Question #29b74

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")))#