Water potential worked example

A zucchini squash was peeled and cut into six identical cubes. After being weighed, each cube was soaked in a different sucrose solution for 24 hours in an open container and at a constant temperature of 21 degrees Celsius. The cubes were then removed from the sucrose solutions, carefully blotted on paper towels, and weighed again. 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 below. I put the graph to the right, right over here.

A straight line is drawn on the graph to help in estimating results from other sucrose concentrations not tested. Using the straight line on the graph above, calculate the water potential in bars of the zucchini squash at 21 degrees Celsius. Give your answer to one decimal place. So pause this video and see if you can have a go at it.

All right, now let's do this together. So we have our six cubes of zucchini squash, and we have each of these data points for each of those cubes. For example, this data point shows a cube that was put into a solution that contains no sucrose. When it was put into a solution that contains no sucrose, it gained 20 percent mass. That mass would be from the water going into the cube.

So one way to think about it is it got put into a solution, and so you had a net inflow of water into that sucrose cube, enough to increase its mass by 20 percent. This second data point right over here was put into a solution that is 0.20 moles per liter of sucrose, and that also had an increase of mass; but it was a smaller increase of mass. Then we could go all the way to the extreme right over here, where we had a zucchini cube put into a solution that had a lot of sucrose in it, one mole per liter, and there you have a net outflow of water.

It decreased the cube's mass by 30 percent. So how do we figure out the water potential of the zucchini squash? Well, one way to think about it is we could figure out the water potential of the zucchini squash if we put it into a solution that has a certain sucrose concentration. If we have no net inflow or outflow of water in either direction, that means that the zucchini squash and the sucrose solution have the same water potential.

We see that happening—at least we can interpolate it—right over here, because if we just follow that line, it looks like if we were to put our zucchini cubes or zucchini squash cubes into a sucrose solution that is… this looks about, let's see, this would be 0.3, so let's just call that 0.36 molar solution. Well then, it has the same water potential as that solution because you have no net inflow or outflow of water.

So the zucchini squash water potential is going to be the same as the water potential of a solution of sucrose that has 0.36 moles per liter. So we just have to figure out the water potential of that. Well, the water potential formula we know—and we've seen this in other videos—is equal to our solute potential plus our pressure potential. We know that these are open containers, so the pressure potential is going to be zero.

We just need to figure out the solute potential of our sucrose solution with this concentration. Well, we know our solute potential is going to be equal to negative i, where this is the ionization constant for every molecule of sucrose. How does it disassociate when it gets put into water? Well, sucrose doesn't disassociate, so it would just be 1 times our molarity times what's sometimes called the pressure constant, but it's really the same thing as the universal gas constant you might have learned in chemistry, times the temperature in Kelvin.

So this is going to be equal to negative i. I already said that this i is 1 because the sucrose just stays sucrose when you put it in water. So negative 1 times 0.36; I'll write the units moles per liter, times r. If you're doing something like an AP test, they actually give you these formulas, and they even give you r as being 0.0831 liter bars per mole Kelvin, and then times the temperature in Kelvin.

Well, if we're dealing with 21 degrees Celsius, we just add 273 to that, so times 294 Kelvin. Now we can just get our calculator out: negative 1 times 0.36 times 0.0831 times 294 is equal to—if I round to one decimal place, it's going to be negative 8.8.

Negative 8.8, and the units work out as well: moles cancel with moles, liters cancel with liters, Kelvin cancels with Kelvin, and we're just left with negative 8.8 bars, and we are done.