Water Potential ProblemsWater Potential Problems

Water Potential ()Water Potential ()

Water potential of pure water in an open container is 0.

Water potential of pure water in an open container is 0.

Water will flow from an area of higher water potential to an area of lower water

potential.

Water will flow from an area of higher water potential to an area of lower water

potential.

Question: Which way will the water flow? Into the cell or out of the cell?

Question: Which way will the water flow? Into the cell or out of the cell?

= -0.5

= -2.5

Answer: Water will flow into the cell.

Explanation: Water will continue to flow in both directions. However, more water will be entering the cell. So the NET MOVEMENT of water will be INTO the cell. Water will flow from an area of higher water potential to an area of lower water potential (from = -0.5 to =-2.5). This will continue until dynamic equilibrium is reached.

Answer: Water will flow into the cell.

Explanation: Water will continue to flow in both directions. However, more water will be entering the cell. So the NET MOVEMENT of water will be INTO the cell. Water will flow from an area of higher water potential to an area of lower water potential (from = -0.5 to =-2.5). This will continue until dynamic equilibrium is reached.

Addition of solutes = a more negative solute potential.

Addition of solutes = a more negative solute potential.

S < 0

In an open container or in an animal cell, the pressure potential will be 0.In an open container or in an animal cell, the pressure potential will be 0.

P = 0

In turgid plant cells, the cell wall can exert positive pressure.

In turgid plant cells, the cell wall can exert positive pressure.

P > 0

Solute Potential (S)Solute Potential (S)

i = ionization constanti = ionization constant

• For sucrose, i = 1 because sucrose does not ionize in water.

• For NaCl, i = 2 because NaCl ionizes when it dissolves and becomes one Na+ and one Cl- ion in an aqueous solution.

• For sucrose, i = 1 because sucrose does not ionize in water.

• For NaCl, i = 2 because NaCl ionizes when it dissolves and becomes one Na+ and one Cl- ion in an aqueous solution.

1 ion + 1 ion = 2 ions

Question: For MgCl2, what would i be?

Question: For MgCl2, what would i be?

Answer: i = 3

Explanation: Mg would ionize into 3 ions if dissolved in water: one Mg2+ ion and two Cl- ions.

Answer: i = 3

Explanation: Mg would ionize into 3 ions if dissolved in water: one Mg2+ ion and two Cl- ions.

C is the molar concentration of the solution.

C is the molar concentration of the solution.

Molarity =

Moles Solute

Volume of SolutionM =

moles

L

R is the pressure constant.R is the pressure constant.

• Always will be 0.0831 L bars / mole K• Always will be 0.0831 L bars / mole K

T = temperature in KelvinT = temperature in Kelvin

• Kelvin is the temperature in degrees Celsius + 273.

• Kelvin is the temperature in degrees Celsius + 273.

Answer: 20 + 273 = 293K Answer: 20 + 273 = 293K

Question: Room temperature is about 20°C. What would this be in Kelvin?

Question: Room temperature is about 20°C. What would this be in Kelvin?

Example: If a cell’s P = 3 bars and S = -4.5 bars, what is the resulting ?

= P + S

Example: If a cell’s P = 3 bars and S = -4.5 bars, what is the resulting ?

= P + S

Answer:= 3 bars + (-4.5 bars)

= -1.5 bars

Answer:= 3 bars + (-4.5 bars)

= -1.5 bars

Question: A cell with a = -1.5 bars is placed in a beaker with a solution of = -4

bars. Will water flow into or out of the cell?

Question: A cell with a = -1.5 bars is placed in a beaker with a solution of = -4

bars. Will water flow into or out of the cell?

Answer: Water will flow out of the cell from an area of higher water potential (-1.5 bars) to an area of lower water potential (-4 bars).

Answer: Water will flow out of the cell from an area of higher water potential (-1.5 bars) to an area of lower water potential (-4 bars).

Question: What is the of a 0.1 M solution of sucrose in at open container at

20°C?

Question: What is the of a 0.1 M solution of sucrose in at open container at

20°C?

= P + S

S = -iCRT

= P + S

S = -iCRT

= ?

Answer: -2.4 barsExplanation: S = -iCRT

S = -(1)(0.1 )(0.0831 )(293K)

S = -2.4 bars

= P + S

= 0 + (-2.4 bars) = -2.4 bars

Answer: -2.4 barsExplanation: S = -iCRT

S = -(1)(0.1 )(0.0831 )(293K)

S = -2.4 bars

= P + S

= 0 + (-2.4 bars) = -2.4 bars

L barmole· K

moleL