Lab Overview Investigation #22: "Determination of an Ionization Constant"

Concepts of the Experiment

In this lab you will determine the ionization constant of a weak acid. "What is the ionization constant," you might ask? A good question and a good place to start. So let's first define and go over some concepts.

Strong Acid – An acid is strong if it completely or almost completely ionizes. An acid's strength has nothing to do with chemical properties such as corrosion however. Remember it only has to do with ionization. This may be expressed in the following generic formula:

Taking this generic equation and solving for the ionization constant, which we shall call the acid ionization constant, we get:

Ka =

OR pKa = -log Ka = pH + log

A strong acid will have a K_a value close to 1 and a negative pK_a value while a weak acid will have a small K_a value <<1. For example, acetic acid has a K_a value of 1.8 x 10^-5 and a pK_a value of +4.76.

You will use these concepts to calculate the K_a of an unknown weak acid. To do this, you will measure out a certain mass of the unknown solid and dissolve it in 300 mL of boiled distilled H₂O. You will then use the given density to determine the volume of the acid. This volume will be V_initial.

Once you know the volume of the unknown acid, you will titrate it with a standardized solution of NaOH. At the endpoint of the titration, the moles of acid will equal the moles of base. In other words, the moles of acid = mol/L NaOH x L NaOH titrated.  Actually though, the previous statement here is wrong.  The endpoint in a titration does not actually mean that the moles of acid = the moles of base.  This is actually the definition of equivalence point.  The endpoint is actually a physical phenomenon associated with a color change at some pH.  For bromophenol blue which we used in the last investigation this color change or endpoint actually occurred at around pH = 4.6.  The endpoint for phenolphthalein is pH = 8.5.  At pH of 7, the moles of acid will equal the moles of base in a neutral solution.  So in the past, we have made it easier for you by assuming that these endpoints were actually very close to the equivalence points.  In this experiment, we will continue to assume this as well.  Are we justified in doing this?  (answer at the bottom of the page)
 

In this experiment, you added some known mass of unknown weak acid to a beaker and titrated with a standardized solution of NaOH.  You should then be able to make a graph like the one above.  From this graph, you can determine the equivalence point and the 1/2 equivalence point.  At the graphically determined equivalence point, the moles of acid = moles of base.  At the endpoint (color change) the moles of acid will almost equal the moles of base.  Compare the two and see how close they are.

To make the graph you will plot your experimentally determined pH values on the ordinate (y-axis) and the total volume of NaOH (V_titration- V_initial) up to each of the pH readings on the abscicca (x-axis).  Once you have your graph made you will use it to determine the equivalence point and the 1/2 equivalence point for the calculations.

When calculating the pH you will have to use some of the equations derived below.  The usefull ones are the equations at the 1/2 equivalence point, the beginning  point and the general K_a equation at any point.  It would help if you used a spreasheet program like excell.  Don't forget to at least show an example of your calculations.
 

You will use a pH meter to determine the pH which will also tell you the hydronium ion concentration, H₃O⁺. This will not be enough however to calculate K_a. We will also need [A^-] and [HA]. So at this point we have three unknown, but only one equation. We need more equations to solve these unknowns so let's consider equations for charge balance and mass balance. These will be:

If you take the charge balance equation and rearrange solving for [A^-] then take this and substitute into both the equations for K_a and mass balance you get:

Ka =

Now take this second equation and substitute it into the equation for K_a.

Ka =

Examining this equation you will see a term for the concentration of [OH^-]. If you are using an acidic solution however, this concentration must be very small and considered negligible. The equation can therefore be rewritten:

Ka =

Examining this equation, your unknowns are [HA]_total and [Na⁺]. You can find these using the same principles we have used throughout titration reactions.

[HA]total =

=

Substituting these into are ongoing large equation gives (general K_a equation):

Ka =

This big mess can thankfully be simplified. At the beginning of the experiment, the V_NaOH is equal to 0 since none of the NaOH has been added. Also, if you compare the concentrations of [HA]_total and [H⁺], the concentration of [H⁺] is negligible in comparison. This big equation therefore simplifies to:

Ka =

The big messy equation can also be simplified at the half-neutralization point, . This simplifies to:

If you look at the volume of titrant which would accompany the difference between the endpoint and the equivalence point, this will usually be very small.  This is especially true when the overall volume of titrant is large.  The small volume difference becomes more negligible with larger volumes of titrant.