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Continuity and change - Water potential

Grade 12IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Water potential (ψ\psi) is a measure of the free energy of water in a system and determines the direction of water movement via osmosis.

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Water always moves from a region of higher water potential (less negative) to a region of lower water potential (more negative) across a partially permeable membrane.

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The water potential of pure water at standard temperature and atmospheric pressure is defined as 0 MPa0\text{ MPa}.

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Solute potential (ψs\psi_s) represents the effect of dissolved solutes on water potential. Adding solutes always lowers the water potential, making ψs\psi_s a negative value.

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Pressure potential (ψp\psi_p) is the physical pressure exerted on a solution. In plant cells, this is usually positive due to the cell wall resisting the expansion of the vacuole (turgor pressure).

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A cell is in dynamic equilibrium with its surroundings when the water potential inside the cell equals the water potential of the external environment, resulting in no net movement of water.

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Plasmolysis occurs when a plant cell is placed in a hypertonic solution (lower ψw\psi_w), causing the cell membrane to pull away from the cell wall as water leaves the vacuole.

📐Formulae

ψw=ψs+ψp\psi_w = \psi_s + \psi_p

ψs=−iCRT\psi_s = -iCRT

💡Examples

Problem 1:

A plant cell has a solute potential ψs=−0.8 MPa\psi_s = -0.8\text{ MPa} and a pressure potential ψp=0.3 MPa\psi_p = 0.3\text{ MPa}. It is placed in a solution with a water potential ψw=−0.9 MPa\psi_w = -0.9\text{ MPa}. Calculate the water potential of the cell and determine the direction of water flow.

Solution:

  1. Calculate the cell water potential: ψw(cell)=ψs+ψp\psi_{w(\text{cell})} = \psi_s + \psi_p ψw(cell)=−0.8 MPa+0.3 MPa=−0.5 MPa\psi_{w(\text{cell})} = -0.8\text{ MPa} + 0.3\text{ MPa} = -0.5\text{ MPa}

  2. Compare cell and solution potentials: ψw(cell)=−0.5 MPa\psi_{w(\text{cell})} = -0.5\text{ MPa} ψw(sol)=−0.9 MPa\psi_{w(\text{sol})} = -0.9\text{ MPa}

Since −0.5>−0.9-0.5 > -0.9, water will move out of the cell into the solution.

Explanation:

Water moves from a higher (less negative) water potential to a lower (more negative) water potential. The cell has a higher potential than the solution, so net osmosis occurs outwards.

Problem 2:

Calculate the solute potential (ψs\psi_s) of a 0.5 M0.5\text{ M} glucose solution at 27∘C27^{\circ}\text{C}. Assume the ionization constant i=1i = 1 and the pressure constant R=0.00831 L⋅MPa/(mol⋅K)R = 0.00831\text{ L}\cdot\text{MPa}/(\text{mol}\cdot\text{K}).

Solution:

  1. Convert temperature to Kelvin: T=27+273=300 KT = 27 + 273 = 300\text{ K}

  2. Use the formula: ψs=−iCRT\psi_s = -iCRT ψs=−(1)(0.5)(0.00831)(300)\psi_s = -(1)(0.5)(0.00831)(300) ψs=−1.2465 MPa\psi_s = -1.2465\text{ MPa}

Explanation:

The solute potential is calculated using the Van't Hoff equation. Since glucose does not ionize in water, i=1i = 1. The resulting value is negative because solutes decrease the free energy of water.