Lab Overview Investigation #23: "The NH3/NH4Cl Buffer System"

Concepts of the Experiment

In this lab you will study a basic buffer solution. You will of course wonder why you should do this. The answer is of course because your grade depends upon it but also because buffer solutions are extremely important to us in both a biological and practical sense. Just imagine how much beer you are allowed to drink because the pH of your blood is buffered. If your blood and other important systems in your body were not buffered from sudden pH changes we would all be in a lot of trouble and probably dead.

Solutions are almost never exactly at a pH of 7. The water that we drink from the faucet around here has a pH of about 5. The reason for this is not necessarily that the city treats the water with chlorine acids, but that the water around here contains a lot of CO₂. That is why you boiled the water in the last experiment. You wanted to get rid of the CO₂ so the solution would be close to 7.

When a solution is at pH = 7, this means that the number of moles of acid = the number of moles of base = 1 x 10^-7 mol/L of [H⁺]. Pretty cool huh. But wait, there is more:
 

In the movie Aliens, the blood of the aliens was an extremely corrosive acid.  Could such an animal actually exist?  (Answer at the bottom)

The buffer that you will use in this lab is the NH₃/NH₄Cl system. The buffer in this case acts in the following manner:

1. When an acid is added:

2. When a base is added:

From these two expressions you can obtain equations for K_a and K_b. How do K_a and K_b relate to each other? (answer at the bottom of the page) By rearranging these equations you can obtain the Henderson-Hasselbach equation:

pH = pKa – log

You need to be aware of how the concentrations of the salt, [NH₄⁺], and the base, [NH₃], affect the pH.
Consider that the ratio, [NH₄⁺]/[NH₃] is equal to one.
That is, the concentration of the salt and base are equal. The log of 1 is 0 so in this case, the pH = pK_a which is 9.26. Now consider the following situations:

1. [NH₄⁺]/[NH₃] changes by the addition of a acid. Looking above you will find that when an acid is added to your buffer, the [NH₄⁺] concentration will increase. So when the [NH₄⁺] concentration increases, the [NH₄⁺]/[NH₃] ratio will also increase. When this ratio increases, the –log [NH₄⁺]/[NH₃] value also increases resulting in a lower pH value.

2. [NH₄⁺]/[NH₃] changes by the addition of a base. When a base is added to our buffer the [NH₃] concentration increases. When this happens, the [NH₄⁺]/[NH₃] ratio decreases. Similarly, the –log [NH₄⁺]/[NH₃] changes and the value will actually turn into a positive resulting into a higher pH value.

All of this makes sense, but how strong is this buffering capability. Buffers are considered to be stable in the range, pH = pK_a 1. But we want to know more. We want to know the buffering capacity which is defined as the number of moles of an added acid required to change the pH of one litter of a buffer solution by one pH unit.

BC =

This is what you will determine in this experiment.

You will make several different buffer solutions and measure the pH for each. You will then plot the moles of acid added vs. pH. From the graph, you will be able to obtain the buffer capacity between any two points. The buffer capacity will equal the slope between those two points.

Above is an example of a table you should have after your lab.  I have made one modification to the table however.  I have added a column for Dmol HCl.  You will see why shortly.

Lets look at Solution 1 first.  In this solution you placed 10 mL of a 0.20 M solution of NH₃ (Ammonia) and 10.0 mL of water.  So in essence all you did was dilute the solution.  Use M₁V₁=M₂V₂ to find [NH₃].  We are told to neglect ionization.  So this concentration is therefore:

Because we are neglecting ionization and there is no acid/base reaction, there is no formation of [NH₄⁺].  The moles of acid added so far are 0 and we cannot yet calculate Dmol HCl/DpH.  The table now looks like:
 

Now we need to look at Solution 2 and this will be where things get a little more interesting.  In Solution 2 you added some acid so we must consider this in our calculations.  How will the acid affect our concentration of NH₃?  If we add a strong acid to a base the two will react with each other.  In this case, all of the HCl reacts with the NH₃ which is in excess.  Remember from above:

To calculate the concentrations of ammonia and ammonium, lets first calculate the number of moles of acid in this reaction that will react with the ammonia.  This will simply be 0.20 M x 0.0005 L = 0.0001 moles of HCl.  Now calculate the number of moles of NH₃ = 0.20 M x 0.010 L = 0.002 moles.  To calculate the number of moles of ammonium (NH₄⁺) we must first understand that all of the acid will react with the ammonia which is in excess.  This means that the above reaction goes to completion.  By completion we mean that all of the reactants continue to react until one of them runs out and cannot react further.  So in this case, all of the moles of HCl will react with the moles of NH₃ .  So let us calculate the number of moles of ammonia left over after this reaction.

Similarly, the moles of ammonium can be calculated.  Remember that the reaction above is 1:1.

Now you can find the molarity of the buffer species by simply dividing by the total volume which is 20 mL or 0.020 L.  The table can then be updated to:
 

Notice that I also went ahead and calculated the change in pH, the change in moles of HCl and the buffer capacity.  Now repeat this for Solution 3 on your own and you should get:
 

Note that in the column, moles of HCl added, this is a running total.  You will use this column along with the pH column to make your graph.

1.  Yes, there are fish that live at such extremely cold temperatures that in order to keep their blood from freezing or becoming too thick to move about in their cardiovascular system they essentially have antifreeze for blood.  I wouldn't recommend eating any of them.

2.  K_w = K_aK_b or K_w = [H⁺][OH^-]