Download water Probs 2009 key

Plant Physiology
Spring 2009
Name:
KEY
Water Potential Problem Set
The following is all you need to use to perform the problems in this set:
Ψw=Ψs+Ψp
Ψs = -CγRT
C = concentration of solute in molalitiy
γ = chemical activity of the solute ≅ 1
R = the gas constant = 0.00831 kg-MPa/mol-°Kelvin
T = temperature in °Kelvin
Molarity = moles solute/liter of solution
molality = moles of solute/ kilogram of H2O
Assume all calculations are for room temperature (RT=20°C, so T = 273°K + 20°C = 293°K)
1. How do you make a 1 liter solution of 0.5 molar sucrose?
M.W. sucrose=342 g/mol
weigh 171 grams of sucrose, place into large beaker add ~800 mL water, stir to dissolve, pour into 1 liter
graduated cylinder, add water to make exactly 1 liter total volume.
2. How do you make a (1 liter) solution of 0.5 molal sucrose?
weigh 171 grams of sucrose, place into large beaker, add exactly 1 liter of water, stir to dissolve.
3. What is the Ψs of a 0.5 molal solution of sucrose?
Ψs = (0.5)(0.00831 kg-MPa/mol-°Kelvin)( 293°K) = Ψs = -1.22 MPa
4. What is the Ψs of a 0.5 molal solution of NaCl?
Ψs = -CγRT
Ψs = -(1.0) (1) (.00831) (293°K)
Ψs = -2.43 MPa
5. What is the Ψs of a plant cell that has a Ψw of -3 MPa and a Ψp of +1 MPa?
Ψw = Ψs + Ψp
-3 MPa = Ψs + 1 MPa
use arithmetic: subtract 1 from both sides to get:
-3 MPa - 1 MPa = Ψs
-4 MPa = Ψs
6.
Cell A has a Ψs of -2.0 MPa and a Ψp of +0.5 MPa
Cell B has Ψs of-4.0 MPa and a Ψp of +0.9 MPa
Which way will H2O flow when the two cells are placed against each other?
A to B or B to A?..............show how you deduced your answer:
First, solve for Ψw of cells A and B
Ψw cell A = Ψs+Ψp = (-2.0 MPa) + (0.5 MPa) = -1.5 MPa
Ψw cell B = Ψs+Ψp = (-4.0 MPa) + (0.9 MPa) = -3.1 MPa
Second, compare Ψw's of cells A and B; which cell has a more negative Ψw?
-since cell B has a more negative Ψw (-3.1 vs -1.5), then H2O flows from cell A to cell B
7.
Cell A has a Ψs of -4.5 MPa and a Ψp of +2.0 MPa
Cell B has Ψs of -4.0 MPa and a Ψp of +1.0 MPa
Which way will H2O flow when the two cells are placed against each other?
A to B or B to A? ..............show how you deduced your answer:
Solve as in #6 above:
Ψw of cell A = -2.5 MPa
Ψw of cell B = -3.0 MPa
B is more negative than A, so H2O flows from A to B
8. A wind borne pollen grain with an Ψs of -3.0 MPa has dried out somewhat after blowing around in the wind.
This has caused its turgor pressure, Ψp, to go to zero. It lands on a flower stigma whose cells have a Ψs of 3.0 MPa and Ψp of +1 MPa.
a. Which way will H2O flow? From the pollen grain to the sigma or the stigma to the pollen grain?
..............show how you deduced your answer:
First, determine the Ψw of the pollen grain and the stigma cells using the Ψw=Ψs+Ψp equation.
Ψw pollen = (-3.0) + (0) = -3.0 MPa
Ψw stigma = (-3.0) + (+1) = -2.0 MPa
Second, compare the two Ψw 's.
-Since the pollen grain has a more negative Ψw than the stigma, water will flow from the stigma cells to the
pollen.
b. What will be the turgor pressure of the pollen grain after it has become equilibrated with the Ψw of the
stigma?
