Water Potential of a Plant Cell in Pure Water: 22-Agric-A1 Applied Plant, Animal or Human Physiology, May 2015, Q3

Question 3: Water Potential of a Plant Cell in Pure Water (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Solute concentration in cell cytoplasm, C	0.3 M
Van't Hoff constant, RT (@ 20°C)	2.436 MPa
Solute potential of distilled water, ψs(water)	0 MPa (by definition)
Cell placed in a volume of pure (distilled) water at STP	—

Find. The water's own ψs and ψp (part a); the system's total water potential ψ once the cell and the surrounding water reach equilibrium (part b); and whether the water is hyper- or hypotonic relative to the cell (part c).

Approach. Apply the Van't Hoff equation to get the cell's solute potential, use the definitions of pure water's ψs and ψp directly, then use the principle that net water movement (and hence equilibrium) occurs when total water potential ψ = ψs + ψp is equal on both sides of the membrane.

a) Solute and pressure potential of the pure water. Distilled water is defined to have zero solute potential, and an open/unconfined volume of water is at atmospheric reference pressure, so its pressure potential is also zero: $$\psi_s(\text{water}) = 0\text{ MPa}, \qquad \psi_p(\text{water}) = 0\text{ MPa}$$ There is no ionization constant to assume here because pure water has no dissolved solute at all — this is the reference state the Van't Hoff equation is defined against.
Solute potential of the cell (needed for parts b and c). The cytoplasm is treated as a single generic non-ionizing solute (Check: assumes i = 1, since the question does not name a specific ionic species), so: $$\psi_s(\text{cell}) = -CiRT = -(0.3)(1)(2.436\text{ MPa}) = \boxed{-0.7308\text{ MPa}}$$
b) Water potential at equilibrium. Water moves from high ψ to low ψ until the two sides equalize. The surrounding pure water is effectively an infinite reservoir, so its water potential stays fixed at ψ(water) = ψs + ψp = 0 + 0 = 0 MPa regardless of how much water leaves it. Equilibrium is reached only when the cell's total water potential rises to meet it: $$\psi(\text{equilibrium}) = \psi(\text{water}) = \boxed{0\text{ MPa}}$$ Substituting into the cell's own ψ = ψs + ψp (holding ψs,cell approximately constant, since only a small fraction of the cell's water changes) shows how that equilibrium is reached mechanically: the cell wall resists expansion and builds up turgor (pressure potential) as water enters, $$\psi_p(\text{cell, eq}) = \psi(\text{eq}) - \psi_s(\text{cell}) = 0 - (-0.7308) = \boxed{+0.7308\text{ MPa}}$$ i.e. the cell reaches equilibrium turgid, not by its solute potential changing, but by wall pressure rising to exactly offset it.
c) Tonicity of the water relative to the cell. Tonicity compares solute concentration: the pure water has C = 0 M, far below the cell's C = 0.3 M. A solution with lower solute concentration than the cell is hypotonic to it (and, equivalently, the cell is hypertonic to the water) — consistent with part (b): water moves down its potential gradient into the more concentrated cell, building turgor pressure rather than the cell losing water. $$\boxed{\text{the water is hypotonic relative to the cell}}$$

Quantity	Result
ψs(water)	0 MPa
ψp(water)	0 MPa
ψs(cell)	−0.7308 MPa
ψ at equilibrium	0 MPa
ψp(cell) at equilibrium (turgor)	+0.7308 MPa
Tonicity of water relative to cell	Hypotonic

Concept & Theory

Topic: Water potential (ψ), the Van't Hoff equation, and osmotic equilibrium across a plant cell membrane.

Key relations: $\psi = \psi_s + \psi_p$ (solute + pressure potential, ignoring gravitational potential for a single cell); Van't Hoff: $\psi_s = -CiRT$; net water flux is proportional to $\Delta\psi$ and stops at $\Delta\psi = 0$.

Why this works: Water moves passively down its free-energy gradient. Dissolving solute lowers a solution's potential energy (more negative ψs), so pure water sits at the top of the gradient (ψ = 0) relative to any solution. In a walled plant cell, water entering down that gradient is opposed by the wall pushing back (positive ψp, turgor) — equilibrium is reached when rising turgor pressure has raised the cell's total ψ to match the surrounding water's ψ, not by the solute concentration itself changing.

Common pitfall: Forgetting the rigid cell wall and treating a plant cell like an animal cell (which would simply keep swelling toward zero solute concentration, i.e. lyse) — the wall is what lets ψp rise and caps water uptake well before the cell's internal solute concentration reaches zero.

Source: Taiz & Zeiger, Plant Physiology and Development, 6th ed., Ch. 3 (Water and Plant Cells).

Question 3 of 6: Water Potential of a Plant Cell in Pure Water

Question 3: Water Potential of a Plant Cell in Pure Water (15 marks)