# What mass of benzoic acid, C_6H_5COOH, would you dissolve in 400.0 mL of water to produce a solution with a pH = 2.95?

## Equation: C_6H_5COOH+H_2O⇌H_3O^++C_6H_5COO^- Ka=6.3×10^(-5) Website response: My work So far, I haven't found any issues...

Mar 10, 2017

Here's what I got.

#### Explanation:

I'll just solve this problem from scratch so that you can follow my reasoning and see where you made the error.

So, you know that

${\text{C"_ 6"H"_ 5"COOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "C"_ 6"H"_ 5"COO"_ ((aq))^(-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

The acid dissociation constant for benzoic acid is defined as

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["C"_ 6"H"_5"COO"^(-)])/(["C"_6"H"_5"COOH}\right]\right)$

Now, you know that

$\left[{\text{H"_3"O}}^{+}\right] = {10}^{- 2.95}$

["H"_3"O"^(+)] = 1.12 * 10^(-3)"M"

As you can see from the ionization equilibrium, every mole of benzoic acid that ionizes produces $1$ mole of hydronium cations and $1$ mole of benzoate anions.

This means that at equilibrium, you will have

$\left[{\text{H"_3"O"^(+)] = ["C"_6"H"_5"COO}}^{-}\right]$

Consequently, you can say that

["C"_6"H"_5"COO"^(-)] = 1.12 * 10^(-3)"M"

Rearrange the equation you have for ${K}_{a}$ to find the equilibrium concentration of the acid

["C"_6"H"_5"COOH"] = (["H"_3"O"^(+)] * ["C"_6"H"_5"COO"^(-)])/K_a

Plug in your values to find

["C"_6"H"_5"COOH"] = (1.12 * 10^(-3))^2/(6.3 * 10^(-5)) = 1.99 * 10^(-2)"M"

Now, you know that this is how much benzoic acid remains in solution after $1.12 \cdot {10}^{- 3} \text{M}$ has ionized to produced hydronium cations and benzoate anions.

This means that the initial concentration of the acid was

["C"_6"H"_5"COOH"]_0 = overbrace(1.99 * 10^(-2)"M")^(color(blue)("what remains unionized")) + overbrace(1.12 * 10^(-3)"M")^(color(purple)("what ionizes"))

["C"_6"H"_5"COOH"]_0 = 2.10 * 10^(-2)"M"

You can assume that the volume of the solution is equal to that of water

400.0 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.400 L"

which means that this solution will contain

0.400 color(red)(cancel(color(black)("L"))) * (2.10 * 10^(-2)color(white)(.)"moles C"_6"H"_5"COOH")/(1color(red)(cancel(color(black)("L solution"))))

$= 8.40 \cdot {10}^{- 3} \textcolor{w h i t e}{.} \text{moles C"_6"H"_5"COOH}$

To find the mass of benzoic acid that will contain this number of moles, use the compound's molar mass

$8.40 \cdot {10}^{- 3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles C"_6"H"_5"COOH"))) * "122.12 g"/(1color(red)(cancel(color(black)("mole C"_6"H"_5"COOH")))) = color(darkgreen)(ul(color(black)("1.0 g}}}}$

The answer is rounded to two sig figs.

As a final note, it's worth pointing out that only one of your proposed solutions was rounded to two significant figures, the others were rounded to three sig figs!

$1.08 , 1.10 , 1.06 , 1.07 \to$ three sig figs

Moreover, I'm not sure where $\text{0.0018 M}$ in

$\text{0.0199 M " + " 0.0018 M}$

came from. The concentration of the hydronium cations was actually $\text{0.00112 M}$, so this is the error that you are looking for.