Le Chatelier’s Principle Lab Part 1: Introduction and Acid-Base Equilibrium
Hi everyone!
Today we will discuss an experiment which shows a rule known as Le Chatelier’s Principle
.
This Principle has to do with how reactions at equilibrium will respond to specific perturbations
or stresses by shifting either to the left (making reactant and consuming product) or
to the right (making product and consuming reactant).
The table below, taken from your lab procedure, shows the shifts that a system at equilibrium
will display, when faced with the specific type of perturbation.
Today, you will see for yourself how various perturbations can be used to shift specific
reactions.
You will apply these stresses to the following reactions: acid-base, dissolution of an insoluble
salt (typically called a solubility equilibrium), and complex ion formation.
In the last part of the experiment, you will combine the first three reactions to observe
the concept of coupled reactions.
Let’s start with part A. Here you will work with an acid base indicator called bromothymol
blue.
You will add or remove certain species to cause the reaction to shift.
The indicator has two chemical forms: HA and A-, each has its own specific color.
You may see two of the colors shown here in your actual experiment.
The equilibrium between the two forms are shown in the equation.
Note that H+ is a colorless species.
How do we shift the reaction?
We must add or remove reagents that are colorless.
The reason is if we add the colorful reagents, that will interfere with the observation.
So we must add a colorless reagents that causes one colored form to switch to the other colored
form as shown in the picture.
A quick note that this picture is just an illustration and may not be what you actually
see in your experiment.
To switch to the color of HA, we must make more HA and consume A- which results in more
intense HA color and less intense A- color.
This means we want a left shift.
Since our only option is to add a colorless reagent, that leaves us with adding H+.
That means you have to think about what reagent will release H+.
It’s more difficult to switch left, to make more A- and consume HA.
One option is to add HA, but as I said earlier, this won’t help us because if we add say
a yellow reagent to shift something to blue, the yellow that we add will mask any blue
color that is formed.
So again, our option is to remove a colorless species, which means removing H+.
The question is how?
We don’t have a special vacuum that will suck out all the H+ ions in the solution.
Well, it turns out we have something that achieves the same goal.
Basically we want to come up with a product-favored reaction that uses H+ as a reactant.
If we can get that reaction to happen, our H+ will be removed and the indicator reaction
will shift to the right.
This concept is known as coupled reaction, where the product of one reaction becomes
the reactant of another reaction.
So what is a reaction that uses H+?
As it happens, in water, there is a reaction called the autoionization of water, shown
here, where a water molecule dissociates to form the H+ and OH- ions.
This reaction has a very small K, called Kw, of 10 to the power of -14.
Since the reaction is reversible, the reverse reaction of H+ combining with OH- must have
a K equal to the reciprocal of Kw or 10 to the power of 14.
This large K means that the reaction goes to completion.
This reaction becomes our tool to remove H+ in the indicator equilibrium.
This means if we add a reagent that produces OH- in solution, that effectively removes
the H+ and causes the indicator reaction to shift right, creating the color of A-.
To summarize this what you will do in Part A:
First start with a bromothymol blue solution and note its color.
Then find a reagent that changes the color of the bromothymol blue.
Last, find a different reagent that shifts the color back to what you saw initially.
This is the end of the first video.
In the next video, I will discuss Parts B and C of the Le Chatelier’s Lab.