Table of Contents
Electric Fields Part 3: Electrophoresis of pigments - testing the model
In last week's lab, you started investigating the electrophoresis of pigments in chromatography paper. You saw that different pigment molecules seem to have different charge to mass ratios that affect how they move in the presence of an electric field. More specifically, these different ratios result in different terminal velocities, which allowed you to use the method to separate a mixture of pigments based on their properties. This is a method commonly used in the study of DNA and other biomolecules.
This week, you will test the model that we derived last week. More specifically, you will estimate the electrophoretic mobility for a pigment of your choosing in a 2g/L solution of baking soda.
Modeling the system
As a refresher, let's consider the system at hand again. If you're comfortable with the model, you can skip this part.
Let us first begin by thinking about the forces that a small particle with charge $q$ suspended in water would experience in the presence of an electric field $\vec{E}$. By Newton's second law, we know that the acceleration due to the electric field will be given by
| $m\vec{a_E} = \vec{F_E} = q\vec{E}.$ | (1) |
But as the particle moves, there will also be a viscous drag force $\vec{f}$ given by the collisions the particle will experience with the water molecules, as well as the electrodynamic interactions between its charge and ions in water. In the case that the velocity and the particle size are small, this drag force will be linearly proportional to the velocity of the particle and the viscosity of water, so that
| $\vec{f} = -\alpha \eta \vec{v}.$ | (2) |
How small and slow is enough?
When we say that the particle is moving slowly and that it is small enough, it means that we are assuming that the system's Reynolds number is small. That is, we are assuming that the inertial forces that the particle exerts on the fluid are not too large compared to the viscous drag forces that the fluid exerts on the particle. When this is the case, the flow of the fluid is said to be laminar instead of turbulent. In this regime, we can apply Stokes' law for drag, which states that the drag force $\vec{f}$ depends linearly on the velocity $\vec{v}$. For a small sphere like those you observed in the demonstration, the geometric constant $\alpha = 6\pi r$, where $r$ is the radius of the sphere.
Here $\alpha$ is some geometric constant, $\eta$ is the viscosity of water, and $\vec{v}$ is the particle's velocity. Newton's second law thus becomes
| $m\vec{a_E} = \vec{F_E} + \vec{f} = q\vec{E} - \alpha \eta \vec{v}$ | (3) |
If we express the acceleration as the rate of change of the velocity, so that $\vec{a_E} = \frac{\text{d}\vec{v}}{\text{d}t}$ and rearrange, the above becomes the following differential equation:
| $\frac{\text{d}\vec{v}}{\text{d}t}=\frac{q}{m}\vec{E}-\frac{\alpha \eta}{m}{\vec{v}}$ | (4) |
Instead of solving it analytically, let's think about the limiting cases:
- At $\mathbf{t = 0}$. Just after the constant electric field is turned on while the particle is stationary, the drag force is zero. The particle will then accelerate to some velocity $\vec{v}$ at a constant rate given by Eq. 1: $\vec{a_E} = \frac{q}{m}\vec{E}$.
- At $\mathbf{0<t<\infty}$. As the particles acquires a velocity, it will start decelerating at a rate given by $-\frac{\alpha \eta}{m}\vec{v}$. The motion here will be governed by the time-dependent Eq. 4.
- At $\mathbf{t \to \infty}$. In the long-time limit, the electric force and the drag force will cancel each other, so that $\vec{a_E} = 0$. At this point,$\frac{\text{d}\vec{v}}{\text{d}t}=\frac{q}{m}\vec{E}-\frac{\alpha \eta}{m}{\vec{v}}=0$. We can then solve for the terminal velocity,
| $\vec{v_t} = \frac{q}{\alpha \eta} \vec{E} = \mu_e \vec{E}$ | (5) |
Show us the math :0 :0 :0
Let us start first with the case where there is only a frictional force $\vec{f} = -\alpha \eta \vec{v}$. Using Newton's second law we then find that $m\frac{\text{d}\vec{v}}{\text{d}t} = -\vec{f}=-\alpha \eta \vec{v}$. Orienting the system in one direction and rearranging we find
| $\frac{\text{d}v}{v}=-\frac{\alpha \eta}{m} \text{ d}t$ |
Integrating,
| $\ln \frac{v}{v_0} = -\frac{\alpha \eta}{m} t $ |
assuming initial time is $t_0 = 0$. From this it follows that
| $\vec{v}(t) = \vec{v_0} e^{-t/\tau}$ |
where $\tau \equiv \frac{m}{\alpha \eta}$ is the time constant of the system (the $1/e$ time, that is, the time at which the particle reaches a velocity given by $v_0/e =0.3678v_0$). We can see that the velocity decreases exponentially toward zero, meaning that friction eventually slows down the particle to a stop in the absence of external forces that keep the particle moving.
