How is the equilibrium reaction between water and strong acids, and water with weak acids, represented?

1 Answer
Jan 4, 2017

Answer:

Consider the equilibrium reaction in which an acid participates in water:

Explanation:

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

For a #"strong acid"# this equilibrium lies to the right as written.

For a #"weak acid"# this equilibrium lies to the left.

Strong acids, #"HX" ("X"!="F"), "H"_2"SO"_4, "HClO"_4# when dissolved in water are stoichiometric in #"H"_3"O"^+# (#"H"_2"SO"_4# gives more than one equiv of #"H"_3"O"^+#, why?).

For weak acids, #"HF", "H"_3"PO"_4, "H"_3"CCO"_2"H", "NH"_4^+ etc.#, the equilibrium lies somewhat to the left, and at equilibrium, significant quantities of the parent acid remain.

As for weak bases, the best example is ammonia:

#"NH"_3(aq) + "H"_2"O"(l) rightleftharpoons "NH"_4^+ + ""^(-)"OH"#

The equilibrium lies to the left.