Water Potential Simulator

Two compartments, LEFT and RIGHT, are separated by a selectively permeable membrane. Each side has its own panel with two sliders, and every value is in megapascals (MPa):

The simulator opens with the left side at Ψ_s = −0.50 and the right side at Ψ_s = −1.20, both with Ψ_p = 0.00.

Under each panel's sliders, a line works out that side's total, for example Ψ_left = Ψ_s + Ψ_p = −0.50 MPa. The same totals appear in large type inside the two compartments.

An arrow across the membrane, labelled water → right or water → left, shows the direction of net water movement. A larger difference in Ψ makes the arrow longer and thicker, up to a difference of 2 MPa. When the two totals match, the arrow is replaced by equilibrium and no net flow.

Below everything, a sentence names the side with the higher Ψ, gives the direction of net movement, and reports ΔΨ, the size of the difference, in MPa. At the default settings it reports that water moves left → right with ΔΨ = 0.70 MPa.

The AP Biology formula sheet gives the water potential equation:

The formula sheet also gives the solute potential of a solution:

With this value of $R$, $-iCRT$ comes out in bars. The simulator uses MPa, and $1 \text{ MPa} = 10 \text{ bar}$, so divide a bar value by 10 before you enter it.

The rule for direction: net water moves from higher Ψ to lower (more negative) Ψ, and it keeps moving until the two water potentials are equal.

Problem: A plant cell has Ψ_s = −0.90 MPa and Ψ_p = +0.30 MPa. It is placed in an open beaker of 0.2 M sucrose at 27 °C. Which way does water move, and what pressure potential will the cell reach at equilibrium? Assume the cell's solute potential does not change.

Step 1: Water potential of the cell. $\Psi = -0.90 + 0.30 = -0.60$ MPa.

Step 2: Solute potential of the sucrose. $T = 27 + 273 = 300$ K and $i = 1$. $\Psi_S = -(1)(0.2)(0.0831)(300) = -4.986$ bar $\approx -0.50$ MPa.

Step 3: Water potential of the beaker. The beaker is open, so $\Psi_P = 0$ and $\Psi = -0.50$ MPa.

Step 4: Direction. The beaker (−0.50 MPa) has a higher Ψ than the cell (−0.60 MPa), so net water moves into the cell. $\Delta\Psi = 0.10$ MPa.

Step 5: Equilibrium. As water enters, the cell wall pushes back and $\Psi_P$ rises. Net flow stops when the cell's Ψ reaches −0.50 MPa: $-0.90 + \Psi_P = -0.50$, so $\Psi_P = +0.40$ MPa.

Check it in the simulator: set the left side (the cell) to Ψ_s = −0.90 and Ψ_p = 0.30, and the right side (the beaker) to Ψ_s = −0.50 and Ψ_p = 0.00. The arrow reads water → left, with ΔΨ = 0.10 MPa. Now raise the left Ψ_p to 0.40. Both sides read −0.50 MPa and the display shows equilibrium. Notice that the solute potentials are still different (−0.90 and −0.50); the pressure potential makes up the difference.

Water potential belongs to Unit 2 (Cell Structure and Function), alongside tonicity and osmoregulation, and both equations are on the formula sheet. Expect calculation questions that combine −iCRT with Ψ = Ψs + Ψp. Lab questions, such as potato cores in sucrose solutions, use the concentration that causes no mass change to find the tissue's water potential.