Osmosis and water potentialAQA A-Level Biology: Revision notes
Section 1
Water potential
Water potential (symbol ψ) is the pressure created by water molecules, measured in kilopascals (kPa). Pure water has the highest water potential, 0 kPa. Dissolving solutes lowers it, so solutions have negative values, and a more concentrated solution has a more negative water potential.
Water moves down a water potential gradient, from a higher (less negative) to a lower (more negative) water potential.
A more negative water potential means a lower water potential. −700 kPa is lower than −400 kPa.
Section 2
Osmosis
Osmosis is the net movement of water from a region of higher water potential to a region of lower water potential, across a partially permeable membrane. It is passive and needs no ATP.
Plant cells in pure water gain water, the vacuole swells and the cell becomes turgid (the cell wall prevents bursting). In a solution of lower water potential, water leaves, the cell becomes flaccid, and in extreme cases the membrane pulls away from the wall (plasmolysis). If the water potentials are equal there is no net movement.
Section 3
Making a dilution series
To make a lower concentration from a stock, use C₁V₁ = C₂V₂, then make up to the final volume with distilled water.
Worked example: 50 cm³ of 0.2 mol dm⁻³ from 1.0 mol dm⁻³ stock: stock volume = (0.2 ÷ 1.0) × 50 = 10 cm³; add 40 cm³ distilled water.
Repeat for each concentration to make a dilution series.
Keep the total volume the same in every tube so that only concentration changes.
Section 4
Required practical 3: water potential of potato tissue
- Prepare a dilution series of sucrose, including pure water.
- Cut potato cylinders of the same size, blot, and record initial mass.
- Place one in each solution (same volume, temperature and time).
- Remove, blot and record final mass.
- Calculate percentage change in mass = (final − initial) ÷ initial × 100.
- Plot a calibration curve of percentage change against concentration. Where the line crosses zero there is no net osmosis, so the water potential of the tissue equals that of the solution.
Percentage change is used because cylinders may start with different masses.
Section 5
Worked example: reading the zero point
Percentage changes of +6.0% at 0.2 mol dm⁻³ and −4.0% at 0.4 mol dm⁻³ give a zero crossing at 0.2 + (6.0 ÷ 10.0) × 0.2 = 0.32 mol dm⁻³.
At this concentration the water potential of the solution equals that of the tissue.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Osmosis and water potential
- Cylinders of potato tissue were placed in sucrose solutions of different concentrations. Pure water has a water potential of 0 kPa and the potato cells have a water potential of −400 kPa.Explain why the mass of a potato cylinder placed in pure water increased.2 marks
- A student has a stock solution of 1.0 mol dm⁻³ sucrose and distilled water. She makes a dilution series and measures the mass of potato cylinders before and after 30 minutes in each solution.Explain how a calibration curve of percentage change in mass against sucrose concentration can be used to find the water potential of the potato tissue.2 marks
- A student recorded the percentage change in mass of potato cylinders after 30 minutes in sucrose solutions. In 0.0 mol dm⁻³ the change was +14.0%, in 0.2 mol dm⁻³ it was +6.0%, in 0.4 mol dm⁻³ it was −4.0%, in 0.6 mol dm⁻³ it was −11.0% and in 0.8 mol dm⁻³ it was −17.0%.Explain why the cylinder in 0.0 mol dm⁻³ sucrose gained mass and the cylinder in 0.8 mol dm⁻³ sucrose lost mass.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).