Question

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

Answer 1

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.