What is the new water potential of a plant cell after immerse into pure water?

Question

There is a question given by my teacher.The question state that , plant cell A contains water potential of -400kPa ,solute potential of -800kPa,pressure potential of 400kPa.When cell A is immersed into a beaker of pure water and allow the water potential to become equalibrium , what is the new water potential of cell A?(Assume that only water molecule can pass through the cell wall )

From the answer give by my teacher , the new water potential is (-400kPa + 0kPa) / 2 = -200kPa.

But,on my view,this is a comparison between pure water and cell A, so that the additional of the cell should not effect the water potential of the pure water.Since "Assume that only water molecule can pass through the cell wall" ,means there is nothing else in the pure water except the cell itself!SO, the water potential of the pure water should be remaining constant which is 0kPa and the new water potential for cell A should be 0kPa in order to become equalibrium .

What's your opinion about this?

Answer 1

If there was only pure water in the cell to begin with it wouldnt be a cell. It would be a membrane bound bag of water. You should know a plant cell is much more than this.

I believe your teacher means that now that the living plant cell has been placed in the pure water, lets assume that only water will move.

This is to ease the solving of the problem. If water AND solutes could move you'd have to worry about how the solutes leave the cell and escape to the pure water outside.