in a large volume of 095 Molar sucrose solution at 20 degree

in a large volume of 0.95 Molar sucrose solution at 20 degrees C. A. Will the water potential of the algal cell change? B. If so, at equilibrium, what will be the new water potential of the algal cell? If pressure potential does not change, what will be the new (Psi subscript P) (ii) After the algal cell had equilibrated to the 0.95 Molar sucrose solution, it was then transferred to a large vlume of a 0.44 Molar NaCl solution A. What will be the new water potential of the algal cell after equilibration? B. If (Psi subscript s) does not change, what will be the new (Psi subscrpt P)?

Solution

Answer:

(i)

A. The total water potential of an algal cell is –1.7 MPa and the solute potential within the cell is –2.6 MPa. This algal cell is placed in a large volume of 0.95 M sucrose solution (20°C).

The solute potential of this sucrose solution can be calculated using the following formula:

s = -iCRT,

where i = Ionization constant (for sucrose this is 1.0 because sucrose does not ionize in water)

C = Molar concentration

R = Pressure Constant (R = 0.0831 liter bars/mole K)

T = Temperature (degree K) (273 + degree C of solution)

A 0.95 M sugar solution at 20 degreeC under standard atmospheric conditions

s = -i * C * R * T

s = -(1)(0.95 mole/liter)(0.0831 liter bar/mole K)(293 K) s = -23.13 bars = -2.313 MPa

The water potential will be equal to the solute potential of the solution.

= 0 + s or = s = -2.313 MPa

Here, the water potential of algal cells = -1.7 MPa, which is lower than the sucrose solution. So, the algal cell becomes hypertonic. Consequently, the cell will gain water as water molecules travel from high water potential to low water potential area.

B. The new water potential of the algal cell will be equivalent to the water potential of 0.95 M sucrose solution, that is -2.313 MPa.

w = p + s  (Assuming that solute potential in the algal cell does not change)

So, p = -2.313 + 2.6 MPa = 0.287 MPa.

(ii)

A. The solute potential of this NaCl solution can be calculated using the following formula:

s = -iCRT, (i=2 for NaCl)

= -(2)(0.44 mole/liter)(0.0831 liter bar/mole K)(293 K) s = -21.426 bars = -2.426 MPa

w = s + p = -2.426 MPa

w for the algal cell is -2.313 MPa, which is less than that of the NaCl solution. Consequently, the cell will gain water as water molecules travel from high water potential to low water potential area.

So, new water potential of the algal cell = -2.426 MPa (Equivalent to NaCl solution)

B. If solute potential does not change, s = -2.6 MPa

w = s + p

-2.426 = -2.6 + p

p = 0.174 MPa