Practical-Psi: Investigating Water Potential of Potato Tissue
- Scense Me
- 5 days ago
- 2 min read
Updated: 4 days ago
🎯 Aim
To determine the water potential of potato tissue by immersing potato cylinders in sucrose solutions of different concentrations and measuring the percentage change in mass.
🧰 Materials
Fresh potato
Cork borer (to cut uniform circular cylinders)
Scalpel and ruler (to trim equal lengths)
Balance (±0.01 g)
Blotting paper/tissue
6 labelled beakers or test tubes
50 mL each of sucrose solutions: 0.0 M, 0.1 M, 0.2 M, 0.3 M, 0.4 M, 0.5 M
Timer
Graph paper or excel spreadsheet (even better, instruct AI to construct your curve!)
🧫 Method
Use a cork borer to cut 6 potato cylinders of equal diameter. Trim each to the same length (e.g., 3 cm).
Blot dry gently and weigh each cylinder. Record the initial mass.
Place each cylinder into a separate container with 50 mL of sucrose solution.
Leave for 1 hour at room temperature.
Remove, blot dry, and weigh again. Record the final mass.
Calculate:
Change in mass = Final mass – Initial mass
% Change in mass =
Plot % change in mass (Y‑axis) against sucrose concentration (X‑axis).
Draw a smooth curve or line of best fit.
Identify the concentration where % change = 0. This is the isotonic point.
Sucrose Concentration (M) Initial Mass (g) Final Mass (g) Change in Mass (g) % Change in Mass Osmotic Potential (Ψ, MPa)
0.0
0.1
0.2
0.3
0.4
0.5
🧮 How to Fill the Osmotic Potential Column
Students use the formula:
Ψ=-iCRT
i=1.0(sucrose does not ionize)
C=molar concentration of sucrose (M)
R=0.0831" " L⋅bar⋅mol^(-1)⋅K^(-1)
T=295" K" (room temperature ~22°C)
Then convert bar → MPa (1 bar = 0.1 MPa).
📌 Example Calculation (for 0.3 M sucrose)
Ψ=-(1.0)(0.3)(0.0831)(295)
Ψ=-7.36" bar"=-0.736" MPa"
So in the Osmotic Potential column, students would enter -0.736 MPa for 0.3 M.
📈 Graph & Interpretation
Example Data Points for the Curve
Sucrose Conc. (M) % Change in Mass Osmotic Potential (MPa)
0.0 +7.5% 0.00
0.1 +4.0% -0.245
0.2 -0.8% -0.491
0.3 -3.4% -0.736
0.4 -5.2% -0.982
0.5 -7.7% -1.227
The curve crosses the X axis between 0.1–0.2 M, which corresponds to about -0.486 MPa.
This is the water potential of the potato tissue.
How the Graph Should Look
Y axis: % change in mass (gain or loss relative to original weight).
Lower X axis: Sucrose concentration in M (Molarity), e.g., 0.0 → 0.3 M.
Upper X axis: Osmotic potential (Ψ) in MPa, calculated using:
Ψ=-iCRT
where i=1(sucrose), R=0.0831" " L⋅bar⋅mol^(-1)⋅K^(-1), T=295K. Convert bar → MPa (1 bar = 0.1 MPa).
Interpretation for Students
At 0.0 M, the potato gains mass because water enters (solution is hypotonic).
At 0.5 M, the potato loses mass because water exits (solution is hypertonic).
The equilibrium point (~0.17 M) is where the potato neither gains nor loses mass.
This corresponds to a water potential of about –0.486 MPa, which is the osmotic potential of the potato tissue.
The Lab Manual in PDF:
Sample report as follows:
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