Water Potential

Water potential?? Potential energy of water per unit area compared to pure water Allows us to figure out where water will flow Measured in Psi Remember Posidon, Greek god of ocean had trident Formula

Water potential What would happen? Pull water out of slug Why Outside of cell Cl- Na+ Cell Membrane Inside of cell NaCl immediately separates Hydration shell formed around Na and Cl Opens up areas and decreased H2O potential outside slug Open area where H2O from slug can move into

We can measure H2O potential on either side of membrane H2O will flow from an area of high water concentration (water potential) to and area of low concentration (water potential) = -40 bar = -5 bar Water always moves from high to low water potential

Water always moves from high to low water potential Water flow into roots vie osmosis Lower water potential in steams Lower yet in leaves Lower more in atm Water flows down water potential gradient

Equation = -5 bars Due to osmosis Pressure Potential Solute Potential Water potential build on two things Pressure potential Physical squeezing Solute potential Water flowing through osmosis = -5 bars Due to osmosis Solute potential is going to drop as I increase the # of solutes in the area If I add two little bits of NaCl to it what will it do? Why?

Equation = 2 bars = -5 bars -3 bar Plant cell Pressure potential and solute potential will determine if water will flow into or out of a cell Pressure Potential Solute Potential = 2 bars + number; pressure is pushing water out of area = -5 bars Plant cell Water continues to flow in and expand cell but because of cell wall cell wont explode Cell wall will exert a pressure to the inside Creates a pressure potential (measured in bar)

Solute potential equation = -iCRT i= ionization constant (factor between 1 and 2) C= concentration (moles/ liter) R= pressure constant (0.0831 liter/bar/mole K) T = temperature in Kalvin (273 + oC)

CYU The molar concentration of a sugar solution in an open beaker has been determined to be 0.2M. Calculate the solute potential at 22 degrees Celsius. = - (1)(.2moles/liter)(.0831 liter/bars/mole K)(295K) = - (1)(.2)(.0831 bars)(295) = - (1)(.2)(.0831)(295) bars = - 4.9029 bars = - 5 bars

What is the overall water potential? We know that = -5 bars In the problem it is given that our solution is in an open container So that makes our what? Zero 0 + -5 bar

AND DONE!