Question #fcf52

1 Answer
Mar 13, 2016

Answer:

#7.2 * 10^(-4)#

Explanation:

The first thing to note here is that the concentration of the acid is said to be #"0.1 F"#, or #"0.1 formal"#.

Formality is simply a measure of concentration that does not take into account the form in which the solute exists in solution.

In this case, you can say that a #"0.1-F"# solution will be equivalent to a #"0.1-M"# solution.

Now, hydrofluoric acid, #"HF"#, is a weak acid that does not ionize completely in aqueous solution. The problem tells you that in your solution, only #8.13%# of the acid is ionized.

If you think about this at a molecular level, a #8.13%# ionization basically tells you that out of #10000#, or #10^4#, molecules of hydrofluoric acid present in solution, only #813# will ionize, i.e. give off their proton.

The equilibrium reaction for the ionization of hydrofluoric acid looks like this

#color(red)("H")"F"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "F"_text((aq])^(-) + "H"_3"O"_text((aq])^(color(red)(+)#

Notice that every molecule of #"HF"# that ionizes forms one molecule of fluoride anions, #"F"^(-)#, the conjugate base of the acid, and one molecule of hydronium ions, #"H"_3"O"^(+)#.

By definition, the acid dissociation constant, #K_a#, will be equal to

#K_a = (["F"^(-)] * ["H"_3"O"^(+)])/(["HF"])#

Now, if you start with #"0.1 M"# of hydrofluoric acid, and you know that #8.13%# of the molecules ionize, you can calculate the equilibrium concentration of the fluoride and hydronium ions by using a proportion

#["F"^(-)] = ["H"_3"O"^(+)] = 8.13/100 * ["HF"]#

In this case, you will have

#["F"^(-)] = ["H"_3"O"^(+)] = 8.13/100 * "0.1 M" = 8.13 * 10^(-3)"M"#

But remember, if #8.13 * 10^(-3)"M"# is ionized, that must mean that #8.13 * 10^(-3)# of the acid is no longer in molecular form. Therefore, the equilibrium concentration of the acid will be

#overbrace("0.1 M")^(color(purple)("what you start with")) - overbrace(8.13 * 10^(-3)"M")^(color(blue)("what ionizes")) ~~ "0.09187 M"#

Now you're ready to plug in your values into the equation for #K_a#

#K_a = (8.13 * 10^(-3) * 8.13 * 10^(-3))/0.09187 = color(green)(|bar(ul(color(white)(a/a)7.2 * 10^(-4)color(white)(a/a)|)))#

The actual value for the acid dissociation constant of hydrofluoric acid is #7.2 * 10^(-4)#, so this is an excellent result!

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