Osmosis and Water Potential

AS · 16 min

Water moves into and out of cells constantly, and whether a cell swells, shrinks, bursts or stays firm depends on the direction of that movement. Biologists describe it using water potential, a single quantity that predicts which way water will move by osmosis. This note defines water potential and osmosis, explains the very different effects of osmosis on animal and plant cells, and covers the potato (or other plant tissue) practical used to estimate the water potential of a tissue. Expect calculations of percentage change in mass, interpretation of graphs, and explanations using "higher" and "lower" water potential, which is exactly where most marks are lost.

Water potential

Water molecules move about randomly. In pure water, all the molecules are water and they are free to move; in a solution, some water molecules are attracted to and cluster around solute molecules (forming hydrogen bonds and hydration shells), so they are less free to move. The more solute there is, the lower the tendency of water to move out of that solution.

Definition

Water potential (symbol $\psi$ , the Greek letter psi) is a measure of the tendency of water molecules to move from one place to another. It is measured in units of pressure, usually kilopascals (kPa).

Pure water has the highest possible water potential: 0 kPa (at atmospheric pressure).
Adding solute lowers the water potential, so all solutions have negative water potentials. The more concentrated the solution, the more negative (lower) its water potential.
Water always moves from a region of higher (less negative) water potential to a region of lower (more negative) water potential.

For example, a solution with $\psi = -200\ \text{kPa}$ has a higher water potential than one with $\psi = -800\ \text{kPa}$ . Water will move from the $-200\ \text{kPa}$ solution to the $-800\ \text{kPa}$ solution. Students often get this backwards: $-200$ is the larger number, so it is the higher water potential.

Osmosis

Definition

Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane, as a result of their random movement. It is a passive process.

A partially permeable membrane allows water molecules through but not some solute molecules (such as sucrose). Cell surface membranes are partially permeable; so is Visking tubing. Water crosses cell membranes partly by slipping between phospholipids and mostly through channel proteins called aquaporins.

Osmosis continues until the water potentials on both sides of the membrane are equal. Then there is no net movement, although water molecules still move in both directions at equal rates.

Watch out

Use water potential, not "water concentration", "solute concentration gradient" or "osmotic pressure". Mark schemes credit "from higher water potential to lower water potential through a partially permeable membrane".
Do not write "high water potential" and "low water potential" as absolute terms when comparing; write "higher" and "lower" (comparatives), and say which region has which.
Solutes do not move in osmosis. It is the water that moves.
"Hypotonic", "hypertonic" and "isotonic" are not used in the Cambridge syllabus. If you use them, the examiner may not credit them; use water potential instead.

Effects of osmosis on animal cells

Animal cells have no cell wall, only a cell surface membrane, which cannot resist much expansion.

External solution compared with cytoplasm	Net movement of water	Effect on a red blood cell
Higher water potential (e.g. pure water or dilute solution)	into the cell by osmosis	cell swells and bursts (haemolysis; in other cells, lysis), because the membrane is not strong enough to resist the pressure
Equal water potential (e.g. blood plasma, or 0.9% sodium chloride)	no net movement	cell stays the same (normal biconcave shape)
Lower water potential (e.g. concentrated salt solution)	out of the cell by osmosis	cell shrinks and becomes wrinkled/shrivelled, described as crenated

This is why the water potential of blood plasma and tissue fluid must be kept constant (by the kidneys), and why intravenous drips use solutions with the same water potential as blood.

Effects of osmosis on plant cells

Plant cells are surrounded by a cellulose cell wall, which is strong and fully permeable (it lets water and solutes through). The living part of the cell (cell surface membrane, cytoplasm, vacuole and nucleus) is called the protoplast. The large central vacuole contains cell sap, a solution of sugars and salts.

In a solution of higher water potential

Water enters the cell by osmosis, through the cell surface membrane and tonoplast, into the cytoplasm and vacuole. The protoplast swells and pushes outwards against the cell wall. The wall is strong and resists further expansion, exerting an inward pressure on the protoplast. This pressure raises the water potential inside the cell. When the water potential inside equals that outside, net entry of water stops.

The cell is now turgid: firm, with the protoplast pressing against the wall. The cell does not burst, because of the cell wall. Turgid cells support non-woody plants and keep leaves spread out for photosynthesis.