What about in the presence of an electric field? In this case, Newton's law states:
| $m\frac{\text{d}\vec{v}}{\text{d}t} = q\vec{E}-\alpha \eta \vec{v}$. |
We see that forces balance out (that is, $\frac{\text{d}\vec{v}}{\text{d}t} = 0$) when it reaches a terminal velocity $\vec{v} \equiv \vec{v}_t = \frac{q\vec{E}}{\alpha \eta}$. We can thus reexpress the law as
| $m\frac{\text{d}\vec{v}}{\text{d}t} = \alpha \eta \vec{v}_t-\alpha \eta \vec{v} = -\alpha \eta \vec{u} $. |
where $\vec{u} = \vec{v}-\vec{v}_t$. Note that $\vec{v}_t$ is constant, so $\frac{\text{d}\vec{v}}{\text{d}t}=\frac{\text{d}\vec{u}}{\text{d}t}$ and the equation becomes $\frac{\text{d}\vec{u}}{\text{d}t} = -\alpha \eta \vec{u}$, which has the solution $\vec{u}(t) = \vec{u}_0 e^{-t/\tau}$ as above. Now recall that $\vec{u} = \vec{v}-\vec{v}_t$, so we have
| $\vec{v}-\vec{v}_t = \vec{u}_0 e^{-t/\tau} = (\vec{v}_0-\vec{v}_t)e^{-t/\tau}$. |
Assuming the particle starts from rest, $\vec{v}_0 = 0$ and we get
| $\vec{v}(t) = \vec{v}_t(1-e^{-t/\tau})$ |
From this we clearly see that in the long time limit, the particle's velocity is given by $\vec{v}_t$.
In Eq. 5, we find that we have introduced the electrophoretic mobility, $\mu_e \equiv \frac{q}{\alpha \eta}$, so that for spheres suspended in a low Reynolds number environment this becomes $\mu_e = \frac{q}{6\pi r \eta}$.
But what, really, is $q$?
When a particle is suspended in water –even deionized water–, it is not in a perfectly electrically neutral environment. In fact, water molecules are constantly dissociating and recombining in the reaction $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$. This means that, when we add, for instance, a negatively charged particle with $q<0$, positive ions in water, such as $\text{H}^+$, will be attracted to it, while negatively charged ions like $\text{OH}^-$ will be repelled by it. Therefore an electric double layer will form around the suspended particle, where just above the particle surface we will find charged ions of opposite sign, which will in turn attract particles of the other polarity. The lengthscale of this double layer is not important for our experiment, but if you're curious, it is known as the Debye length, which is given by $1/\kappa$ in the figure below.
What does this mean for the purposes of electrophoresis? The total charge $q < 0$ will be screened by the positive ions, which is to say, that the total enclosed charge over the region of space that is defined by the particle will be effectively smaller, ie $|q_{\text{effective}}| < |q|$, and thus the field that accelerates it will be felt as weaker. From equation 5 above, this means that the terminal velocity should be smaller in solutions with more ions compared to their terminal velocity in solutions with fewer ions. This translates to a smaller electrophoretic mobility $\mu_e$.
More generally, we can say
| $\vec{v_t} = \frac{q_{\text{eff}}}{\alpha \eta} \vec{E} = \mu_e \vec{E}$ | (6) |
Where $q_{\text{eff}} = q_{\text{eff}}(c)$, is a function of the ion concentration $c$.
 |
| Figure 2: Illustration of the electric double layer for a positively charged sphere in suspension. Taken from Ohshima, H. (2013). Electrical Double Layers. In: Tadros, T. (eds) Encyclopedia of Colloid and Interface Science. Springer, Berlin, Heidelberg. https://doi.org/10.1007/978-3-642-20665-8_15 |
| Figure 3: This video illustrates what happens in suspension for the electric double layer. Link: https://www.youtube.com/watch?v=HDQ8ct4md-8 |
Experimental Procedure
This week we want to test Equation 5 for the baking soda solution we were using last week. That is, we will be estimating the electrophoretic mobility $\mu_e$ for a pigment of your choosing in a 2 g/L baking soda solution.
