Question #48597
1 Answer
Here's what I got.
Explanation:
The idea here is that the acid dissociation constant,
When an acid is added to water, an ionization equilibrium is established
#"HA"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "A"_ ((aq))^(-)#
The position of the equilibrium can be determined by looking at the value of the acid dissociation constant, which is defined as
#K_a = (["H"^(+)] * ["A"^(-)])/(["HA"])#
For
For
You can thus say that the smaller the value of
For really small values of
In other words, you can say that the smaller the value of
In your case, you have
#1.0 * 10^(-3) > 1.0 * 10^(-4) > 1.0 * 10^(-6)#
#color(white)(aa)["H"^(+)]color(white)(aa) > color(white)(aa)["H"^(+)]color(white)(a) > color(white)(aa)["H"^(+)]#
The acid with the smallest
Keep in mind that all three acids will have equilibrium concentrations of hydrogen ions that are lower than the initial concentration of the acid, but the acid with the smallest value of
In fact, you could probably use this approximation
#["HA"]_ "initial" - ["H"^(+)] ~~ ["HA"]_ "initial"#
for all three acids, but this approximation will hold best, i.e. it will be most valid, for the acid with the smallest value of