In a solution of lower water potential

Water leaves the cell by osmosis. The vacuole and cytoplasm shrink, and the protoplast stops pushing against the wall: the cell becomes flaccid (soft). Plants with flaccid cells wilt.

If water continues to leave, the protoplast shrinks so much that the cell surface membrane pulls away from the cell wall. This is plasmolysis, and the cell is plasmolysed. Because the cell wall is fully permeable, the space between the wall and the membrane fills with the external solution.

The point at which the membrane is just beginning to pull away from the wall is called incipient plasmolysis. In a sample of tissue, incipient plasmolysis is taken to be when 50% of cells are plasmolysed: at this point the water potential of the external solution is taken to equal the water potential of the cells.

Key result

	Animal cell (red blood cell)	Plant cell
External solution has higher $\psi$	water enters; cell swells and bursts (haemolysis)	water enters; protoplast swells; pushes against wall; cell becomes turgid; does not burst (cell wall)
Equal $\psi$	no net movement; no change	no net movement
External solution has lower $\psi$	water leaves; cell shrinks (crenated)	water leaves; protoplast shrinks; cell becomes flaccid, then plasmolysed (membrane pulls away from wall)

Beyond the syllabus: solute potential and pressure potential

The water potential of a plant cell is the sum of a solute potential (negative, due to dissolved solutes) and a pressure potential (positive, due to the wall pushing on the protoplast). Knowledge of these terms is not expected in the current syllabus: explain the effects of water movement using water potential alone, and describe the cell wall's role in words ("the cell wall resists expansion, so the cell does not burst").

Direction of water movement (routine)

Three adjacent plant cells have the following water potentials: cell A $-450\ \text{kPa}$ , cell B $-700\ \text{kPa}$ , cell C $-300\ \text{kPa}$ .

(a) Which cell has the highest water potential? (b) Describe the net movement of water between the cells.

Solution

(a) Cell C ( $-300\ \text{kPa}$ is the least negative value).

(b) Water moves by osmosis from higher to lower water potential:

from C to A ( $-300$ to $-450$ );
from C to B ( $-300$ to $-700$ );
from A to B ( $-450$ to $-700$ ).

Cell B, with the lowest water potential, gains water from both of the others.

Red blood cells (moderate, 4 marks)

Samples of red blood cells were placed in (i) distilled water and (ii) a $3\%$ sodium chloride solution, which has a lower water potential than blood plasma. Describe and explain what happens to the cells in each.

Solution

(i) Distilled water has a higher water potential than the cytoplasm of the red blood cells. Water moves into the cells by osmosis through the partially permeable cell surface membrane. The cells swell and burst (haemolysis) because there is no cell wall to resist the increased volume / pressure.

(ii) The salt solution has a lower water potential than the cytoplasm. Water moves out of the cells by osmosis, so the cells shrink and become crenated (wrinkled).

Estimating the water potential of a plant tissue

The principle: if a piece of tissue is placed in a solution with the same water potential as its cells, there will be no net movement of water and no change in mass (or length). Find that solution and you know the tissue's water potential.

Practical skills

Estimating the water potential of potato tissue

Apparatus: potato, cork borer, ruler, scalpel, white tile, top-pan balance (to $0.01\ \text{g}$ ), boiling tubes, sucrose solutions (e.g. $0.0$ , $0.2$ , $0.4$ , $0.6$ , $0.8$ , $1.0\ \text{mol dm}^{-3}$ , made by serial or proportional dilution of a stock solution), paper towels, stopwatch.
Method: cut cylinders of potato with the cork borer and trim them to the same length (e.g. $3\ \text{cm}$ ), removing peel. Blot each gently and record its initial mass. Place each in the same volume of a different sucrose solution (e.g. $20\ \text{cm}^3$ ), ensuring it is fully covered. Leave for the same time (e.g. 30 minutes or longer, at room temperature or in a water bath). Remove, blot dry in the same way, and reweigh. Repeat with three cylinders per concentration.
Calculation: percentage change in mass $= \dfrac{\text{final mass} - \text{initial mass}}{\text{initial mass}} \times 100$ . Percentage change is used, rather than change in mass, because the cylinders do not all have exactly the same starting mass.
Graph: plot percentage change in mass (y-axis, positive and negative values) against sucrose concentration (x-axis). Draw a line or curve of best fit. The point where it crosses the x-axis (zero change in mass) gives the sucrose concentration with the same water potential as the potato cells.
Convert the concentration to a water potential using a table supplied in the question.
Standardised variables: same potato (cells of a similar water potential); same size and surface area of cylinders; volume of solution; time; temperature; blotting technique.
Sources of error: inconsistent blotting (too much or too little surface water); cylinders of different surface area; solution evaporating; not enough time for equilibrium. Improvements: standardise blotting (e.g. roll twice on the same paper), cover tubes, leave longer, use more concentrations near the x-intercept.
Alternative: measure length instead of mass, or use onion epidermis or red onion cells and count the percentage of cells plasmolysed under a microscope (the solution giving 50% plasmolysis has the same water potential as the cells).

