Water Potential (Ψ)

The two equations

Sign conventions (graders look for these)

• Pure water at 1 atm: Ψ = 0.
• Adding solute always decreases Ψs (it becomes more negative).
• Increasing pressure increases Ψp (positive). A pulling/tension force makes Ψp negative (e.g., xylem tension during transpiration).
• Living plant cells therefore typically have negative or zero total Ψ.

Plant-cell scenarios

Animal cells (no wall): hypotonic → lyse, hypertonic → crenate.

Worked example #1: solution Ψ from scratch

A 0.5 M sucrose solution sits open to the atmosphere at 25 °C. What is its water potential?

Step 1 · Convert to Kelvin
  T = 25 + 273 = 298 K

Step 2 · Identify i
  Sucrose does not dissociate → i = 1.0

Step 3 · Solve Ψs = −iCRT
  Ψs = −(1.0)(0.5)(0.0831)(298)
  Ψs = −12.38 bars

Step 4 · Solve Ψp
  Open to atmosphere → Ψp = 0

Step 5 · Add
  Ψ = Ψp + Ψs = 0 + (−12.38) = −12.38 bars

Worked example #2: predict water flow

A potato cube is placed in the 0.5 M sucrose solution above. The potato cells have Ψ = −7 bars. Which way does water flow?

Solution Ψ = −12.38 bars  ← lower (more negative)
Cell Ψ      = −7    bars  ← higher (less negative)

Water flows from higher → lower Ψ
   = from cell → solution
∴ The potato loses water and shrinks (plasmolyzes).

Example FRQs

At zero net mass change → potato Ψ = solution Ψ.

T = 22 + 273 = 295 K
i = 1 (sucrose)
C = 0.3 M
R = 0.0831 L·bar/(mol·K)

Ψs (solution) = −iCRT
              = −(1)(0.3)(0.0831)(295)
              = −7.35 bars

Solution is open to air → Ψp = 0
∴ Solution Ψ = −7.35 bars
∴ Potato Ψ ≈ −7.35 bars

Assuming Ψp of the potato cells is small (≈ 0 in this experimental setup),
Ψs of potato ≈ −7.35 bars.

Drill flashcards