Water Potential Problems

Name: ______Date: ______Period ______

The solute potential (ψS) = – iCRT, where i is the ionization constant, C is the molar concentration, R is the pressure constant (R = 0.0831 liter bars/mole-K), and T is the temperature in K (273 + °C).

Example 1: A 0.15 M solution of sucrose at atmospheric pressure (ψP = 0) and 25°C.

(ψS) = – iCRT

C= 0.15 M T= (273 +25) R= 0.0831 i= 1 (this is because sucrose doesn’t disassociate in water)

ψS= - (1)(0.15)(0.0831)(273+25) = -3.7 bars

This example has an osmotic potential of -3.7 bars

Ψs + Ψp = Ψ

Ψ = 0 + -3.7 bars = -3.7 bars

Example 2: A 0.15 M NaCl solution contains 2 ions, Na+ and Cl-; therefore i = 2. Room Temperature of 25°C.

(ψS) = – iCRT

i=2 C= 0.15 M R= 0.0831 T= (273 +25)

(ψS)= -(2)(0.15)(0.0831)(298) = -7.4 bars

Ψ = 0 + -7.4 bars =-7.4 bars

Careful when answering these questions:

Which of the above has MORE water potential? Example 1 Which of the above is MORE concentrated? Example 2

1. If a cell’s ΨP = 3 bars and its ΨS = -4.5 bars, what is the resulting Ψ?

Ψ= 3 + -4.5 = -1.5

2. The cell from question #1 is placed in a beaker of sugar water with ΨS = -4.0 bars. In which direction will the net flow of water be?

Remember pressure potential in this beaker will be 0, so the Ψs of -4 is also the water potential. Water will leave the cell and enter the sugar solution.

3. The original cell from question # 1 is placed in a beaker of sugar water with ΨS = -0.15 MPa (megapascals). We know that 1 MPa = 10 bars. In which direction will the net flow of water be? Need to convert MPa to bars so Ψs of -0.15 MPa = -1.5 bars. Now we see that the cell from question 1 is -1.5 and our solution is -1.5 so these are isotonic, no net flow of water. We can use Ψs because the beaker is open to the atmosphere.

4. 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?

ΨS = -(1)(0.1 mol/L)(0.0831 L*bars/mol*K)(293 K) = -2.43 bars Ψ = ΨP + ΨS = 0 bars + -2.43 bars = -2.43 bars The Ψ of the root tissue is -3.3 bars and the Ψ of the sucrose solution is -2.43 bars. Water will flow into the root tissue because free water always moves towards the lower overall water potential.

5. 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?

ΨS = -(2)(0.1 mol/L)(0.0831 L*bars/mol*K)(293 K) = -4.87 bars Ψ = ΨP + ΨS = 0 bars + -4.87 bars = -4.87 bars The Ψ of the root tissue is -3.3 bars and the Ψ of the NaCl solution is -4.87 bars. Water will flow out of the root tissue and into the salt solution because free water always moves towards the lower overall water potential.

6. A plant cell with a Ψs of -7.5 bars keeps a constant volume when immersed in an open- beaker

solution that has a Ψs of -4 bars. What is the cell’s ΨP? We know that the solution has a ΨS = -4 so this is actually the Ψ also because the beaker is open to the atmosphere. We will set our equation equal to the ΨS of the solution Ψ = ΨP + ΨS -4 = ____ + -7.5 ΨP = 3.5 (the pressure potential of the cell)

7. At 20°C, a cell containing 0.6M glucose is in equilibrium with its surrounding solution

containing 0.5M glucose in an open container. What is the cell’s ΨP?

Surrounding solution: Ψs = - (1)(0.5 mol/L)(0.0831 L*bars/mol*K)(293 K) = -12.2 bars Cell at equilibrium: Ψs = - (1)(0.6 mol/L)(0.0831 L*bars/mol*K)(293 K) = -14.6 bars Ψ = ΨP + ΨS -12.2 (solution) = ____ + -14.6 (cell) ΨP = 2.4 (the pressure potential of the cell)

8. At 20°C, a cell with ΨP of 3 bars is in equilibrium with the surrounding 0.4M solution of sucrose in an open beaker. What is the molar concentration of sucrose in the cell?

Start by determining the ΨS for the solution using –iCRT. ΨS = - (1)(0.4 mol/L)(0.0831 L*bars/mol*K)(293 K) = -9.7 bars

Next determine the ΨS of the cell using the solute potential above and the pressure potential you are given. Ψ = ΨP + ΨS -9.7 (solution) = 3 + ______(cell) ΨS = -12.7 (the solute potential of the cell) Solve for molar concentration

-12.7 = - (1)(X mol/L)(0.0831 L*bars/mol*K)(293 K) -12.7 = -(24.34)(X mol/L) -12.7/-24.34 = X mol/L 0.5 mol/L = X