Typical values of water potential for sucrose solutions (at about $20\ ^\circ\text{C}$ ) are:

Sucrose concentration / mol dm⁻³	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8
Water potential / kPa	−260	−540	−820	−1120	−1450	−1800	−2180	−2580

Percentage change and the x-intercept (moderate)

A student obtained these results for potato cylinders.

Sucrose concentration / mol dm⁻³	0.0	0.2	0.4	0.6	0.8	1.0
Initial mass / g	2.50	2.48	2.52	2.46	2.55	2.50
Final mass / g	2.96	2.65	2.41	2.12	2.04	1.94

(a) Calculate the percentage change in mass for each concentration. (b) Estimate the sucrose concentration with the same water potential as the potato cells. (c) Use the table of water potentials above to estimate the water potential of the potato cells.

Solution

(a) Using $\dfrac{\text{final} - \text{initial}}{\text{initial}} \times 100$ :

Sucrose concentration / mol dm⁻³	0.0	0.2	0.4	0.6	0.8	1.0
Percentage change in mass	+18.4	+6.9	−4.4	−13.8	−20.0	−22.4

For example, $0.0\ \text{mol dm}^{-3}$ : $\dfrac{2.96 - 2.50}{2.50} \times 100 = +18.4\%$ ; $0.4\ \text{mol dm}^{-3}$ : $\dfrac{2.41 - 2.52}{2.52} \times 100 = -4.4\%$ .

y = 64.6exp(-1.1x) - 45.45 (0, 18.4) (0.2, 6.9) (0.4, -4.4) (0.6, -13.8) (0.8, -20.0) (1.0, -22.4)

(Horizontal axis: sucrose concentration / mol dm⁻³; vertical axis: percentage change in mass. The points are the student's results and the curve is a line of best fit. It falls from about +19% to about −24% and crosses zero change in mass at about 0.32 mol dm⁻³.)

(b) The curve of best fit crosses zero change in mass between $0.2$ and $0.4\ \text{mol dm}^{-3}$ , at about $0.32\ \text{mol dm}^{-3}$ . Without a graph, linear interpolation between the two results either side of zero gives the same answer:

0.2 + 0.2 \times \frac{6.9}{6.9 + 4.4} = 0.2 + 0.12 = 0.32\ \text{mol dm}^{-3}

(c) $0.32\ \text{mol dm}^{-3}$ lies between $0.3$ ( $-820\ \text{kPa}$ ) and $0.4$ ( $-1120\ \text{kPa}$ ), one-fifth of the way along:

\psi \approx -820 - 0.2 \times 300 = -880\ \text{kPa}

So the water potential of the potato cells is about $-880\ \text{kPa}$ .

Explaining the results (moderate, 5 marks)

Explain the results in the previous example for the cylinders in $0.0\ \text{mol dm}^{-3}$ and $1.0\ \text{mol dm}^{-3}$ sucrose, and suggest why the percentage decrease in mass levels off at high concentrations.

Solution

In $0.0\ \text{mol dm}^{-3}$ (distilled water) the solution has a higher water potential than the potato cells;
water enters the cells by osmosis through the partially permeable cell surface membrane, so mass increases; the cells become turgid; the cell wall stops them bursting.
In $1.0\ \text{mol dm}^{-3}$ the solution has a lower water potential than the cells;
water leaves the cells by osmosis, so mass decreases; cells become flaccid / plasmolysed.
The decrease levels off because once cells are fully plasmolysed, little more water can be lost / the cell walls do not shrink further / the space between wall and membrane fills with sucrose solution, which adds mass; so mass changes little at even lower water potentials.

