1Tonicity: always a comparison

Hypotonic, hypertonic and isotonic compare two solutions. A solution is never hypotonic on its own; it is hypotonic to something.

• Hypotonic to the cell: lower solute concentration outside. Net water movement into the cell.
• Hypertonic to the cell: higher solute concentration outside. Net water movement out of the cell.
• Isotonic: equal solute concentrations. Water moves both ways at equal rates; no net movement.

So water moves from the hypotonic side to the hypertonic side. Osmolarity is the total concentration of dissolved solute particles; water moves from low osmolarity to high osmolarity.

What happens next depends on whether the cell has a wall. Figure 1 shows animal cells (top) and plant cells (bottom) in the three kinds of solution.

• In a hypotonic solution an animal cell swells and may burst (lysis); a plant cell swells until its membrane presses hard on the wall and becomes turgid, which is its healthy state.
• In an isotonic solution an animal cell keeps its normal shape; a plant cell is flaccid, soft, and the plant wilts.
• In a hypertonic solution an animal cell shrinks; in a plant cell the membrane and cytoplasm pull away from the wall: plasmolysis.

2Water potential

Water potential ($\Psi$, the Greek letter psi) measures how likely water is to move out of a region. It is measured in bars (or megapascals). Pure water at normal atmospheric pressure has $\Psi = 0$ by definition. From the formula sheet:

$$\Psi = \Psi_P + \Psi_S$$

• $\Psi_P$, pressure potential: the physical pressure on the water. Positive when the water is pushed, as inside a turgid plant cell pressing on its wall. In an open beaker it is zero.
• $\Psi_S$, solute potential: the effect of dissolved solutes. Solute particles hold water molecules around them, so adding solute always lowers it. $\Psi_S$ is zero for pure water and negative for every solution.

Water moves from higher $\Psi$ to lower $\Psi$: toward more solute, away from more pressure.

Because $\Psi_P = 0$ in an open container, the water potential of a solution in a beaker is just its solute potential.

3Calculating solute potential

The formula sheet gives:

$$\Psi_S = -iCRT$$

• $i$, the ionization constant: the number of particles each unit of solute makes when it dissolves. Sucrose does not ionize, so $i = 1.0$. NaCl splits into Na⁺ and Cl⁻, so $i = 2$.
• $C$, the molar concentration, in mol L⁻¹ (M).
• $R$, the pressure constant, $0.0831\ \text{L bar mol}^{-1}\,\text{K}^{-1}$.
• $T$, the temperature in kelvin: °C + 273.

Worked example 1. A 0.30 M sucrose solution in an open beaker at 22 °C.

$$T = 22 + 273 = 295\ \text{K}$$ $$\Psi_S = -(1)(0.30\ \text{mol L}^{-1})(0.0831\ \text{L bar mol}^{-1}\,\text{K}^{-1})(295\ \text{K}) = -7.35\ \text{bar}$$

The units cancel to bars. The beaker is open, so $\Psi_P = 0$ and $\Psi = -7.35$ bar.

Worked example 2. A 0.15 M NaCl solution at 22 °C. Now $i = 2$:

$$\Psi_S = -(2)(0.15)(0.0831)(295) = -7.35\ \text{bar}$$

Half the concentration, the same solute potential, because each NaCl makes two particles. What matters is the number of dissolved particles.

Worked example 3: which way will water move? A plant cell has $\Psi_S = -8.0$ bar and, because it is pressing on its wall, $\Psi_P = +3.0$ bar.

$$\Psi_{\text{cell}} = 3.0 + (-8.0) = -5.0\ \text{bar}$$

It is placed in the 0.30 M sucrose solution from example 1, $\Psi = -7.35$ bar. Water moves from higher to lower: from the cell ($-5.0$) to the solution ($-7.35$). The cell loses water, its pressure potential falls, and if it loses enough it becomes flaccid and then plasmolysed.

4Finding the water potential of a tissue

The classic experiment puts equal pieces of a plant tissue, such as potato, into a series of sucrose solutions and measures the percentage change in mass:

$$\text{percentage change} = \frac{\text{final mass} - \text{initial mass}}{\text{initial mass}} \times 100$$

Percentage change, not grams, because the pieces start at slightly different masses. An invented data set (mean of 5 cores each, 23 °C):

Where the tissue neither gains nor loses mass, the solution is isotonic to it, so their water potentials are equal. The change crosses zero between 0.2 and 0.4 M. Reading along a straight line between those points:

$$C = 0.2 + \frac{6.2}{6.2 + 4.1} \times 0.2 = 0.32\ \text{M}$$

So the potato's water potential equals that of 0.32 M sucrose at 23 °C (296 K):

$$\Psi = \Psi_S = -(1)(0.32)(0.0831)(296) = -7.87\ \text{bar}$$

On a graph you find the same point where the best-fit line crosses the x-axis (section 7).

5Osmoregulation

Osmoregulation is how an organism keeps its water balance and controls its internal solute composition and water potential. Growth and homeostasis depend on molecules moving across membranes all the time, so this control never stops.

• Contractile vacuoles in freshwater protists. A single-celled organism such as Paramecium living in fresh water is hypertonic to its surroundings, so water floods in by osmosis all the time. It has no wall to stop it bursting. A contractile vacuole collects the excess water and expels it from the cell, using energy. The more dilute the water outside, the faster the vacuole has to work.
• The central vacuole in plant cells. The large vacuole holds solutes, giving it a low water potential, so water moves into it. The swelling vacuole pushes the cytoplasm against the wall; the wall pushes back, raising $\Psi_P$ until water stops entering. That pressure, turgor pressure, holds leaves and stems up. Lose it and the plant wilts.
• Animal cells have neither wall nor contractile vacuole, so the body keeps the fluid around them close to isotonic. Body fluids are regulated by organs such as the kidneys.