First, assume that if the pollen fully equilibrates with the stigma cells, then the pollen grain will have THE
SAME Ψw as the stigma cells. Thus, the new Ψw of pollen is -2.0 MPa.
Second, assume that the Ψs, the solute potential, does not change in the pollen grain IN THE SHORT period
of the equilibration process, then the pollen grain must have equilibrated with the stigma cells by adjusting its
Ψp (turgor pressure).
Then, use Ψw=Ψs+Ψp equation to determine the NEW turgor pressure value for the pollen grain:
-2.0 MPa = (-3.0 MPa) +Ψp
-2.0 + 3.0 = Ψp
+1.0 MPa = Ψp of pollen grain after equilibration
c. Suppose a botanist places some freshly harvested undessicated (non-dried) pollen from another flower that
has a Ψs of -3.0 MPa and Ψp of +1.5 MPa, to this same stigma (Ψs of -3.0 MPa and Ψp of +1 MPa). Which
way will the H2O flow? From the pollen grain to the sigma or the stigma to the pollen grain? ..............show
how you deduced your answer:
Solve as in (a.) above:
fresh pollen Ψw = Ψs+Ψp
Ψw = (-3.0) + (+1.5) = -1.5 MPa
when placed on the stigma with a Ψw of -2.0 MPa, then water will flow from the pollen to the stigma;
9. In which of the following is the chemical free energy of water, µ H2O, greatest (most positive)?
(no calculations are intended for these problems, just apply what you know about solute particles and water
potential)
a. a beaker of pure H2O or a beaker of 1 molal sucrose?
pure H2O
b. a beaker of 1 molal sucrose or a beaker of 0.1 molal sucrose?
0.1 molal sucrose
c. a beaker of 1.1 molal sucrose or a beaker of 1 molal NaCl?
1.1 molal sucrose...............when NaCl is added to a water, it separates into two molecules (Na+ and Cl-)
that behave independently as osmotic particles........thus, a beaker with 1 mole of NaCl crystals added
to becomes 2 moles of solute particles and hence more negative water potential than a 1.1 molal
solution of sucrose.
10. A plant physiologist wishes to determine the average water potential of a soybean leaf cell from a wellwatered field plant. Using a pressure probe instrument, she obtains an average cell pressure of +0.5 MPa. She
then prepares a cell sap extract from the leaves and places a drop of the clarified cell sap in a freezing point
depression apparatus. She then determines that the freezing point of the sap was minus 1.0°C [see standard
curve below] .
a. Using the attached fpd chart that represents the standard curve for the instrument she was using, together
with the pressure probe measurement, determine the Ψw of the leaf cells.
part I:
From the fpd chart, a –1.0°C freezing point depression corresponds to a 0.7 molal solution of solute particles.
Now, insert this in the Ψ s equation to obtain the solute potential component of water potential:
Ψ s = -CγRT...............- (0.7 molal)(1)( 0.00831 kg-MPa/mol-°Kelvin)( 293°K)
Ψ s = -1.7 MPa
Part II: Now solve for Ψw using the standard Ψw equation:
Ψ w=Ψs+Ψp.................................Ψ w= (-1.7 MPa) + (+0.5 MPa) = -1.2 MPa
b. After a long dry spell, the plant physiologist's returns to the same soybean field and repeats the same
measurements. This time she records an average cell pressure of +0.3 MPa and a fpd of minus 1.7 °C.
What is the Ψw of the leaf cells now?
part I:
From the fpd chart, a –1.7 °C freezing point depression corresponds to a 1.2 molal solution of solute particles.
Now, insert this in the Ψ s equation to obtain the solute potential component of water potential:
Ψ s = - CγRT...............- (1.2 molal)(1)( 0.00831 kg-MPa/mol-°Kelvin)( 293°K)
Ψ s = - 2.92 MPa
Part II: Now solve for Ψw using the standard Ψw equation:
Ψ w= Ψs+Ψp........................Ψ w = (-2.92 MPa) + (+0.3 MPa) = -2.62 MPa