Acids and Bases in Equilibrium Reactions

When an acid (let’s use the generic acid HA) is placed in water, it reacts as Brønsted and Lowry explained to make hydronium. Here is the reaction:

HA + H2O ⇄ H3O+1 + A-1

If we write out the equilibrium expression for this reaction we will get:

The water is not included in the equilibrium expression, since it is a liquid and the solvent. 

Note: Reactions of this type (an acid in water) are so common that we have given the constant a special subscript “a”. This is important - any time we see a constant written as Ka we know that the reaction involves an acid added to water. This means that you may face an equilibrium problem in which you are NOT given the reaction but are expected to know what it is.

What all of this means is that you can determine the [H3O+1] (and therefore the pH) when given the initial concentration of the acid and the value of Ka by setting up an ICE table and solving the problem from there.

Here’s an example:

What is the [H3O+1] in a solution of 0.55 M acetic acid? (The Ka for HC2H3O2 = 1.8x10-5)

Our ICE table will look like this:

That will give us the equation

which we can see is a "little x" problem. Solving the problem gives us x=0.0031, so [H3O+1] = 0.0031M.

There are two lessons here:

1. The concentration of acid given in the problem is the “I” value for the acid and we assume that the hydronium and conjugate concentrations are 0

2. The concentration of hydronium at equilibrium will depend on the initial concentration AND on the value of Ka.

Everything we have discussed here apples equally to bases. So, a generic base (B) can be added to water and will react to make hydroxide (instead of hydronium) according to the reaction:

B + H2O → HB+1 + OH-1  

That will give the equilibrium expression:

using Kb for the reaction of a base with water.