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Osmosis Review

1. What is osmosis?
2. What is happening in the picture below?

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Water Balance

Tonicity: the ability of an extracellular solution to cause a cell to gain or lose water

• Depends on the concentration of solutes that cannot pass through the cell membrane
• Cells can be in three types of solutions:
• Isotonic
• Hypertonic
• Hypotonic
• Osmoregulation: cells must be able to regulate their solute concentrations and maintain water balance
• Animal cells will react differently than cells with cell walls, like plants, fungi, and some protists

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Tonicity: Isotonic Solutions

Cells immersed in an isotonic solution have no net movement of water

• The concentration of nonpenetrating solutes inside the cell is equal to that outside the cell.
• Water diffuses into the cell at the same rate water moves out of the cell

Animal Cell

Plant Cell

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Tonicity: Hypertonic Solutions

Cells immersed in a hypertonic solution lose water to their extracellular surroundings

• The concentration of nonpenetrating solutes is higher outside of the cell
• Water will move to the extracellular fluid
• Cells shrivel and die

Animal Cell

Plant Cell

Plasmolysis: vacuole shrinks and the plasma membrane pulls away from the cell wall

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Tonicity: Hypotonic Solutions

Cells immersed in a hypotonic solution gain water

• The concentration of nonpenetrating solutes is lower outside of the cell
• The cell will gain water
• Animal cells swell and lyse

Hemolysis – saline IV

• Plant cells work optimally

Animal Cell

Plant Cell

Maintain turgor pressure

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Water Potential�clip by Bozeman Science

Water potential: a physical property that predicts the direction water will flow

• Includes the effects of solute concentration and physical pressure
• Water will flow from areas of:
• High water potential to areas of low water potential
• Low solute to areas of high solute concentration
• High pressure to areas of low pressure

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Water Potential

Water potential formula:

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𝚿 = 𝚿_s + 𝚿_p

Greek letter “psi” represents water potential

• Measured in megapascals (MPa) or bars
• 1MPa = 10 bars

Solute potential (osmotic potential)

• An increase in solutes causes binding to more free water
• This reduces water potential
• Expressed as a negative number
• Pure water is 0 MPa

Pressure potential

• The physical pressure on a solution
• Can be + or - relative to atmospheric pressure
• “Open air” means 𝚿_p= 0 MPa

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Calculating Solute Potential

Solute potential formula:

𝚿_s = -iCRT

Solute potential

• Measured MPa or bars

Ionization constant

• Number of particles formed
• If no ions are formed the ionization constant:
• 1 for sucrose
• 2 for salt

Molar concentration

Pressure constant

• 0.0831 liter bars/mol-K
• 0.00831 liter MPA/mol K

Temperature in K

• K = 273 + ℃

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Practice Problems

1. If the concentration of NaCl inside a plant cell is 0.15M, which way will water diffuse if the cell is placed in a 0.3M NaCl solution?

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Answer: Water will leave the cell.

Water will move from the low solute concentration to the high solute concentration

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Practice Problems

2. The molar concentration of a sugar solution is 0.3M. This beaker is sitting on your desk in the open air. Calculate the solute potential at 27℃. Find your answer in bars

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𝚿_s = - (1) (0.3) (0.0831) (300)

𝚿_s = -iCRT

-i= -1

C = 0.3 mol/L

R = 0.0831 liter bar/mol K

T = 27 + 273 = 300 K

𝚿_s = -7.48 bars

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Practice Problems

3. Using the information from practice problem #2, now solve for the water potential of this beaker of liquid.

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𝚿 = (-7.48) + 0

𝚿_s = -7.48 bars

𝚿_p = 0 bars

𝚿 = -7.48 bars

𝚿 = 𝚿_s + 𝚿_p

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Practice Problems

4. Calculate the solute potential of a 0.2M NaCl solution at 30℃. Find your answers in bars.

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𝚿_s = -iCRT

-i= -2

C = 0.2 mol/L

R = 0.0831 liter bar/mol K

T = 30 + 273 = 303 K

𝚿_s = - (2) (0.2) (0.0831) (303)

𝚿_s = -10.07 bars

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Practice Problems

5. The value for 𝚿 in root tissue of a plant has found to be -4.6 bars. If you place this root tissue in a 0.2M sucrose solution at 20℃ in an open beaker, what is the 𝚿 of the solution? Which direction will the net flow of water be?

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𝚿_s = -iCRT

-i= -1

C = 0.2 mol/L

R = 0.0831liter bar/mol K

T = 20 + 273 = 293K

𝚿_s = - (1)(0.2)(0.0831)(293)

𝚿_s = -4.87 bars

𝚿 = 𝚿_s + 𝚿_p

𝚿= -4.87 + 0

𝚿= -4.87 bars

𝚿 of root = -4.6 bars

𝚿 of solution = -4.87 bars

Water will move into the solution

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