Water potential example | Cell structure and function | AP Biology | Khan Academy

We're told that six identical potato core cubes were isolated from a potato. The initial weight of each cube was recorded. Each cube was then placed in one of six open beakers, each containing a different sucrose solution. The cubes remained in the beakers for 24 hours at a constant temperature of 23 degrees Celsius.

After 24 hours, the cubes were removed from the beakers, blotted, and re-weighed. The percent change in mass due to a net gain or loss of water was calculated for each cube, and the results are shown in the graph to the right.

So, this graph right over here, a straight line is drawn on the graph to help estimate results from other sucrose concentrations not tested. Using the straight line on the graph, calculate the water potential in bars of the potato core cubes at 23 degrees Celsius. Give your answer to one decimal place. So pause this video and see if you can work that out.

All right, so first let's make sure we're understanding what's going on here. There was a potato, we took six cubes from that potato, and we stuck those six cubes into six different sucrose molarity solutions. And so this data point right over here, this was a situation where we took one of the cubes. So this was a sucrose solution. This was a solution, actually, that contained no sucrose.

When we put the cube in that solution, we saw a net gain of mass. It looks like it's about a 22 percent gain in mass, and so that would have happened because water would have flowed into the cube. Now the other extreme right over here, this is a solution that has a lot of sucrose in it, a very high sucrose concentration. When we put a cube in there, we see that the mass of that cube went down by 25 percent, and that would have been because of a net outflow of water from that cube.

So how do we figure out the water potential of the core cubes at 23 degrees Celsius? Well, we could think about a situation where there's some sucrose concentration where if the cube and the sucrose solution have the same water potential, then you're not going to have any net inflow or outflow. And so where do we see that on the graph?

Well, what we'd want to do, we have that line where they're trying to fit the data points. So where would we expect to see zero percent change in mass? So we would go right over here to zero percent change in mass. We would go to the line right over there, and then we see that this line would say that there's a zero percent change in mass.

See if this is 0.4 right over here, this is 0.5 right over here, so this is about a 0.44 molar sucrose solution, 0.44 molar solution. So if we can figure out the water potential of this 0.44 molar sucrose solution, well that's also going to be the water potential of the potato cubes.

Well, how do we do that? Well, we've seen the equations before where we introduced ourselves to the idea of water potential. That water potential, using the Greek letter psi, is going to be equal to the solute potential plus the pressure potential. Now we're dealing with all open containers; we don't have anything that's some piston or something that's pressing down on these containers.

And so because of that, the pressure potential is going to be equal to zero. And so we just have to figure out the solute potential. The solute potential, we have introduced ourselves to this formula in previous videos. It's negative i times c times r times t.

This i right over here, this is our ionization constant. Since we're dealing with sucrose solution, it says, "Okay, if I took sucrose and put it into water, does every one of those sucrose molecules stay one molecule or does it disassociate?" Well, sucrose just disassociates, doesn't dissociate at all; it just stays one molecule. So this would be one. If we're dealing with, say, sodium chloride, each sodium chloride molecule would disassociate into a sodium ion and a chloride ion, and so then this would be 2. But this was 1 for sucrose.

C is the molarity of our solution, and so we estimated that to be 0.44. So let me write this down: our solute water potential is going to be equal to negative 1 times 0.44. I should say, and that's going to be moles. I'll write down all the units: moles per liter times, it's sometimes called the pressure constant, the pressure constant in this context, but this is also the universal gas constant.

If you were doing something like the AP exam, they would give you what this is. So this is 0.0831 liters times bars, all of that over mole Kelvin. If you're used to seeing other values of this, it's probably because they're dealing with other units right over here. But this is the universal gas constant.

And then we have to multiply that times the temperature that we're dealing with in Kelvin. Now it's 23 degrees Celsius; to convert to Kelvin, we just add 273. So 273 plus 23 is going to be 296.

296 Kelvin. And so this is going to be equal to... We have a negative here, and we could look at the units; we have the liter canceling out, liters; moles canceling out with moles; kelvin canceling out with kelvin. So we're going to get something in bars, which makes sense; that is the unit for our water potential.

And then we get the calculator out. So we have 0.44 times 0.0831 times 296. 296 is equal to... And they want us to round our answer to one decimal place. So approximately -10.8, and we already had that negative out front, so negative 10.8 bars, and we're done.