What is the water potential?
Enter a solution’s concentration, its van ’t Hoff factor (ionization constant) and the temperature to get the solute potential Ψs = −iCRT, then the pressure potential of the cell. The calculator adds them (with gravity and matric potentials if you have them) to give the water potential in megapascals and bars.
- Answer
- Ψ = −0.231215 MPa (−2.31215 bar)
Ψ = −0.231215 MPa (−2.31215 bar).
- Water potential Ψ (MPa)
- −0.231215
- Water potential Ψ (bar)
- −2.31215
- Solute potential Ψs (MPa)
- −0.731215
- Solute potential Ψs (bar)
- −7.31215
Answer: Ψ = −0.231215 MPa (−2.31215 bar). Ψ = −0.231215 MPa (−2.31215 bar).
How to calculate
Finds the water potential Ψ of a plant cell or solution as the sum of the solute, pressure, gravity and matric potentials, with the solute potential from Ψs = −iCRT, in megapascals and bars.
Example with the default inputs (Solute potential from Concentration, van ’t Hoff factor i 1, Molar concentration C (mol/L) 0.3, Temperature 20 °C, Pressure potential Ψp 0.5, Potentials in MPa): Ψ = −0.231215 MPa (−2.31215 bar).
Method: Ψ = Ψs + Ψp + Ψg + Ψm; Ψs = −i × C × R × T with R = 0.00831446261815324 L·MPa/(mol·K) and T in kelvins; 1 MPa = 10 bar.
- An ideal dilute solution: Ψs = −iCRT, with i the particles per formula unit (a whole number for full ionization).
- Pure water at atmospheric pressure, at the reference height, is 0 MPa; water moves from higher to lower water potential.
Worked examples
Each example is checked against the calculator on every build.
- Solute potential from Concentration, van ’t Hoff factor i 1, Molar concentration C (mol/L) 1, Temperature 22 °C, Pressure potential Ψp 0, Potentials in bar gives Solute potential Ψs (bar) −24.5401, Water potential Ψ (MPa) −2.45401, Answer Ψ = −24.5401 bar (−2.45401 MPa).Source: OpenStax, Biology 2e, §30.5 Transport of Water and Solutes in Plants (Ψ = Ψs + Ψp + Ψg + Ψm; Ψs = −MiRT; pure water 0 MPa), https://openstax.org/books/biology-2e/pages/30-5-transport-of-water-and-solutes-in-plants; NIST, CODATA value: molar gas constant R = 8.314 462 618… J/(mol·K), exact, https://physics.nist.gov/cgi-bin/cuu/Value?r
- Solute potential from Concentration, van ’t Hoff factor i 2, Molar concentration C (mol/L) 0.15, Temperature 25 °C, Pressure potential Ψp 0.3, Potentials in MPa gives Solute potential Ψs (MPa) −0.743687, Water potential Ψ (MPa) −0.443687.Source: OpenStax, Biology 2e, §30.5 Transport of Water and Solutes in Plants (Ψ = Ψs + Ψp + Ψg + Ψm; Ψs = −MiRT; pure water 0 MPa), https://openstax.org/books/biology-2e/pages/30-5-transport-of-water-and-solutes-in-plants; NIST, CODATA value: molar gas constant R = 8.314 462 618… J/(mol·K), exact, https://physics.nist.gov/cgi-bin/cuu/Value?r
- Solute potential from A value, Solute potential Ψs -0.7, Pressure potential Ψp 0.6, Potentials in MPa gives Water potential Ψ (MPa) −0.1, Water potential Ψ (bar) −1, Answer Ψ = −0.1 MPa (−1 bar).Source: OpenStax, Biology 2e, §30.5 Transport of Water and Solutes in Plants (Ψ = Ψs + Ψp + Ψg + Ψm; Ψs = −MiRT; pure water 0 MPa), https://openstax.org/books/biology-2e/pages/30-5-transport-of-water-and-solutes-in-plants
- Solute potential from Concentration, van ’t Hoff factor i 1, Molar concentration C (mol/L) 0, Temperature 20 °C, Pressure potential Ψp 0, Potentials in MPa gives Water potential Ψ (MPa) 0, Answer Ψ = 0 MPa (0 bar).Source: OpenStax, Biology 2e, §30.5 Transport of Water and Solutes in Plants (Ψ = Ψs + Ψp + Ψg + Ψm; Ψs = −MiRT; pure water 0 MPa), https://openstax.org/books/biology-2e/pages/30-5-transport-of-water-and-solutes-in-plants: pure water’s potential is 0 MPa by definition
How it works
Water potential (OpenStax Biology 2e §30.5).
