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

Jan 4, 2017

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

#### Explanation:

${\text{HA"(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^+ + "A}}^{-}$

For a $\text{strong acid}$ this equilibrium lies to the right as written.

For a $\text{weak acid}$ this equilibrium lies to the left.

Strong acids, ${\text{HX" ("X"!="F"), "H"_2"SO"_4, "HClO}}_{4}$ when dissolved in water are stoichiometric in ${\text{H"_3"O}}^{+}$ (${\text{H"_2"SO}}_{4}$ gives more than one equiv of ${\text{H"_3"O}}^{+}$, why?).

For weak acids, ${\text{HF", "H"_3"PO"_4, "H"_3"CCO"_2"H", "NH}}_{4}^{+} e t c .$, 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:

$\text{NH"_3(aq) + "H"_2"O"(l) rightleftharpoons "NH"_4^+ + ""^(-)"OH}$

The equilibrium lies to the left.