<p>Name: ______Date: ______Period ______</p><p>Water Potential Problems</p><p>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).</p><p>Example 1: A 0.15 M solution of sucrose at atmospheric pressure (ψP = 0) and 25°C.</p><p>(ψS) = – iCRT</p><p>C= 0.15 M T= (273 +25) R= 0.0831 i= 1 (this is because sucrose doesn’t disassociate in water)</p><p>ψS= - (1)(0.15)(0.0831)(273+25) = -3.7 bars</p><p>This example has an osmotic potential of -3.7 bars </p><p>Ψs + Ψp = Ψ</p><p>Ψ = 0 + -3.7 bars = -3.7 bars</p><p>Example 2: A 0.15 M NaCl solution contains 2 ions, Na+ and Cl-; therefore i = 2. Room Temperature of 25°C.</p><p>(ψS) = – iCRT</p><p> i=2 C= 0.15 M R= 0.0831 T= (273 +25)</p><p>(ψS)= -(2)(0.15)(0.0831)(298) = -7.4 bars</p><p>Next:</p><p>Ψ = 0 + -7.4 bars =-7.4 bars</p><p>Careful when answering these questions:</p><p>Which of the above has MORE water potential? Example 1 Which of the above is MORE concentrated? Example 2</p><p>1. If a cell’s ΨP = 3 bars and its ΨS = -4.5 bars, what is the resulting Ψ?</p><p>Ψ= 3 + -4.5 = -1.5</p><p>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? </p><p>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.</p><p>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.</p><p>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?</p><p>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.</p><p>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?</p><p>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.</p><p>6. A plant cell with a Ψs of -7.5 bars keeps a constant volume when immersed in an open- beaker </p><p> 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)</p><p>7. At 20°C, a cell containing 0.6M glucose is in equilibrium with its surrounding solution </p><p> containing 0.5M glucose in an open container. What is the cell’s ΨP? </p><p>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)</p><p>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?</p><p>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 </p><p>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</p><p>-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</p>