fluid dynamicshome-experimentsurface-tension
I recently did an experiment.
I have a cup of tap water, and a cup of tap water mixed with soap.
I used an eyedropper to slowly drip drops of tap water onto a penny.
Then I cleaned the penny, and dripped soapy water onto the same side of the penny.
My results are:
Tap water: 32 drops
Soapy water: 20 drops
Can someone tell me why that is?
Did the soap in the soapy water break the surface tension?
What happened?
Surface tension occurs because water molecules attract on another. That means that water prefers to form compact shapes with little surface area. Creating a large, extended area, as you do in a soap bubble, is actually opposed by surface tension. This is why you can't get stable bubbles with pure water: the bubble wants to collapse into a compact shape.
Soap changes the situation. Soap molecules have one end that is attracted to water molecules (the hydrophilic end); the other is not (the hydrophobic end). In water, the soap molecules will try to arrange themselves so their hydrophilic ends are in the bulk of the water but the hydrophobic ends are at the surface of the water (or clustered together to minimize contact with the water - these clusters are called micelles). This means it is no longer so costly for the water to form large surfaces, because the hydrophobic soap ends are the parts that are exposed at the surface instead of the water itself. In a very soapy solution, it's easy to form bubbles: you will have two layers of soap molecules, one on the inside and one on the outside. In between, a layer of self-attracting water molecules will hold the bubble together.
We have to make a few assumptions here. The first would be that the soap solution mixes with the water evenly and instantaneously in the capillary - this would be a good approximation, as it's a fairly small volume in the tube, on a similar order of magnitude to the volume of the drop. The next is the form of the meniscus:
Please excuse my crude diagram! You can see that the meniscus is approximated to have a flat bottom, with the sloped sides making an angle $\theta$ - this is symmetrical. The difference in the liquid height from the surface of the body to the meniscus will be $h$. The capillary will have a radius of $r$.
We can take the difference in the surface free energy by changing the height as
$2\pi rh(\gamma_{sl}-\gamma_{sg})$;
$\gamma_{sl}$ is the surface free energy of the solid-liquid interface, and $\gamma_{sg}$ is the surface free energy of the solid-gas interface.
Then, we also take the difference in the gravitational potential energy in changing the height of the liquid:
$\pi r^2h\rho\times\frac{h}{2}g$,
where $\rho$ is the density of the liquid, and we take the centre of mass of the liquid in the capillary to be at $h/2$.
So, the free energy of the meniscus is given by
$\frac{\pi}{2}r^2h^2\rho g+2\pi rh(\gamma_{sl}-\gamma_{sg})$.
We have a stable meniscus when this is minimised, so we differentiate it with respect to $h$, and set that equal to zero, giving
$h=\frac{-2(\gamma_{sl}-\gamma_{sg})}{\rho gr}$.
We can then substitute in the Young Equation, finally giving
$h=\frac{2\gamma_{lg}cos\theta}{\rho gr}$.
The surface free energy of the liquid-gas interface, $\gamma_{lg}$, will be determined by the soap, which would also alter $\rho$. Also, note that $\theta$ can be negative - in this case the level in the capillary would fall, which would be the case with soap using the numbers you gave.
With these assumptions, the water level would change near instantaneously - though it may require some mixing.
Best Answer
When you put a drop of liquid on a solid surface the surface of the liquid will meet the solid at a characteristic angle called the contact angle.
The contact angle is measured through the liquid so a high contact angle means the drop tends to ball up while a low contact angle means it tends to spread out. I've labelled the contact angle as $\theta$. I don't know what the contact angle of water ona coin is, but I'm guessing it's quite high so the drop would look a bit like the drawing on the left.
When you add a surfactant this usually lowers the contact angle, so the drop will look more like the drawing on the right.
Now suppose you keep adding water until the drop covers the whole coin. The result will be a magnified version of the diagram above:
I've kept the contact angles the same as they were in the first diagram, and it should be immediately obvious that the lower contact angle diagram on the right doesn't contain as much water as on the left. That's why you were able to add more drops of pure water than of soapy water.
In practice the difference wouldn't be as big as my diagram suggests because the weight of the drop will deform it. So the pure water drop shown on the left would squish down a bit and this will reduce its volume. However even taking this into account the higher contact angle drop will contain more water than the lower contact angle drop.