Osmosis and Diffusion Lab Part 1 Unit 4 Day 3 Osmosis and Diffusion Lab Part 1

Opener Get out the lab pre questions and the reading

Agenda 10 min: Explanation of lab 15 min: students set up lab 30 min: wait for results, discuss pre-lab questions 10 min: collect and share data, clean up

I can explain how the cell membrane controls the internal environment of the cell in order to maintain dynamic homeostasis Learning Target #1

Osmosis v Diffusion Diffusion is the movement of molecules from high to low concentration Osmosis is the movement of water through a semipermeable membrane

Lab: part 1 Purpose: to determine unknown molarities of solutions Materials: 6 cups labeled correctly, 6 pieces of 20 cm dialysis tubing, 10-15 mL of different solutions, scale, syringe Procedure: soak dialysis tubing in water for 30 seconds, tie a knot on one end, use syringe to add the correct solution ½ of the tubing, tie other end, take the mass, record, place in correct cup, add distilled water, wait 30 minutes, dry off, reweigh, share data with class

Pre-Lab questions

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-i= ionization constant #4 Calculate the water potential of a 0.1 M NaCl solution at 25°C. Assume that the pressure potential is zero, and the solute potential=-iCRT. Ψw = Ψs + Ψp = -(2)(.1 mol/L)(.0831 L-bar/mol-K)(298 K) + 0 Ψw = -4.95 bar Ψs = -iCRT -i= ionization constant C= concentration (M) R= pressure constant T= temp (K)

-i= ionization constant #5 If the concentration of NaCl inside of a plant cell is 0.15 M, which way will the water diffuse if the cell is placed into the 0.1 M NaCl solution? First step: calculate water potential for solution INSIDE cell Ψw = Ψs + Ψp = -(2)(.15 mol/L)(.0831 L-bar/mol-K)(298 K) + 0 Ψw = -7.43 bar Ψs = -iCRT -i= ionization constant C= concentration (M) R= pressure constant T= temp (K)

#5 Ψw = -4.95 bar .1 M NaCl Cell Solution Ψw = -7.43 bar .15 M NaCl Water will flow (by osmosis) INTO the cell, from HIGHER water potential, to LOWER water potential

#6 Turgid: the pressure from the vacuole swells and pushes against the cell wall

#6 Goal to make Ψ of cell = Ψ of solution (-4.95) What must Turgor Pressure (ΨP ) equal if there is no net diffusion between the solution and the cell? Goal to make Ψ of cell = Ψ of solution (-4.95) Ψ = ΨP + ΨS (of cell) Ψ = 2.49 + (-7.43) Ψ = -4.95 ΨP in cell must equal 2.49