Question #fcf52
1 Answer
Explanation:
The first thing to note here is that the concentration of the acid is said to be
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
Now, hydrofluoric acid,
If you think about this at a molecular level, a
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
By definition, the acid dissociation constant,
#K_a = (["F"^(-)] * ["H"_3"O"^(+)])/(["HF"])#
Now, if you start with
#["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
#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 = (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