Chapter 36, Problem 1DA

What would ψ_w, ψ_s, and ψ_p of the cell in figure 36.5a be at equilibrium if it had been placed in a solution with a ψ_s of −0.5 MPa?

Summary Introduction

To determine: The ${\text{φ}}_W$, $\text{φs}$ and $\text{φp}$ of a cell at equilibrium if it has been placed in a solution with ${\text{φ}}_W=-0.5\text{MPa}$.

Introduction: Water potential defines the ability of water to move from one are to another through osmosis under the influence of gravity, pressure and capillary action. Water potential constitutes of pressure potential and solute potential.

Explanation of Solution

The solute potential of the solution and the cell before equilibrium is -0.5 MPa and -0.2 MPa respectively. The solution has a higher concentration of solute than the cell and hence, water will leave the cell, causing the cell to plasmolyze. Initially, the turgor pressure of the cell and the solution $\text{φp}$, is -0.5 MPa and 0 MPa respectively.

Thus, at equilibrium, $\text{φw}$ for both the cell and the solution would be equal. Before attaining equilibrium, the solute potential of the cell would be $\text{φ}cell=-0.2MPa+0.5\text{MPa=0}\text{.3MPa}$. At equilibrium, $\text{φ}cell=\text{φsolution=-0}\text{.5MPa}$ and $\text{φw}\left(cell\right)=-0.5\text{MPa}$. Thus, at equilibrium, the plasmolyzed cell would have $\text{φp=0MPa}$.

By using the relationship, $\text{φw}\left(cell\right)=\text{φp+φs}$ and $\text{φw}\left(cell\right)=-0.5\text{MPa}$, the ${\text{φ}}_W$, $\text{φs}$ and $\text{φp}$ of the cell is determined to be -0.5 MPa, -0.5 MPa and 0 MPa.