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Calculate ψ (psi) instantly — osmotic potential, pressure potential, and total water potential.
Water potential (ψ) = Osmotic potential (ψs) + Pressure potential (ψp). For a solution, ψs = −iCRT, where i is the ionisation constant, C is molar concentration, R = 0.00831 L·MPa/mol·K, and T is temperature in Kelvin. Pure water has a water potential of 0 MPa.
Sucrose does not ionize (i = 1). Ionic compounds like NaCl dissociate (i ≈ 2).
Enter the molarity of the solution (e.g., 0.15 M, 0.3 M, 0.4 M).
Standard room temperature is 25°C. Will be converted to Kelvin (K = °C + 273).
For open containers (beakers), ψp = 0. For turgid cells, ψp > 0. For xylem tension, ψp < 0.
Osmotic Potential: ψs = −iCRT
Total Water Potential: ψ = ψs + ψp
Where R = 0.00831 L·MPa/mol·K (gas constant)
The standard water potential formula (AP Bio & A-Level) is:
Where the osmotic potential formula (also called solute potential) is:
| Symbol | Meaning | Unit | Value / note |
|---|---|---|---|
| ψ | Total water potential | MPa | Negative for solutions |
| ψs | Osmotic (solute) potential | MPa | Always ≤ 0 |
| ψp | Pressure potential (turgor) | MPa | +ve (turgor), −ve (tension) |
| i | Ionisation constant | — | Sucrose = 1, NaCl ≈ 2 |
| C | Molar concentration | mol/L | From experiment or given |
| R | Gas constant (pressure form) | L·MPa/mol·K | 0.00831 |
| T | Absolute temperature | K | °C + 273 |
Caused by dissolved solutes. More solutes = more negative ψs. Drives water into cells.
Turgor pressure pushes water out (+). Tension in xylem creates negative values.
Adhesion of water to soil particles or cell walls. Important in soil water potential.
Height above a reference point. Matters in tall trees (up to +1 MPa difference).
Pure water at standard conditions = 0 MPa. All solutions are negative relative to this.
Problem: Calculate the water potential of a 0.15 M sucrose solution at 25 °C in an open container.
1. Write the formula: ψ = ψs + ψp
2. Identify values: i = 1 (sucrose does not ionise), C = 0.15 mol/L, R = 0.00831, T = 25 + 273 = 298 K, ψp = 0 (open system)
3. Calculate ψs = −(1)(0.15)(0.00831)(298) = −0.37 MPa
4. Calculate ψ = −0.37 + 0 = −0.37 MPa
Interpretation: water moves from a region with ψ > −0.37 MPa into this solution.
Problem: Potato cells placed in a 0.3 M sucrose solution reach equilibrium. Find the water potential of the potato cells.
1. At equilibrium, ψcell = ψsolution
2. Calculate ψsolution: ψs = −(1)(0.3)(0.00831)(298) = −0.74 MPa
3. ψp of solution = 0 (open beaker), so ψsolution = −0.74 MPa
Therefore, ψpotato cells = −0.74 MPa at equilibrium.
Problem: How does temperature affect the water potential of a 0.2 M NaCl solution?
At 20 °C: ψs = −(2)(0.2)(0.00831)(293) = −0.975 MPa
At 37 °C: ψs = −(2)(0.2)(0.00831)(310) = −1.031 MPa
Higher temperature → more negative osmotic potential → stronger driving force for water entry.
In AP Biology and A-Level labs, you calculate water potential experimentally by placing tissue (like potato or zucchini cores) in sucrose solutions and observing mass changes.
| Sucrose concentration (M) | % change in mass | Interpretation |
|---|---|---|
| 0.0 (pure water) | +8% | Water enters cells — solution ψ > cell ψ |
| 0.2 | +3% | Net water entry — still gaining mass |
| 0.4 | 0% | Equilibrium — ψcell = ψsolution |
| 0.6 | −4% | Water leaves cells — solution ψ < cell ψ |
| 0.8 | −9% | Strong plasmolysis — large water loss |
At the equilibrium point (0.4 M, 0% change), you calculate the cell's water potential using ψs = −iCRT with C = 0.4 M, giving ψcell ≈ −0.99 MPa at 25 °C.
A fully turgid plant cell has a high positive ψp (turgor pressure) that partially offsets the negative ψs. A flaccid cell has ψp ≈ 0. A plasmolysed cell can have negative ψp. This is why plants wilt when water is scarce — ψp drops and turgor is lost.
Soil water potential includes matric potential (ψm), which comes from water adhesion to soil particles. Dry soil has a very negative matric potential (−1 MPa to −10 MPa). Roots absorb water because their osmotic potential is more negative than the surrounding soil water potential.
Water always moves by osmosis from a region of higher (less negative) water potential to a region of lower (more negative) water potential. This simple rule governs absorption in roots, transport in xylem, and exchange in leaf cells.
| Curriculum | Formula used | Notation | Notes |
|---|---|---|---|
| AP Biology | ψ = ψs + ψp | ψs for solute, ψp for pressure | R = 0.00831 L·MPa/mol·K |
| A-Level Biology | ψ = ψs + ψp | Same notation | Usually measured in kPa; same principle |
| Advanced / Soil Science | ψ = ψs + ψp + ψm + ψg | Adds matric and gravitational | Used for soil water potential analysis |
Always convert T = °C + 273 before plugging into the formula.
Sucrose = 1, NaCl = 2, CaCl₂ = 3. Forgetting this under-estimates the osmotic effect of ionic solutes.
In a closed or pressurised cell, ψp ≠ 0. An open beaker has ψp = 0.
Osmotic potential is always negative. If you calculate a positive ψs, you missed the negative sign in −iCRT.
Water moves from higher (less negative) to lower (more negative) ψ, not from lower solute to higher solute concentration.