Exam-hard: incipient plasmolysis (6 marks)

Strips of red onion epidermis were placed in sucrose solutions for 30 minutes, then viewed under a microscope. The number of plasmolysed cells in a sample of 50 cells was counted.

Sucrose concentration / mol dm⁻³	0.30	0.40	0.50	0.60	0.70
Plasmolysed cells (out of 50)	0	6	19	37	48

(a) Calculate the percentage of plasmolysed cells at each concentration. (b) Estimate the sucrose concentration at incipient plasmolysis and use the table above to estimate the water potential of the onion cells. (c) Explain why the cells are not all plasmolysed at the same concentration. (d) Describe how a plasmolysed cell would look under the microscope.

Solution

(a) $0\%$ , $12\%$ , $38\%$ , $74\%$ , $96\%$ .

(b) Incipient plasmolysis = 50% plasmolysed, which lies between $0.50$ (38%) and $0.60$ (74%):

0.50 + 0.10 \times \frac{50 - 38}{74 - 38} = 0.50 + 0.10 \times \frac{12}{36} = 0.533\ \text{mol dm}^{-3}

That is about $0.53\ \text{mol dm}^{-3}$ . From the table, between $0.5$ ( $-1450\ \text{kPa}$ ) and $0.6$ ( $-1800\ \text{kPa}$ ): $\psi \approx -1450 - 0.33 \times 350 = -1567$ , so about $-1570\ \text{kPa}$ (accept about $-1550$ to $-1600\ \text{kPa}$ ).

(c) Different cells have slightly different water potentials (different concentrations of solutes in their vacuoles); cells with a higher (less negative) water potential lose water and plasmolyse in more dilute solutions.

(d) The coloured protoplast (cytoplasm and vacuole, with red pigment in the vacuole) has shrunk and pulled away from the cell wall; there is a colourless space (filled with external sucrose solution) between the cell surface membrane and the wall; the cell wall keeps its shape.

Practical skills

Osmosis with Visking tubing: an osmometer

Fill a length of Visking tubing with concentrated sucrose solution, tie one end, and attach the open end to a capillary tube with a known bore. Stand the tubing in a beaker of distilled water.
Water moves into the tubing by osmosis (from higher water potential outside to lower water potential inside), so the liquid rises up the capillary tube. Record the height at regular intervals.
Rate of osmosis $= \dfrac{\text{distance moved}}{\text{time}}$ , or, if the radius of the capillary is known, volume per unit time $= \pi r^2 \times \text{distance} / \text{time}$ .
Variables to investigate: concentration of sucrose inside (water potential gradient), temperature, surface area of tubing.
The rise eventually slows because water entering dilutes the sucrose (reducing the gradient) and the weight of the column of liquid opposes further entry.

Exam tip

Always say where the water potential is higher and where it is lower: "the solution has a higher water potential than the cytoplasm".
Always include "through a partially permeable membrane" (or name the membrane: cell surface membrane) and "by osmosis" when explaining water movement.
In plant cell questions, mention the cell wall: it prevents bursting (turgid) and stays in place when the protoplast shrinks (plasmolysis).
In the potato practical, the answers examiners want: use percentage change (different starting masses); blot to remove surface solution; x-intercept gives equal water potential; repeat and take means; same potato.
When converting between concentration and water potential, interpolate from the table given; show your working.

Watch out

In plasmolysis the cell wall does not move; the cell surface membrane pulls away from it. Do not say "the cell wall shrinks" or "the cell wall pulls away from the membrane".
Plant cells do not burst in pure water. They become turgid.
In red blood cells, "the cell bursts" is credited; "the cell becomes turgid" is not (turgidity requires a cell wall).
A more concentrated solution has a lower water potential. Do not mix "concentration" and "water potential" in the same comparison.

