Water Potential

What you need to know! The role of diffusion (osmosis), active transport, and bulk flow in the movement of water and nutrients in plants. How water potential explains transpiration.

Water Potential Ψ Potential Energy of Water Determines direction of water flow Determined by two factors: pressure potential (ΨP) and solute potential (ΨS). The most common unit for water potential problems is bars. The mathematical equation is: Ψ = ΨP + ΨS

Pressure Potential ΨP Pressure potential will always be provided for you Open systems are systems w/out pressure potential Often cell walls will provide pressure to a cellular example When not given you may assume it is an open system ΨP = 0 bars

Solute Potential ΨS ΨS = -iCRT Or ΨS = -CRiT for my gamers! i = ionization constant Measure of how easily the solute breaks into ions in water The value is always given and is normally either 1 or 2 C = molar concentration of solute This is a measurement of the density of molecules of the solute R = Pressure constant This will always be the same no matter what: 0.0831 L bars/M K T = temperature in Kelvin The temperature will always be given Sometimes it will be given in Celsius and you will need to convert it into Kelvin: K = 273 + °C

Water Potential Ψ Once you have both pressure potential and solute potential simply add them together: Ψ = ΨP + ΨS Make sure to include the units with your answer Water will always flow from higher water potential to lower water potential

Practice: What is the water potential of a 0.5 M sucrose solution at 20°C in an open system? The ionization constant of sucrose is 1. Round your answer to the tenths place. Ψ = ΨP + ΨS Pressure Potential: Open system ΨP = 0 bars Solute Potential: ΨS = -iCRT i = 1, C = .5, R = 0.0831, T = (20 + 273) ΨS = -(1)(.5)(.0831)(273+20) Ψs = -12.17 bars Ψ = 0 bars + -12.17 bars = -12.17 bars rounded to -12.2 bars

Practice: What is the water potential of a plant cell with 0.3 M sucrose solution at 20°C? The cell wall applies 2 bars of pressure on the cellular solution. Round your answer to the tenths place. Ψ = ΨP + ΨS Pressure Potential: Cell Wall ΨP = 2 bars Solute Potential: ΨS = -iCRT i = 1, C = .3, R = 0.0831, T = (20 + 273) ΨS = -(1)(.3)(.0831)(273+20) Ψs = -7.302 bars Ψ = 2 bars + -7.302 bars = -5.302 bars rounded to -5.3 bars

Practice: Determine the net flow of water if the plant cell from the second practice problem is placed into the sucrose solution from the first practice problem what will be the net flow of water? Justify your answer. Ψ plant cell = -5.3 bars Ψ sucrose solution = -12.2 bars Remember: Water will always flow from higher water potential to lower water potential Water will flow out of the cell because the water potential inside the cell is higher than the water potential in the sucrose solution and water always flows from higher to lower potential.