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### Course: Biology library>Unit 11

Lesson 1: Cell membranes

# Water potential example

Example calculating the water potential of potato squares based on placing them in various concentrations of sucrose solutions.

## Want to join the conversation?

• That lesson comes out of absolutely nowhere.
Perhaps the course was updated and it all got taken out except for that practical question part?

On the biology course up to this point 'water potential' has never even been mentioned and now suddenly a video about solving a question on it and full of 'as we have talked before' when even the theme is completely new how much less all the supposedly already-known equations, symbols and concept.
• I know this comment was a year ago but, for anyone else having the same issue, I found the video explaining water potential on YouTube: https://youtu.be/O_eFNOz5WtY. Not sure why it's not in this playlist, though.
• why did he choose 0% change as the point of determining the molarity?
• Im confused as well so forgive me if I end up being wrong but I tried to think about it like this: the molarity is supposed to represent the molarity of all the potatoes. it doesn't make sense to use the molarity of a potato that changed mass, because a change in mass means that water went in or out, which means that its molarity changed in the beaker. so that means that in the beakers, all the potatoes have the same molarity because they are in the same sucrose solution, so they will go through different degrees of mass changes to gain an equilibrium of concentration in the soln. we need to find the potato with the molarity that had no mass change because that represents the molarity of all the potatoes after changing mass; in other words, after adjusting their molarities.
• If a student cut three potato pieces weighing 10.0g each and placed each in a different beaker containing different salt concentrations. Beaker A has 5% salt, Beaker B has 20% salt, and beaker C just has distilled water. The physiological concentration of the potato cell is 4.6%.
After 2 hours, the student removed one of the potato pieces form one of the beakers and weighed it.
The potato piece weighed 12.2g. How do I calculate which beaker was this piece taken from?
• there's no calculation involved here. you can immediately see that beakers A and B are more concentrated than the potato (leading to water osmosizing OUT of the cube, and losing mass), whereas beaker C is LESS concentrated (leading to water osmoizing INTO the cube). since the new mass of the cube is higher than the starting mass, it must be C
• If there was a pressure potential, would it be provided in the problem? Can I assume that problems without indication of container status are open? If not, how would pressure potential be calculated?
• How do you calculate water potential of the cytoplasm?
• The label on the y-axis of the graph says "% change in Mass of Potato Cubes." How can you have a negative % change?
• A % change indicates what percentage of the potato's mass has changed. The change can be positive or negative because the potato can gain or lose mass.
• Why did we calculate and use the % change in mass as the dependent variable rather than using mass difference
• why students calculate the percentage change in mass
• what is hypotonic in a cell
• I am getting my terminology mixed up. What are the differences between tonicity, water potential, solute potential, and osmolarity? Are there different calculations for each one?

## Video transcript

- [Instructor] 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 reweighed. 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 just make sure we're understanding what's going on here. So, 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 the situation where we took one of the cubes, so, this was a sucrose solution. This was a solution, actually, that contained no sucrose, and so, when we put the cube in that solution, we saw a net gain of mass. It looks like it's about 22% gain in mass, and so, that would have happened because water would have flowed into the cube. Now, at the other extreme, right over here, this is a solution that has a lot of sucrose. It had a very high sucrose concentration, and when we put a cube in there, we see that the mass of that cube went down by 25%, and that would have been because of the 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 wanna do, we have that line where they're trying to fit the data points, and so, where would we expect to see 0% change in mass? So, we would go right over here to 0% 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 0% 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 equation 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. So, 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. This is, since we're dealing with sucrose solutions, it says okay, if I took sucrose and put it into water, every one of those sucrose molecules, does it stay one molecule, or does it disassociate? Well, sucrose doesn't disassociate at all. It just stays one molecule, so, this would be one. If we were 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 two, but this was one 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 one times 0.4, 0.44, I should say, and that's going to be moles. I'll write out all the units. Moles per liter times, it's sometimes called the pressure constant in this context, but this is also the universal gas constant, and 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 to 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 0.831 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.