Remember that you will be using last week's notebook. Clearly mark which part of the notebook belongs to last week's lab, and which one belongs to today's. If for some reason you are with a different group today, let your TA know and make it clear in the notebook.
As a reminder, the electrophoresis procedure will be done on the strips of chromatography paper. Each strip will be placed on the tray so that it lays flat on the surface. Both ends of the strip will go over the well so that, by carefully inserting the graphite in the well, the strip remains pinched and stationary. To prepare this:
- Hold the strip of paper parallel to the surface and over the well.
- Insert the graphite electrode, letting it pull the paper into the well.
- Pull the paper softly so that its surface is flush with the surface of the tray.
- Holding the paper fixed to the table by pressing on it, place the other electrode in its well; it should pull paper from the other end with it inside the well.
- When both electrodes are placed in the wells and the paper is held taut and flush with the surface of the tray, tighten the screws with the bars pressing on the electrodes. These keep the electrodes in place and in electrical contact with the paper, and also serve as a point of contact with the power supply electrodes.
- Using the pipette, wet the paper with the baking soda solution and place the ends of the strip in a baking soda solution in a petri dish.
For images of what it should look like, consult last week's lab.
Determining $\mu_e$
To perform this experiment, you will need to choose a pigment from the mixture you studied last week. They were Blue-1, Red-40, and Yellow-5 and you are free to choose whichever: based on your data from last week, which would be the most convenient or interesting and why?
For the pigment you choose, measure the displacement as a function of time for 3-4 voltages (note that 0V is a valid voltage). Make sure to track more than one particle at each voltage (have them race each other).
- From the plots of displacement vs time, what are the average terminal velocities at the different voltage?
- Plot terminal velocity vs field. What does the slope of this graph tell you?
- Ask groups who chose a different pigment what they get for the electrophoretic mobility. Are their results different from yours when you consider experimental uncertainties?
Testing for significant disagreement
We want you to compare the values for $\mu_e = v_t/E$ for two different pigments. To test whether or not there is a significant difference between the two values, we will be using the $t'$ test. Recall from last quarter that $t'$ is defined as
| $t' = \dfrac{A - B}{\sqrt{(\delta A)^2 + (\delta B)^2}}.$ |
Where $A$ and $B$ are the two values we're comparing, and $\delta A$ and $\delta B$ are their respective uncertainties.
Agreement
If the values are within one uncertainty of each other, it is possible that the difference is due only random chance. We will consider this to constitute agreement: $ |t'| \leq 1$.
Note that agreement might turn into disagreement if more data is taken and the size of the uncertainties shrink. Remember that we can never prove something to be true… we can only say that the current data supports agreement.
Disagreement
If the values are more than three uncertainties away from each other, it is statistically unlikely that the difference is due only random chance. We will consider this to constitute disagreement: $ |t'| \geq 3$.
Inconclusive
If the values are between one and three uncertainties of each other, we cannot say with certainty if the difference is random chance or a real disagreement. We will consider this to be inconclusive: $1 < |t'| < 3$.
- What would be the values $A$ and $B$ in this case?
- How are you determining their uncertainties $\delta A$ and $\delta B$?
Post-lab assignment
Answer the questions/prompts below in a new document (not your lab notebook) and submit that as a PDF to the appropriate assignment on Canvas when you are done. You should write the answers to these questions by yourself (not in a group), though you are welcome to talk to your group mates to ask questions or to discuss.
Conclusions
The conclusion is your interpretation and discussion of your data.
- What do your data tell you?
- How do your data match the model (or models) you were comparing against, or to your expectations in general? (Sometimes this means using the $t^{\prime}$ test, but other times it means making qualitative comparisons.)
- Were you able to estimate uncertainties well, or do you see room to make changes or improvements in the technique?
- Do your results lead to new questions?
- Can you think of other ways to extend or improve the experiment?
In about one or two paragraphs, draw conclusions from the data you collected today. Address both the qualitative and quantitative aspects of the experiment and feel free to use plots, tables or anything else from your notebook to support your words. Don't include throw-away statements like “Looks good” or “Agrees pretty well”; instead, try to be precise.
REMINDER: Your post-lab assignment is due 24 hours after your lab concludes. Submit a single PDF on Canvas.