Question

What would be the water potential (Ψw), solute potential (Ψs) and pressure potential (Ψp) at equilibrium when a cell with Ψs= – 0.7 MPa and Ψp= 0.7 MPa is placed in a solution with Ψs= – 0.5 MPa? (Source)

a. Ψw= – 0.2 MPa, Ψs= – 0.5 MPa, Ψp= – 0.5 MPa

b. Ψw= – 0.5 MPa, Ψs= – 0.7 MPa, Ψp= 0.2 MPa

c. Ψw= – 0.5 MPa, Ψs= – 0.5 MPa, Ψp= 0 MPa

d. Ψw= – 0.3 MPa, Ψs= – 0.3 MPa, Ψp= 0 MPa

Answer 1

The water potential (Ψw) at equilibrium would be -0.2 MPa, with a solute potential (Ψs) and pressure potential (Ψp) both at -0.5 MPa.

Step-by-step explanation:

Water potential (Ψw) is the measure of the ability of water to move from one area to another due to different factors. Ψw is the sum of solute potential (Ψs) and pressure potential (Ψp). In this scenario, the cell has a solute potential (Ψs) of -0.7 MPa and a pressure potential (Ψp) of 0.7 MPa. The solution it is placed in has a solute potential (Ψs) of -0.5 MPa.

To determine the water potential (Ψw) at equilibrium, we add the solute potential (Ψs) and pressure potential (Ψp) of both the cell and the solution:

Ψw = Ψs + Ψp
Ψw = (-0.7 MPa) + (0.7 MPa) + (-0.5 MPa) + (0 MPa)
Ψw = -0.2 MPa

Therefore, the water potential (Ψw) at equilibrium is -0.2 MPa, the solute potential (Ψs) is -0.5 MPa, and the pressure potential (Ψp) is -0.5 MPa.