Water Potential

Ψ = Ψp + Ψs Water potential = pressure potential + solute potential *Bozeman science video

Water Potential Water Potential - In osmosis, the tendency for a system (a cell or solution) to take up water from pure water, through a differentially permeable membrane. Water will ALWAYS move from a region of HIGH water potential to an area of LOW water potential.

Ψp (PRESSURE POTENTIAL) An increase in pressure INCREASES water potential Normal atmospheric pressure: Ψp = 0

Ψs (SOLUTE POTENTIAL) Relative concentration of solutes Addition of solutes LOWERS water potential *Make MORE NEGATIVE!

Ψp and Ψs are inversely related Water Potential Ψp and Ψs are inversely related As pressure increases, Ψp increases As solutes increase, Ψs decreases (value become more NEGATIVE)

Finding Ψs Solute potential = –iCRT i = The number of particles the molecule will make in water; for sucrose or glucose, this number is 1 (NaCl= Na+1 and Cl-1 so = 2 particles) C = Molar concentration (from your experimental data) R = Pressure constant = 0.0831 liter bar/mole K T = Temperature in degrees Kelvin = 273 + °C of solution

Finding Ψs The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the solute potential at 27 °C degrees. Round your answer to the nearest hundredth.

Problem #1 The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the solute potential at 27 degrees. Round your answer to the nearest hundredth. Ψs = -(1)(.3 mole/liter)(0.0831)(273+27) Ψs = -7.48

-2.43 bars Problem #2 -3.3 bars The value for Ψ in root tissue was found to be -3.3 bars. If you take the root tissue and place it in a 0.1 M solution of sucrose at 20°C in an open beaker, what is the Ψ of the solution, and in which direction would the net flow of water be? Ψ s= -iCRT -(1)(0.1)(0.0831)(293)= -2.43 bars Yp=0, so Y=-2.43. The movement will be into the cell. Higher to lower.

Problem #3 -4.86 bars -3.3 bars NaCl dissociates into 2 particles in water: Na+ and Cl-. If the solution in question 4 contained 0.1M NaCl instead of 0.1M sucrose, what is the Ψ of the solution, and in which direction would the net flow of water be? -(2)(0.1)(0.0831)(293) + 0= Ψ= -4.86 Into the environment