Ψ = Ψs + Ψp + Ψg + Ψm
- Ψs, the solute potential: from the concentration, Ψs = −i × C × R × T, or typed (0 or below).
- Ψp, the pressure potential: typed, either sign.
- Ψg (gravity) and Ψm (matric): optional; an empty box counts as 0. Ψm is 0 or below.
Solute potential. C in mol/L, T in kelvins (°C + 273.15; °F converts exactly), i the particles per formula unit, and R = 8.31446261815324 J/(mol·K) exactly (NIST), which is 0.00831446261815324 L·MPa/(mol·K). Pure water (C = 0) has Ψs = 0.
Units. Pick MPa or bar for the typed potentials and the answer; 1 MPa = 10 bar exactly. The rows show Ψ and Ψs in both units.
Rules the page checks. i from 1 to 10; C from 0 to 20 mol/L; T from 1 K to 1,000 K; a typed Ψs from −10,000 to 0; Ψp and Ψg from −10,000 to 10,000; Ψm from −10,000 to 0 (in the chosen unit). A Ψ or Ψs that is not 0 but smaller in size than 2⁻¹⁰²² MPa (about 2.2 × 10⁻³⁰⁸ MPa) has no answer, because a number cannot keep its digits.
How results are written. Every result is worked exactly from the typed decimals (temperatures included) and rounded once: 6 significant figures, trailing zeros dropped, thousands commas, the minus sign −, and m × 10ⁿ below 0.000001 or from 10²¹. The sentence reads “Ψ = −0.231215 MPa (−2.31215 bar)” in MPa, or “Ψ = −24.5401 bar (−2.45401 MPa)” in bars.
Assumptions. An ideal dilute solution (Ψs = −iCRT); pure water at atmospheric pressure at the reference height is 0 MPa.
Worked examples by hand
1.0 M sucrose at 22 °C in an open beaker. T = 22 + 273.15 = 295.15 K. Ψs = −1 × 1.0 × 0.0831446261815324 × 295.15 = −24.5401 bar (−2.45401 MPa); with Ψp = 0, Ψ = Ψs. (AP Biology’s rounded R = 0.0831 and T = 295 give −24.5 bar.)
0.15 M NaCl in a cell at 25 °C with Ψp = 0.3 MPa. Ψs = −2 × 0.15 × 0.00831446261815324 × 298.15 = −0.743687 MPa; Ψ = −0.743687 + 0.3 = −0.443687 MPa.
Typed values. Ψs = −0.7 MPa, Ψp = 0.6 MPa: Ψ = −0.1 MPa (−1 bar).
Pure water. C = 0 and Ψp = 0: Ψ = 0 MPa.
The page’s default. 0.3 M sucrose at 20 °C with Ψp = 0.5 MPa: Ψs = −0.3 × 0.00831446261815324 × 293.15 = −0.731215 MPa; Ψ = −0.231215 MPa (−2.31215 bar).
Other questions people ask
How do I calculate water potential?
Add its parts: Ψ = Ψs + Ψp (+ Ψg + Ψm when gravity or dry surfaces matter). A cell with a solute potential of −0.7 MPa and a turgor pressure of 0.6 MPa has Ψ = −0.7 + 0.6 = −0.1 MPa.
How do I calculate solute potential?
Use Ψs = −iCRT: i is the van ’t Hoff factor, the particles per formula unit (1 for sucrose, 2 for NaCl), C the molar concentration, R the gas constant (0.00831446 L·MPa/(mol·K), or 0.0831446 L·bar/(mol·K)) and T the temperature in kelvins. 1.0 M sucrose at 22 °C: −1 × 1.0 × 0.0831446 × 295.15 = −24.54 bar.
Why is my answer slightly different from my AP Biology worksheet?
AP materials round R to 0.0831 L·bar/(mol·K) and add 273 for kelvins: −1 × 1.0 × 0.0831 × 295 = −24.51 bar, which they write as −24.5 bar. The page uses R = 0.0831446 and adds 273.15, giving −24.54 bar.
Which way does water move?
From higher (less negative) to lower (more negative) water potential. Pure water is 0 MPa, so water moves from pure water into a cell whose water potential is below 0, until the pressure potential rises enough to balance it.
Can solute potential be positive?
No. It is 0 in pure water and negative once anything is dissolved. Pressure potential can be positive (a turgid cell) or negative (tension in the xylem), so the total can be either sign.
What units is water potential in?
Units of pressure: biologists use megapascals (MPa), and AP Biology often uses bars. 1 MPa = 10 bar. The solute potential of cell cytoplasm is typically −0.5 to −1.0 MPa, and turgor pressure 0.6 to 0.8 MPa (OpenStax).