Summary

Water potential ( $\psi$ ) is the tendency of water to move; pure water has $\psi = 0$ ; solutes make it negative.
Osmosis: net movement of water from higher to lower water potential through a partially permeable membrane.
Animal cells in higher $\psi$ swell and burst (haemolysis); in lower $\psi$ they shrink (crenation).
Plant cells in higher $\psi$ become turgid (wall prevents bursting); in lower $\psi$ they become flaccid then plasmolysed (membrane pulls away from wall).
Incipient plasmolysis (50% cells plasmolysed): external $\psi$ equals cell $\psi$ .
Potato practical: percentage change in mass against concentration; x-intercept gives the concentration with equal water potential; convert using a table.
Solute potential and pressure potential are not required at this level.

Practice questions

Question

Define osmosis.
State the water potential of pure water.
Solution X has a water potential of $-350\ \text{kPa}$ and solution Y of $-1200\ \text{kPa}$ . They are separated by a partially permeable membrane. State the direction of net water movement and explain.
Explain why a plant cell placed in distilled water does not burst. (3 marks)
Describe what happens to a plant cell placed in a concentrated sucrose solution. (4 marks)
Explain why percentage change in mass, rather than change in mass, is calculated in the potato cylinder investigation.
A potato cylinder had an initial mass of $3.20\ \text{g}$ and a final mass of $2.88\ \text{g}$ . Calculate the percentage change in mass.
State three variables that should be controlled in the potato investigation, and explain why the cylinders are blotted before weighing.
Percentage changes in mass of beetroot tissue were $+5.2\%$ in $0.2\ \text{mol dm}^{-3}$ sucrose and $-3.8\%$ in $0.3\ \text{mol dm}^{-3}$ sucrose. Estimate the concentration at which there would be no change in mass, and use the table in this note to estimate the water potential of the tissue.
A plant is watered with a concentrated fertiliser solution and wilts. Explain, in terms of water potential, why the plant wilts, and describe what has happened to its root and leaf cells. (6 marks)

Answers

The net movement of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane (as a result of their random movement).
0 kPa.
From X to Y: X has the higher water potential ( $-350$ is less negative than $-1200$ ), and water moves by osmosis from higher to lower water potential through the partially permeable membrane.
Distilled water has a higher water potential than the cell; water enters by osmosis, so the protoplast swells and pushes against the cell wall; the cellulose cell wall is strong and resists expansion (exerts pressure on the protoplast), raising the cell's water potential until it equals that outside, so net water entry stops; the cell becomes turgid but does not burst.
The solution has a lower water potential than the cell (cytoplasm/vacuole); water leaves by osmosis through the partially permeable cell surface membrane (and tonoplast); the vacuole and cytoplasm shrink; the cell becomes flaccid; the cell surface membrane pulls away from the cell wall (plasmolysis); the space between wall and membrane fills with the external solution, as the wall is fully permeable.
The cylinders have different initial masses; percentage change allows a fair comparison of the change relative to starting mass.
$(2.88 - 3.20)/3.20 \times 100 = -10\%$ (a decrease of 10%).
Any three: same potato; same dimensions/surface area of cylinder; volume of solution; time in solution; temperature. Blotting removes solution from the surface, which would add to the mass and is not water that has entered the cells; blotting must be done the same way each time.
Zero lies between 0.2 and 0.3: $0.2 + 0.1 \times \dfrac{5.2}{5.2 + 3.8} = 0.2 + 0.058 = 0.258$ , about $0.26\ \text{mol dm}^{-3}$ . Water potential: between $-540$ (0.2) and $-820$ (0.3): $-540 - 0.58 \times 280 = -702$ , about $-700\ \text{kPa}$ .
The fertiliser solution in the soil has a lower water potential than the cells of the root (root hair cells); so water does not enter the roots, and instead moves out of root cells by osmosis into the soil; the plant cannot replace water lost by transpiration from the leaves; the root cells (and eventually leaf cells) lose water, their vacuoles and cytoplasm shrink, and they become flaccid (and may plasmolyse); flaccid cells no longer press against their cell walls, so they do not provide support; leaves and stems droop (wilt).

Water potential

Osmosis

Effects of osmosis on animal cells

Effects of osmosis on plant cells

In a solution of higher water potential

In a solution of lower water potential

Estimating the water potential of a plant tissue

Practice questions

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