a = 10^(-2"pH") + 10^(-"pH}\right) \cdot {K}_{a}$

$\left[\text{HA"]_0 = 10^(-2"pH")/K_a + 10^(-"pH}\right)$

Finally, plug in your values to get

["HA"]_ 0 = (10^(-2 * 1.90))/(1.8 * 10^(-4)) + 10^(-1.90) = color(darkgreen)(ul(color(black)("0.89 M")))

The answer is rounded to two sig figs, the number of decimal places you have for the $\text{pH}$ of the solution.

Oct 20, 2017

The concentration is 0.89 mol/L.

Explanation:

Step 1. Calculate $\left[\text{H"_3"O"^"+}\right]$

["H"_3"O"^"+"] = 10^"-pH"color(white)(l) "mol/L" = 10^"-1.9"color(white)(l)"mol/L" = "0.0126 mol/L"

Step 2. Calculate the concentration of formic acid

We can use an ICE table to organize our calculations.

Let $c$ be the initial concentration of the acid.

$\textcolor{w h i t e}{m m m m m m l l} \text{HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m l} c \textcolor{w h i t e}{m m m m m m m} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mll)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m} c - x \textcolor{w h i t e}{m m m m m m} x \textcolor{w h i t e}{m m m} x$

In this problem, $x = \text{0.015 85}$.

So, our ICE table becomes

$\textcolor{w h i t e}{m m m m m m m m l l} \text{HA" + "H"_2"O" ⇌ "H"_3"O"^"+" color(white)(ll)+color(white)(mll) "A"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m m m} c \textcolor{w h i t e}{m m m m m m m l} 0 \textcolor{w h i t e}{m m m m m l} 0$
$\text{C/mol·L"^"-1":color(white)(llm)"-0.0126"color(white)(mmmml)"+0.0126"color(white)(mm)"+0.0126}$
$\text{E/mol·L"^"-1":color(white)(m)c-"0.0126"color(white)(mmmml)"0.0126"color(white)(mmm)"0.0126}$

K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = ("0.0126 × 0.0126")/(c-"0.0126") = 1.8 × 10^"-4"

1.585 × 10^"-4" = 1.8 × 10^"-4"c - 2.268 × 10^"-6"

$158.5 = 180 c + 2.268$

$x = \frac{158.5 + 2.268}{180} = 0.89$

["HA"] = "0.89 mol/L"

Check:

K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = ("0.0126 × 0.0126")/(0.89-0.0126) = 1.8 × 10^"-4"

It checks!