Question

Question #29b74

Answer 1

`"3.2 mL"`

Explanation:

Start by looking up the acid dissociation constant for acetic acid

`K_a = 1.76 * 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.

`"CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)`

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

`["CH"_ 3"COO"^(-)] = ["H"_ 3"O"^(+)]`

You should also know that

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

This implies that

`["H"_3"O"^(+)] = 10^(-"pH")`

In your case, you will have

`["H"_3"O"^(+)] = 10^(-3.0) = 1.0 * 10^(-3)` `"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

`["CH"_ 3"COOH"]_ "equil" = ["CH"_ 3"COOH"]_ 0 - ["H"_3"O"^(+)]`

Here `["CH"_3"COOH"]_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)` `"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

`"DF" = V_"diluted"/V_"stock" implies V_"stock" = V_"diluted"/"DF"`

which, in your case, is equal to

`V_"stock" = "1.0 L"/color(blue)(310.7) = "0.003226 L"`

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

`color(darkgreen)(ul(color(black)(V_"stock" = "3.2 mL")))`