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# Water potential example

AP.BIO:
ENE‑2 (EU)
,
ENE‑2.H (LO)
,
ENE‑2.H.1 (EK)
,
ENE‑2.I (LO)
,
ENE‑2.I.1 (EK)
,
ENE‑2.I.2 (EK)
,
ENE‑2.J (LO)
,
ENE‑2.J.1 (EK)

## Video transcript

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 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 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 a 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 percent gain and mass and so that would have happened because water would have flowed into into the cube now the other extreme right over here this is a solution that has a lot of sucrose it at 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 a net outflow of water from that cube so how do we figure out the water potential of the core cubes 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 and so where would we expect to see 0 percent change in mass so we would go right over here to 0 percent change in mass we would go to the line right over there and then we see that this line would 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 0.4 for molar sucrose solution 0.4 for molar solution so if we can figure out the water potential of this 0.4 for 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 Sai 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 solution it says okay if I took the sucrose and put it into water every one of those sucrose molecules does it stay one molecule or does it disassociate well sucrose just disassociate 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 two but this was one for sucrose see is the molarity of our solution and so we estimated that to be zero point four four so let me write this down our solute water potential is going to be equal to negative one times zero point four zero point four four I should say and that's going to be moles all right on all the units moles per liter times it's sometimes called the pressure contact 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 zero point zero eight three one 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 20 23 degrees Celsius to convert to Kelvin we just add two 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 unit's we have the Leader canceling out liters moles cancel hanging 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 0.4 four times point zero eight three one times 296 296 is equal to and they want us to round our answer to one decimal place so approximately ten point eight and we already had that negative out front so negative ten point eight bars and we're done
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