How does the strength of an acid relate to the magnitude of the equilibrium constant #K_a#?

1 Answer
Jan 7, 2017

Answer:

#"HA"(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^+ + "A"^-#

Explanation:

The forward reaction represents the ionization of the acid #"HA"# in aqueous solution. This is an equilibrium reaction, and represented numerically by the expression:

#"K"_a# #=# #(["H"_3"O"^+]["A"^(-)(aq)])/(["HA"(aq)])#,

where #"K"_a# is the so-called #"acid dissociation constant"#.

Strong acids, #"HX" ("X"!="F"), "H"_2"SO"_4, "HClO"_4#, have high values of #"K"_a# so that there tends to be minimal concentration of free acid #"HA"# at equilibrium.

And a base in water competes for the proton:

#"A"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HA"(aq) + "HO"^-#

#"K"_b# #=# #(["HA"]["HO"^(-)(aq)])/(["A"^(-)(aq)])#

#"K"_b# is the so-called #"base association constant"#.

#"K"_a# is generally quoted, because it can be shown that under standard conditions, #p"K"_a+p"K"_b=14#