# Question 29b74

Jun 5, 2017

$\text{3.2 mL}$

#### Explanation:

Start by looking up the acid dissociation constant for acetic acid

${K}_{a} = 1.76 \cdot {10}^{- 5}$

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

So, acetic acid is a weak acid that only partially ionizes in aqueous solution to produce hydronium cations and acetate anions.

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

As you can see, every $1$ mole of acetic acid that ionzies produces $1$ mole of acetate anions and $1$ mole of hydronium cations.

This means that, at equlibrium, you have

$\left[{\text{CH"_ 3"COO"^(-)] = ["H"_ 3"O}}^{+}\right]$

You should also know that

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))

This implies that

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

In your case, you will have

$\left[{\text{H"_3"O}}^{+}\right] = {10}^{- 3.0} = 1.0 \cdot {10}^{- 3}$ $\text{M}$

Now, notice that the reaction consumes $1$ mole of acetic acid in order to produce $1$ mole of acetate anions and $1$ mole of hydronium cations.

This means that, at equilibrium, the concentration of the acid will be equal to

$\left[{\text{CH"_ 3"COOH"]_ "equil" = ["CH"_ 3"COOH"]_ 0 - ["H"_3"O}}^{+}\right]$

Here ${\left[\text{CH"_3"COOH}\right]}_{0}$ is the initial concentration of the acid.

By definition, the acid dissociation constant will be equal to

K_a = (["CH"_ 3"COO"^(-)] * ["H"_ 3"O"^(+)])/(["CH"_ 3"COOH"]_"equil")

Rearrange to solve for the equilibrium concentration of the acid

["CH"_ 3"COOH"]_ "equil" =(["CH"_ 3"COO"^(-)] * ["H"_ 3"O"^(+)])/K_a

Plug in your values to find

["CH"_ 3"COOH"]_ "equil" = (1.0 * 10^(-3) * 1.0 * 10^(-3))/(1.76 * 10^(-5))

["CH"_ 3"COOH"]_ "equil" = 5.7 * 10^(-2) $\text{M}$

This means that the iniital concentration of the acid was equal to

["CH"_3"COOH"]_0 = 5.7 * 10^(-2) color(white)(.)"M" - 1.0 * 10^(-3)color(white)(.)"M"

["CH"_3"COOH"]_0 = "0.056 M"

Finally, to find the volume of glacial acetic acid needed to make this solution, use the fact that the dilution factor is equal to

"DF" = (17.4 color(red)(cancel(color(black)("mL"))))/(0.056color(red)(cancel(color(black)("mL")))) = color(blue)(310.7)#

This means that you have

$\text{DF" = V_"diluted"/V_"stock" implies V_"stock" = V_"diluted"/"DF}$

which, in your case, is equal to

${V}_{\text{stock" = "1.0 L"/color(blue)(310.7) = "0.003226 L}}$

Expressed in milliliters and rounded to two sig figs, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{V}_{\text{stock" = "3.2 mL}}}}}$