Question #62a5a

1 Answer
Dec 3, 2017

Answer:

Here's my take on this.

Explanation:

For starters, I'm not really sure why you need the molarity of the solution if you know the percent ionization of the acid and the total number of molecules of acetic acid.

The acid's percent ionization essentially tells you the number of molecules of acetic acid that give off their acidic proton to produce hydronium cations and acetate anions for every #100# molecules of acetic acid present in the solution.

So if the acid is #1.33%# ionized, that means that out of every #100# molecules of acetic acid present in the solution, #1.33# are ionized.

This implies that if your sample contains #"1,000"# molecules of acetic acid, only #13.3# will ionize to produce hydronium cations and acetate anions in a #1:1# mole ratio.

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

So the number of molecules of acetic acid that ionize will give you the number of hydronium cations.

Keep in mind that the sample given to you doesn't make much physical sense because you can't have fractions of a molecule.

So you can't really say that the solution contains #13.3# hydronium cations because you can't have #0.3# hydronium cations. You can, however, have #"1,000"# moles of acetic acid in the sample, which would give you #13.3# moles of acetic acid that ionize to produce #13.3# moles of hydronium cations.

Alternatively, you can say that in this solution, for every #"10,000"# molecules of acetic acid, #133# are ionized. This means that at equilibrium, a sample that contains #"10,000"# molecules of acetic acid will have

#overbrace("CH"_ 3"COOH"_ ((aq)))^(color(blue)("9867 unionized molecules ")) + "H"_ 2"O"_ ((l)) rightleftharpoons overbrace("CH"_ 3"COO"_ ((aq))^(-))^(color(blue)("133 acetate anions")) + overbrace("H"_ 3"O"_ ((aq))^(+))^(color(blue)("133 hydronium cations"))#