Question #62a5a
1 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
So if the acid is
This implies that if your sample contains
#"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
Alternatively, you can say that in this solution, for every
#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"))#