When strong acids are placed in water, they completely dissociate. That is, all of the acid (HA) separates into protons (H^{+}) and their companion anions (A¯).

In contrast, weak acids placed in aqueous solution do not completely dissociate. The extent to which they do separate is described by the dissociation constant K_{a}:

**K _{a} = ([H^{+}] [A¯]) ÷ [HA]**

The quantities in square brackets are the concentrations of the protons, anions and intact acid (HA) in solution.

K_{a} is useful for calculating the percent of a given weak acid that is dissociated in a solution with a known acidity, or pH.

## The Dissociation Constant Across Equations

Recall that pH is defined as the negative logarithm of the proton concentration in solution, which is the same as 10 raised to the negative power of the proton concentration:

pH = -log_{10}[H^{+}] = 10^{-[H+]}

[H^{+}] = 10^{-pH}

K_{a} and pK_{a} are related in a similar way:

pK_{a} = -log_{10}K_{a} = 10^{-Ka}

K_{a} = 10^{-pKa}

If given the pK_{a} and pH of an acid solution, calculating the percent of the acid that is dissociated is straightforward.

## Sample Dissociation Calculation

A weak acid, HA, has a pK_{a} of 4.756. If the solution pH is 3.85, what percentage of the acid is dissociated?

First, convert pK_{a} to K_{a} and pH to [H+]:

K_{a} = 10^{-4.756} = 1.754 x 10^{-5}

[H^{+}] = 10^{-3.85} = 1.413 x 10^{-4}

Now use the equation K_{a} = ([H^{+}] [A¯]) ÷ [HA], with [H^{+}] = [A¯]:

1.754 x 10^{-5} = [(1.413 x 10^{-4} M)(1.413 x 10^{-4} M)] ÷ [HA]

[HA] = 0.0011375 M

Percent dissociation is therefore given by 1.413 x 10^{-4} ÷ 0.0011375 = 0.1242 = 12.42%.

References

About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.